Elementary Statistics Tenth Edition

Elementary Statistics Tenth Edition
and the Triola Statistics Series by Mario F. Triola

Table Of Contents 1.Introduction to Statistics 2. Summarizing and Graphing Data 54 3. Statistics for Describing, Exploring, and Comparing 94 4. Probability 5. Probability Distributions 6. Normal Probability Distributions 7. Estimates and Sample Sizes 8. Hypothesis Testing 9. Inferences from Two Samples 10. Correlation and Regression 11. Multinomial Experiments and Contingency Tables 792 12. Analysis of Variance 13. Nonparametric Statistics 14. Statistical Process Control 15. Projects, Procedures, Perspectives

Chapter 1 Introduction to Statistics
1-1 Overview 1-2 Types of Data 1-3 Critical Thinking 1-4 Design of Experiments

Section 1-1 Overview

Overview A common goal of studies and surveys and other data collecting tools is to collect data from a small part of a larger group so we can learn something about the larger group. In this section we will look at some of the ways to describe data.

Definition Data observations (such as measurements, genders, survey responses) that have been collected

Definition Statistics
a collection of methods for planning studies and experiments, obtaining data, and then organizing, summarizing, presenting, analyzing, interpreting, and drawing conclusions based on the data

Definition Population
the complete collection of all elements (scores, people, measurements, and so on) to be studied; the collection is complete in the sense that it includes all subjects to be studied

Definitions Census Collection of data from every member of a population Sample Subcollection of members selected from a population

Chapter Key Concepts Sample data must be collected in an appropriate way, such as through a process of random selection. If sample data are not collected in an appropriate way, the data may be so completely useless that no amount of statistical torturing can salvage them.

Section 1-2 Types of Data

Key Concept The subject of statistics is largely about using sample data to make inferences (or generalizations) about an entire population. It is essential to know and understand the definitions that follow.

Definition population parameter Parameter
a numerical measurement describing some characteristic of a population. population parameter

Definition sample statistic Statistic
a numerical measurement describing some characteristic of a sample. sample statistic

Definition Quantitative data
numbers representing counts or measurements. Example: The weights of supermodels

Definition Qualitative (or categorical or attribute) data
can be separated into different categories that are distinguished by some nonnumeric characteristic Example: The genders (male/female) of professional athletes

Working with Quantitative Data
Quantitative data can further be described by distinguishing between discrete and continuous types.

Definition Discrete data (i.e. the number of possible values is
result when the number of possible values is either a finite number or a ‘countable’ number (i.e. the number of possible values is 0, 1, 2, 3, . . .) Example: The number of eggs that a hen lays

Definition Continuous (numerical) data
result from infinitely many possible values that correspond to some continuous scale that covers a range of values without gaps, interruptions, or jumps Example: The amount of milk that a cow produces; e.g gallons per day

Levels of Measurement Another way to classify data is to use levels of measurement. Four of these levels are discussed in the following slides.

Nominal level of measurement
Definition Nominal level of measurement characterized by data that consist of names, labels, or categories only, and the data cannot be arranged in an ordering scheme (such as low to high) Example: Survey responses yes, no, undecided

Ordinal level of measurement
Definition Ordinal level of measurement involves data that can be arranged in some order, but differences between data values either cannot be determined or are meaningless Example: Course grades A, B, C, D, or F

Definition Interval level of measurement
like the ordinal level, with the additional property that the difference between any two data values is meaningful, however, there is no natural zero starting point (where none of the quantity is present) Example: Years 1000, 2000, 1776, and 1492

Definition Ratio level of measurement
the interval level with the additional property that there is also a natural zero starting point (where zero indicates that none of the quantity is present); for values at this level, differences and ratios are meaningful Example: Prices of college textbooks ($0 represents no cost)

Summary - Levels of Measurement
Nominal - categories only Ordinal - categories with some order Interval - differences but no natural starting point Ratio - differences and a natural starting point

Recap In this section we have looked at:
Basic definitions and terms describing data Parameters versus statistics Types of data (quantitative and qualitative) Levels of measurement

Section 1-3 Critical Thinking

Key Concepts Success in the introductory statistics course typically requires more common sense than mathematical expertise. This section is designed to illustrate how common sense is used when we think critically about data and statistics.

Misuses of Statistics

Misuse # 1- Bad Samples Voluntary response sample
(or self-selected sample) one in which the respondents themselves decide whether to be included In this case, valid conclusions can be made only about the specific group of people who agree to participate.

Misuse # 2- Small Samples
Conclusions should not be based on samples that are far too small. Example: Basing a school suspension rate on a sample of only three students

Misuse # 3- Graphs To correctly interpret a graph, you must analyze the numerical information given in the graph, so as not to be misled by the graph’s shape.

Misuse # 4- Pictographs Part (b) is designed to exaggerate the difference by increasing each dimension in proportion to the actual amounts of oil consumption.

Misuse # 5- Percentages Misleading or unclear percentages are sometimes used. For example, if you take 100% of a quantity, you take it all. 110% of an effort does not make sense.

Other Misuses of Statistics
Loaded Questions Order of Questions Refusals Correlation & Causality Self Interest Study Precise Numbers Partial Pictures Deliberate Distortions

Recap In this section we have: Reviewed 13 misuses of statistics
Illustrated how common sense can play a big role in interpreting data and statistics

Created by Tom Wegleitner, Centreville, Virginia
Section 1-4 Design of Experiments Created by Tom Wegleitner, Centreville, Virginia

Key Concept If sample data are not collected in an appropriate way, the data may be so completely useless that no amount of statistical tutoring can salvage them.

Definition Observational study
observing and measuring specific characteristics without attempting to modify the subjects being studied

Definition Experiment
apply some treatment and then observe its effects on the subjects; (subjects in experiments are called experimental units)

Definitions Cross sectional study data are observed, measured, and collected at one point in time Retrospective (or case control) study data are collected from the past by going back in time Prospective (or longitudinal or cohort) study data are collected in the future from groups (called cohorts) sharing common factors

Definition Confounding
occurs in an experiment when the experimenter is not able to distinguish between the effects of different factors

Controlling Effects of Variables
Blinding subject does not know he or she is receiving a treatment or placebo Blocks groups of subjects with similar characteristics Completely Randomized Experimental Design subjects are put into blocks through a process of random selection Rigorously Controlled Design subjects are very carefully chosen

Replication and Sample Size
repetition of an experiment when there are enough subjects to recognize the differences from different treatments Sample Size use a sample size that is large enough to see the true nature of any effects and obtain that sample using an appropriate method, such as one based on randomness

Definitions Random Sample Simple Random Sample (of size n)
members of the population are selected in such a way that each individual member has an equal chance of being selected Simple Random Sample (of size n) subjects selected in such a way that every possible sample of the same size n has the same chance of being chosen

individual member has an equal chance of being selected
Random Sampling selection so that each individual member has an equal chance of being selected

Systematic Sampling Select some starting point and then
select every k th element in the population

use results that are easy to get
Convenience Sampling use results that are easy to get

subdivide the population into at
Stratified Sampling subdivide the population into at least two different subgroups that share the same characteristics, then draw a sample from each subgroup (or stratum)

divide the population into sections
Cluster Sampling divide the population into sections (or clusters); randomly select some of those clusters; choose all members from selected clusters

Methods of Sampling - Summary
Random Systematic Convenience Stratified Cluster

Definitions Sampling error Nonsampling error
the difference between a sample result and the true population result; such an error results from chance sample fluctuations Nonsampling error sample data incorrectly collected, recorded, or analyzed (such as by selecting a biased sample, using a defective instrument, or copying the data incorrectly)

Recap In this section we have looked at:
Types of studies and experiments Controlling the effects of variables Randomization Types of sampling Sampling errors

Chapter 2 Summarizing and Graphing Data
2-1 Overview 2-2 Frequency Distributions 2-3 Histograms 2-4 Statistical Graphics

Important Characteristics of Data
Overview Important Characteristics of Data 1. Center: A representative or average value that indicates where the middle of the data set is located. 2. Variation: A measure of the amount that the values vary among themselves. 3. Distribution: The nature or shape of the distribution of data (such as bell-shaped, uniform, or skewed). 4. Outliers: Sample values that lie very far away from the vast majority of other sample values. 5. Time: Changing characteristics of the data over time.

Section 2-2 Frequency Distributions

Key Concept When working with large data sets, it is often helpful to organize and summarize data by constructing a table called a frequency distribution, defined later. Because computer software and calculators can generate frequency distributions, the details of constructing them are not as important as what they tell us about data sets.

Definition Frequency Distribution (or Frequency Table)
lists data values (either individually or by groups of intervals), along with their corresponding frequencies or counts

Frequency Distribution Ages of Best Actresses
Original Data Frequency Distribution

Frequency Distributions
Definitions

Lower Class Limits Lower Class Limits
are the smallest numbers that can actually belong to different classes Lower Class Limits

Upper Class Limits Upper Class Limits
are the largest numbers that can actually belong to different classes Upper Class Limits

Class Boundaries Class Boundaries
are the numbers used to separate classes, but without the gaps created by class limits Editor: Substitute Table 2-2 Class Boundaries 20.5 30.5 40.5 50.5 60.5 70.5 80.5

Class Midpoints Class Midpoints
can be found by adding the lower class limit to the upper class limit and dividing the sum by two Class Midpoints 25.5 35.5 45.5 55.5 65.5 75.5

Class Width Class Width
is the difference between two consecutive lower class limits or two consecutive lower class boundaries Editor: Substitute Table 2-2 Class Width 10

Reasons for Constructing Frequency Distributions
1. Large data sets can be summarized. 2. We can gain some insight into the nature of data. 3. We have a basis for constructing important graphs.

Constructing A Frequency Distribution
1. Decide on the number of classes (should be between 5 and 20). 2. Calculate (round up). class width  (maximum value) – (minimum value) number of classes 3. Starting point: Begin by choosing a lower limit of the first class. Using the lower limit of the first class and class width, proceed to list the lower class limits. 5. List the lower class limits in a vertical column and proceed to enter the upper class limits. 6. Go through the data set putting a tally in the appropriate class for each data value.

Relative Frequency Distribution
includes the same class limits as a frequency distribution, but relative frequencies are used instead of actual frequencies relative frequency = class frequency sum of all frequencies

Relative Frequency Distribution
28/76 = 37% 30/76 = 39% etc. Total Frequency = 76

Cumulative Frequency Distribution
Frequencies

Frequency Tables

Critical Thinking Interpreting Frequency Distributions
In later chapters, there will be frequent reference to data with a normal distribution. One key characteristic of a normal distribution is that it has a “bell” shape. The frequencies start low, then increase to some maximum frequency, then decrease to a low frequency. The distribution should be approximately symmetric.

Recap In this Section we have discussed
Important characteristics of data Frequency distributions Procedures for constructing frequency distributions Relative frequency distributions Cumulative frequency distributions

Section 2-3 Histograms

Key Concept A histogram is an important type of graph that portrays the nature of the distribution.

Histogram A bar graph in which the horizontal scale represents the classes of data values and the vertical scale represents the frequencies

Relative Frequency Histogram
Has the same shape and horizontal scale as a histogram, but the vertical scale is marked with relative frequencies instead of actual frequencies

Critical Thinking Interpreting Histograms
One key characteristic of a normal distribution is that it has a “bell” shape. The histogram below illustrates this.

Recap In this Section we have discussed Histograms
Relative Frequency Histograms

Section 2-4 Statistical Graphics
Created by Tom Wegleitner, Centreville, Virginia

Key Concept This section presents other graphs beyond histograms commonly used in statistical analysis. The main objective is to understand a data set by using a suitable graph that is effective in revealing some important characteristic.

Frequency Polygon Uses line segments connected to points directly above class midpoint values

Ogive A line graph that depicts cumulative frequencies
Insert figure 2-6 from page 58

Dot Plot Consists of a graph in which each data value is plotted as a point (or dot) along a scale of values

Stemplot (or Stem-and-Leaf Plot)
Represents data by separating each value into two parts: the stem (such as the leftmost digit) and the leaf (such as the rightmost digit)

Pareto Chart A bar graph for qualitative data, with the bars arranged in order according to frequencies

Pie Chart A graph depicting qualitative data as slices of a pie

Scatter Plot (or Scatter Diagram)
A plot of paired (x,y) data with a horizontal x-axis and a vertical y-axis

Time-Series Graph Data that have been collected at different points in time

Other Graphs

Recap In this section we have discussed graphs that are pictures of distributions. Keep in mind that a graph is a tool for describing, exploring and comparing data.

Chapter 3 Statistics for Describing, Exploring, and Comparing Data
3-1 Overview 3-2 Measures of Center 3-3 Measures of Variation 3-4 Measures of Relative Standing 3-5 Exploratory Data Analysis (EDA)

Overview Descriptive Statistics Inferential Statistics
summarize or describe the important characteristics of a known set of data Inferential Statistics use sample data to make inferences (or generalizations) about a population

Section 3-2 Measures of Center

Key Concept When describing, exploring, and comparing data sets, these characteristics are usually extremely important: center, variation, distribution, outliers, and changes over time.

Definition Measure of Center
the value at the center or middle of a data set

Definition Arithmetic Mean (Mean)
the measure of center obtained by adding the values and dividing the total by the number of values

Notation  denotes the sum of a set of values.
x is the variable usually used to represent the individual data values. n represents the number of values in a sample. N represents the number of values in a population.

Notation x is pronounced ‘x-bar’ and denotes the mean of a set of sample values x = n  x µ is pronounced ‘mu’ and denotes the mean of all values in a population N µ =  x

Median Definitions often denoted by x (pronounced ‘x-tilde’)
the middle value when the original data values are arranged in order of increasing (or decreasing) magnitude often denoted by x (pronounced ‘x-tilde’) ~ is not affected by an extreme value

Finding the Median If the number of values is odd, the median is the number located in the exact middle of the list. If the number of values is even, the median is found by computing the mean of the two middle numbers.

(in order - even number of values – no exact middle shared by two numbers) MEDIAN is 0.915 2 (in order - odd number of values) exact middle MEDIAN is 0.73

Definitions Mode the value that occurs most frequently
Mode is not always unique A data set may be: Bimodal Multimodal No Mode Mode is the only measure of central tendency that can be used with nominal data

Mode - Examples Mode is 1.10 Bimodal - 27 & 55 No Mode
c Mode is 1.10 Bimodal & 55 No Mode

Definition Midrange = Midrange
the value midway between the maximum and minimum values in the original data set Midrange = maximum value + minimum value 2

Round-off Rule for Measures of Center
Carry one more decimal place than is present in the original set of values.

Mean from a Frequency Distribution
Assume that in each class, all sample values are equal to the class midpoint.

use class midpoint of classes for variable x
Mean from a Frequency Distribution use class midpoint of classes for variable x

 (w • x) x =  w Weighted Mean
In some cases, values vary in their degree of importance, so they are weighted accordingly. x = w  (w • x) 

Best Measure of Center

Definitions Symmetric distribution of data is symmetric if the left half of its histogram is roughly a mirror image of its right half Skewed distribution of data is skewed if it is not symmetric and if it extends more to one side than the other

Skewness

Recap In this section we have discussed: Types of measures of center
Mean Median Mode Mean from a frequency distribution Weighted means Best measures of center Skewness

Section 3-3 Measures of Variation

Key Concept Because this section introduces the concept of variation, which is something so important in statistics, this is one of the most important sections in the entire book. Place a high priority on how to interpret values of standard deviation.

Definition The range of a set of data is the difference between the maximum value and the minimum value. Range = (maximum value) – (minimum value)

Definition The standard deviation of a set of sample values is a measure of variation of values about the mean.

Sample Standard Deviation Formula
 (x - x)2 n - 1 s =

Sample Standard Deviation (Shortcut Formula)
n (n - 1) s = n(x2) - (x)2

Standard Deviation - Important Properties
The standard deviation is a measure of variation of all values from the mean. The value of the standard deviation s is usually positive. The value of the standard deviation s can increase dramatically with the inclusion of one or more outliers (data values far away from all others). The units of the standard deviation s are the same as the units of the original data values.

Population Standard Deviation
 (x - µ) 2  = N This formula is similar to the previous formula, but instead, the population mean and population size are used.

Definition The variance of a set of values is a measure of variation equal to the square of the standard deviation. Sample variance: Square of the sample standard deviation s Population variance: Square of the population standard deviation

} s  Variance - Notation standard deviation squared Notation 2 2
Sample variance Notation 2 Population variance

Round-off Rule for Measures of Variation
Carry one more decimal place than is present in the original set of data. Round only the final answer, not values in the middle of a calculation.

Estimation of Standard Deviation Range Rule of Thumb
For estimating a value of the standard deviation s, Use Where range = (maximum value) – (minimum value) Range 4 s 

Estimation of Standard Deviation Range Rule of Thumb
For interpreting a known value of the standard deviation s, find rough estimates of the minimum and maximum “usual” sample values by using: Minimum “usual” value (mean) – 2 X (standard deviation) = Maximum “usual” value (mean) + 2 X (standard deviation) =

Definition Empirical (68-95-99.7) Rule
For data sets having a distribution that is approximately bell shaped, the following properties apply: About 68% of all values fall within 1 standard deviation of the mean. About 95% of all values fall within 2 standard deviations of the mean. About 99.7% of all values fall within 3 standard deviations of the mean.

The Empirical Rule

Definition Chebyshev’s Theorem
The proportion (or fraction) of any set of data lying within K standard deviations of the mean is always at least 1-1/K2, where K is any positive number greater than 1. For K = 2, at least 3/4 (or 75%) of all values lie within 2 standard deviations of the mean. For K = 3, at least 8/9 (or 89%) of all values lie within 3 standard deviations of the mean.

Rationale for using n-1 versus n
The end of Section 3-3 has a detailed explanation of why n – 1 rather than n is used. The student should study it carefully.

Definition The coefficient of variation (or CV) for a set of sample or population data, expressed as a percent, describes the standard deviation relative to the mean. Sample Population m CV = s 100% s x CV = 100%

Recap In this section we have looked at: Range
Standard deviation of a sample and population Variance of a sample and population Range rule of thumb Empirical distribution Chebyshev’s theorem Coefficient of variation (CV)

Section 3-4 Measures of Relative Standing

Key Concept This section introduces measures that can be used to compare values from different data sets, or to compare values within the same data set. The most important of these is the concept of the z score.

z Score (or standardized value)
Definition z Score (or standardized value) the number of standard deviations that a given value x is above or below the mean

Sample Population x - µ x - x z = z =  s Measures of Position z score
Round z to 2 decimal places

Interpreting Z Scores Whenever a value is less than the mean, its corresponding z score is negative Ordinary values: z score between –2 and 2 Unusual Values: z score < -2 or z score > 2

Definition Q1 (First Quartile) separates the bottom 25% of sorted values from the top 75%. Q2 (Second Quartile) same as the median; separates the bottom 50% of sorted values from the top 50%. Q1 (Third Quartile) separates the bottom 75% of sorted values from the top 25%.

Quartiles Q1, Q2, Q3 divide ranked scores into four equal parts 25% Q3 Q2 Q1 (minimum) (maximum) (median)

Percentiles Just as there are three quartiles separating data into four parts, there are 99 percentiles denoted P1, P2, P99, which partition the data into 100 groups.

Finding the Percentile of a Given Score
Percentile of value x = • 100 number of values less than x total number of values

Converting from the kth Percentile to the Corresponding Data Value
Notation n total number of values in the data set k percentile being used L locator that gives the position of a value Pk kth percentile L = • n k 100

Converting from the kth Percentile to the Corresponding Data Value

Some Other Statistics Interquartile Range (or IQR): Q3 - Q1 Q3 - Q1
Semi-interquartile Range: 2 Q3 - Q1 Midquartile: 2 Q3 + Q1 Percentile Range: P90 - P10

Recap In this section we have discussed: z Scores
z Scores and unusual values Quartiles Percentiles Converting a percentile to corresponding data values Other statistics

Section 3-5 Exploratory Data Analysis (EDA)

Key Concept This section discusses outliers, then introduces a new statistical graph called a boxplot, which is helpful for visualizing the distribution of data.

Definition Exploratory Data Analysis (EDA)
the process of using statistical tools (such as graphs, measures of center, and measures of variation) to investigate data sets in order to understand their important characteristics

Definition An outlier is a value that is located very far away from almost all of the other values.

Important Principles An outlier can have a dramatic effect on the mean. An outlier can have a dramatic effect on the standard deviation. An outlier can have a dramatic effect on the scale of the histogram so that the true nature of the distribution is totally obscured.

Definitions For a set of data, the 5-number summary consists of the minimum value; the first quartile Q1; the median (or second quartile Q2); the third quartile, Q3; and the maximum value. A boxplot ( or box-and-whisker-diagram) is a graph of a data set that consists of a line extending from the minimum value to the maximum value, and a box with lines drawn at the first quartile, Q1; the median; and the third quartile, Q3.

Boxplots

Boxplots - cont

Modified Boxplots Some statistical packages provide modified boxplots which represent outliers as special points. A data value is an outlier if it is … above Q3 by an amount greater than 1.5 X IQR or below Q1 by an amount greater than 1.5 X IQR

Modified Boxplot Construction
A modified boxplot is constructed with these specifications: A special symbol (such as an asterisk) is used to identify outliers. The solid horizontal line extends only as far as the minimum data value that is not an outlier and the maximum data value that is not an outlier.

Modified Boxplots - Example

Recap In this section we have looked at: Exploratory Data Analysis
Effects of outliers 5-number summary Boxplots and modified boxplots

Chapter 4 Probability 4-1 Overview 4-2 Fundamentals 4-3 Addition Rule
4-4 Multiplication Rule: Basics 4-5 Multiplication Rule: Complements and Conditional Probability 4-6 Probabilities Through Simulations 4-7 Counting

Overview Rare Event Rule for Inferential Statistics:
If, under a given assumption, the probability of a particular observed event is extremely small, we conclude that the assumption is probably not correct. Statisticians use the rare event rule for inferential statistics.

Section 4-2 Fundamentals

Key Concept This section introduces the basic concept of the probability of an event. Three different methods for finding probability values will be presented. The most important objective of this section is to learn how to interpret probability values.

Definitions Event any collection of results or outcomes of a procedure
Simple Event an outcome or an event that cannot be further broken down into simpler components Sample Space for a procedure consists of all possible simple events; that is, the sample space consists of all outcomes that cannot be broken down any further

P - denotes a probability. A, B, and C - denote specific events.
Notation for Probabilities P - denotes a probability. A, B, and C - denote specific events. P (A) - denotes the probability of event A occurring.

P(A) = Basic Rules for Computing Probability
Rule 1: Relative Frequency Approximation of Probability Conduct (or observe) a procedure, and count the number of times event A actually occurs. Based on these actual results, P(A) is estimated as follows: P(A) = number of times A occurred number of times trial was repeated

s P(A) = n = Basic Rules for Computing Probability - cont
Rule 2: Classical Approach to Probability (Requires Equally Likely Outcomes) Assume that a given procedure has n different simple events and that each of those simple events has an equal chance of occurring. If event A can occur in s of these n ways, then s P(A) = number of ways A can occur = n number of different simple events

Computing Probability - cont
Basic Rules for Computing Probability - cont Rule 3: Subjective Probabilities P(A), the probability of event A, is estimated by using knowledge of the relevant circumstances.

Law of Large Numbers As a procedure is repeated again and again, the relative frequency probability (from Rule 1) of an event tends to approach the actual probability.

Probability Limits The probability of an impossible event is 0.
The probability of an event that is certain to occur is 1. For any event A, the probability of A is between 0 and 1 inclusive. That is, 0  P(A)  1.

Possible Values for Probabilities

Definition The complement of event A, denoted by A, consists of all outcomes in which the event A does not occur.

Rounding Off Probabilities
When expressing the value of a probability, either give the exact fraction or decimal or round off final decimal results to three significant digits. (Suggestion: When the probability is not a simple fraction such as 2/3 or 5/9, express it as a decimal so that the number can be better understood.)

Definitions The actual odds against event A occurring are the ratio P(A)/P(A), usually expressed in the form of a:b (or “a to b”), where a and b are integers having no common factors. The actual odds in favor of event A occurring are the reciprocal of the actual odds against the event. If the odds against A are a:b, then the odds in favor of A are b:a. The payoff odds against event A represent the ratio of the net profit (if you win) to the amount bet. payoff odds against event A = (net profit) : (amount bet)

Recap In this section we have discussed:
Rare event rule for inferential statistics. Probability rules. Law of large numbers. Complementary events. Rounding off probabilities. Odds.

Section 4-3 Addition Rule

Key Concept The main objective of this section is to present the addition rule as a device for finding probabilities that can be expressed as P(A or B), the probability that either event A occurs or event B occurs (or they both occur) as the single outcome of the procedure.

any event combining 2 or more simple events
Definition Compound Event any event combining 2 or more simple events Notation P(A or B) = P (in a single trial, event A occurs or event B occurs or they both occur)

General Rule for a Compound Event
When finding the probability that event A occurs or event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find the total in such a way that no outcome is counted more than once.

Compound Event Formal Addition Rule
P(A or B) = P(A) + P(B) – P(A and B) where P(A and B) denotes the probability that A and B both occur at the same time as an outcome in a trial or procedure. Intuitive Addition Rule To find P(A or B), find the sum of the number of ways event A can occur and the number of ways event B can occur, adding in such a way that every outcome is counted only once. P(A or B) is equal to that sum, divided by the total number of outcomes in the sample space.

Definition Events A and B are disjoint (or mutually exclusive) if they cannot occur at the same time. (That is, disjoint events do not overlap.) Venn Diagram for Events That Are Not Disjoint Venn Diagram for Disjoint Events

P(A) and P(A) are disjoint
Complementary Events P(A) and P(A) are disjoint It is impossible for an event and its complement to occur at the same time.

Rules of Complementary Events
P(A) + P(A) = 1 = 1 – P(A) P(A) = 1 – P(A) P(A)

Venn Diagram for the Complement of Event A

Recap In this section we have discussed: Compound events.
Formal addition rule. Intuitive addition rule. Disjoint events. Complementary events.

Section 4-4 Multiplication Rule: Basics

Key Concept If the outcome of the first event A somehow affects the probability of the second event B, it is important to adjust the probability of B to reflect the occurrence of event A. The rule for finding P(A and B) is called the multiplication rule.

P(event A occurs in a first trial and
Notation P(A and B) = P(event A occurs in a first trial and event B occurs in a second trial) To differentiate between the addition rule and the multiplication rule, the addition rule will use the word ‘or’ - P(A or B) - , and the multiplication rule will use the word ‘and’ - P(A and B). Page 159 of Elementary Statistics, 10th Edition

Tree Diagrams A tree diagram is a picture of the possible outcomes of a procedure, shown as line segments emanating from one starting point. These diagrams are helpful if the number of possibilities is not too large. This figure summarizes the possible outcomes for a true/false followed by a multiple choice question. Note that there are 10 possible combinations.

Key Point – Conditional Probability
The probability for the second event B should take into account the fact that the first event A has already occurred.

Notation for Conditional Probability
P(B A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B A as “B given A.”)

Definitions Independent Events
Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. (Several events are similarly independent if the occurrence of any does not affect the probabilities of occurrence of the others.) If A and B are not independent, they are said to be dependent. Page 162 of Elementary Statistics, 10th Edition

Formal Multiplication Rule
P(A and B) = P(A) • P(B A) Note that if A and B are independent events, P(B A) is really the same as P(B).

Intuitive Multiplication Rule
When finding the probability that event A occurs in one trial and event B occurs in the next trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B takes into account the previous occurrence of event A. Many students find that they understand the multiplication rule more intuitively than by using the formal rule and formula. Example on page 162 of Elementary Statistics, 10th Edition

Applying the Multiplication Rule

Small Samples from Large Populations
If a sample size is no more than 5% of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so they are technically dependent). Since independent events where the probability is the same for each event is much easier to compute (using exponential notation) than dependent events where each probability fraction will be different, the rule for small samples and large population is often used by pollsters. Page 163 of Elementary Statistics, 10th Edition

Summary of Fundamentals
In the addition rule, the word “or” in P(A or B) suggests addition. Add P(A) and P(B), being careful to add in such a way that every outcome is counted only once. In the multiplication rule, the word “and” in P(A and B) suggests multiplication. Multiply P(A) and P(B), but be sure that the probability of event B takes into account the previous occurrence of event A.

Recap In this section we have discussed: Notation for P(A and B).
Tree diagrams. Notation for conditional probability. Independent events. Formal and intuitive multiplication rules.

Section 4-5 Multiplication Rule:
Complements and Conditional Probability

Key Concept In this section we look at the probability of getting at least one of some specified event; and the concept of conditional probability which is the probability of an event given the additional information that some other event has already occurred.

Complements: The Probability of “At Least One”
“At least one” is equivalent to “one or more.” The complement of getting at least one item of a particular type is that you get no items of that type.

Key Principle To find the probability of at least one of something, calculate the probability of none, then subtract that result from 1. That is, P(at least one) = 1 – P(none).

Definition P(A and B) P(B A) = P(A)
A conditional probability of an event is a probability obtained with the additional information that some other event has already occurred. P(B A) denotes the conditional probability of event B occurring, given that event A has already occurred, and it can be found by dividing the probability of events A and B both occurring by the probability of event A: P(B A) = P(A and B) P(A)

Intuitive Approach to Conditional Probability
The conditional probability of B given A can be found by assuming that event A has occurred and, working under that assumption, calculating the probability that event B will occur.

Recap In this section we have discussed: Concept of “at least one.”
Conditional probability. Intuitive approach to conditional probability.

Section 4-6 Probabilities Through Simulations

Key Concept In this section we introduce a very different approach for finding probabilities that can overcome much of the difficulty encountered with the formal methods discussed in the preceding sections of this chapter.

Definition A simulation of a procedure is a process that behaves the same way as the procedure, so that similar results are produced. For students in a classroom with limited time, a simulation procedure is a convenient way to determine probabilities. Page 174 of Elementary Statistics, 10th Edition

Simulation Example Gender Selection When testing techniques of gender selection, medical researchers need to know probability values of different outcomes, such as the probability of getting at least 60 girls among 100 children. Assuming that male and female births are equally likely, describe a simulation that results in genders of 100 newborn babies.

Simulation Examples Solution 1: Solution 2:
Flipping a fair coin 100 times where heads = female and tails = male H H T H T T H H H H Generating 0’s and 1’s with a computer or calculator where = male 1 = female female female male female male male male female female female Solution 2: Page 174 of Elementary Statistics, 10th Edition male male female male female female female male male male

Random Numbers In many experiments, random numbers are used in the simulation of naturally occurring events. Below are some ways to generate random numbers. A table of random of digits STATDISK Minitab Excel TI-83 Plus calculator

Random Numbers - cont STATDISK Minitab

Random Numbers - cont Excel TI-83 Plus calculator

The definition of a simulation. How to create a simulation. Ways to generate random numbers.

Section 4-7 Counting

Key Concept In many probability problems, the big obstacle is finding the total number of outcomes, and this section presents several methods for finding such numbers without directly listing and counting the possibilities.

Fundamental Counting Rule
For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m n ways.

Notation The factorial symbol ! denotes the product of decreasing positive whole numbers. For example, By special definition, 0! = 1.

Factorial Rule A collection of n different items can be arranged in order n! different ways. (This factorial rule reflects the fact that the first item may be selected in n different ways, the second item may be selected in n – 1 ways, and so on.)

Permutations Rule n! P n r = (n - r)! (when items are all different)
Requirements: There are n different items available. (This rule does not apply if some of the items are identical to others.) We select r of the n items (without replacement). We consider rearrangements of the same items to be different sequences. (The permutation of ABC is different from CBA and is counted separately.) If the preceding requirements are satisfied, the number of permutations (or sequences) of r items selected from n available items (without replacement) is (n - r)! n r P = n!

Permutations Rule n! n1! . n2! .. . . . . . . nk!
(when some items are identical to others) Requirements: There are n items available, and some items are identical to others. We select all of the n items (without replacement). We consider rearrangements of distinct items to be different sequences. If the preceding requirements are satisfied, and if there are n1 alike, n2 alike, nk alike, the number of permutations (or sequences) of all items selected without replacement is n1! . n2! nk! n!

Combinations Rule n! nCr = (n - r )! r! Requirements:
There are n different items available. We select r of the n items (without replacement). We consider rearrangements of the same items to be the same. (The combination of ABC is the same as CBA.) If the preceding requirements are satisfied, the number of combinations of r items selected from n different items is (n - r )! r! n! nCr =

Permutations versus Combinations
When different orderings of the same items are to be counted separately, we have a permutation problem, but when different orderings are not to be counted separately, we have a combination problem.

The fundamental counting rule. The factorial rule. The permutations rule (when items are all different). The permutations rule (when some items are identical to others). The combinations rule.

Chapter 5 Probability Distributions
5-1 Overview 5-2 Random Variables 5-3 Binomial Probability Distributions 5-4 Mean, Variance and Standard Deviation for the Binomial Distribution 5-5 The Poisson Distribution

Overview discrete probability distributions
This chapter will deal with the construction of discrete probability distributions by combining the methods of descriptive statistics presented in Chapter 2 and 3 and those of probability presented in Chapter 4. Probability Distributions will describe what will probably happen instead of what actually did happen. Emphasize the combination of the methods in Chapter 3 (descriptive statistics) with the methods in Chapter 4 (probability). Chapter 3 one would conduct an actual experiment and find and observe the mean, standard deviation, variance, etc. and construct a frequency table or histogram Chapter 4 finds the probability of each possible outcome Chapter 5 presents the possible outcomes along with the relative frequencies we expect Page 200 of Elementary Statistics, 10th Edition

Combining Descriptive Methods
and Probabilities In this chapter we will construct probability distributions by presenting possible outcomes along with the relative frequencies we expect. page 200 of Elementary Statistics, 10th Edition

Section 5-2 Random Variables

Key Concept This section introduces the important concept of a probability distribution, which gives the probability for each value of a variable that is determined by chance. Give consideration to distinguishing between outcomes that are likely to occur by chance and outcomes that are “unusual” in the sense they are not likely to occur by chance.

Definitions Random variable
a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure Probability distribution a description that gives the probability for each value of the random variable; often expressed in the format of a graph, table, or formula page 201 of Elementary Statistics, 10th Edition

Definitions Discrete random variable
either a finite number of values or countable number of values, where “countable” refers to the fact that there might be infinitely many values, but they result from a counting process Continuous random variable infinitely many values, and those values can be associated with measurements on a continuous scale in such a way that there are no gaps or interruptions This chapter deals exclusively with discrete random variables - experiments where the data observed is a ‘countable’ value. Give examples. Following chapters will deal with continuous random variables. Page 201 of Elementary Statistics, 10th Edition

Graphs The probability histogram is very similar to a relative frequency histogram, but the vertical scale shows probabilities. page 202 of Elementary Statistics, 10th Edition Probability Histograms relate nicely to Relative Frequency Histograms of Chapter 2, but the vertical scale shows probabilities instead of relative frequencies based on actual sample results Observe that the probabilities of each random variable is also the same as the AREA of the rectangle representing the random variable. This fact will be important when we need to find probabilities of continuous random variables - Chapter 6.

Requirements for Probability Distribution
P(x) = 1 where x assumes all possible values.  0  P(x)  1 for every individual value of x. Page 203 of Elementary Statistics, 10th Edition

Standard Deviation of a Probability Distribution
Mean, Variance and Standard Deviation of a Probability Distribution µ =  [x • P(x)] Mean 2 =  [(x – µ)2 • P(x)] Variance 2 = [ x2 • P(x)] – µ Variance (shortcut)  =  [x 2 • P(x)] – µ Standard Deviation In Chapter 3, we found the mean, standard deviation,variance, and shape of the distribution for actual observed experiments. The probability distribution and histogram can provide the same type information. These formulas will apply to ANY type of probability distribution as long as you have have all the P(x) values for the random variables in the distribution. In section 4 of this chapter, there will be special EASIER formulas for the special binomial distribution. The TI-83 and TI-83 Plus calculators can find the mean, standard deviation, and variance in the same way that one finds those values for a frequency table. With the TI-82, TI-81, and TI-85 calculators, one would have to multiply all decimal values in the P(x) column by the same factor so that there were no decimals and proceed as usual. Page 204 of Elementary Statistics, 10th Edition

Roundoff Rule for µ, , and 2
Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round µ, ,and 2 to one decimal place. If the rounding results in a value that looks as if the mean is ‘all’ or none’ (when, in fact, this is not true), then leave as many decimal places as necessary to correctly reflect the true mean. Page 205 of Elementary Statistics, 10th Edition

Identifying Unusual Results
Range Rule of Thumb According to the range rule of thumb, most values should lie within 2 standard deviations of the mean. We can therefore identify “unusual” values by determining if they lie outside these limits: Maximum usual value = μ + 2σ Minimum usual value = μ – 2σ Page of Elementary Statistics, 10th Edition

Identifying Unusual Results
Probabilities Rare Event Rule If, under a given assumption (such as the assumption that a coin is fair), the probability of a particular observed event (such as 992 heads in 1000 tosses of a coin) is extremely small, we conclude that the assumption is probably not correct. Unusually high: x successes among n trials is an unusually high number of successes if P(x or more) ≤ 0.05. The discussion of this topic in the text includes some difficult concepts, but it also includes an extremely important approach used often in statistics. Page of Elementary Statistics, 10th Edition Unusually low: x successes among n trials is an unusually low number of successes if P(x or fewer) ≤ 0.05.

E =  [x • P(x)] Definition
The expected value of a discrete random variable is denoted by E, and it represents the average value of the outcomes. It is obtained by finding the value of  [x • P(x)]. E =  [x • P(x)] Also called expectation or mathematical expectation Plays a very important role in decision theory page 208 of Elementary Statistics, 10th Edition

Combining methods of descriptive statistics with probability. Random variables and probability distributions. Probability histograms. Requirements for a probability distribution. Mean, variance and standard deviation of a probability distribution. Identifying unusual results. Expected value.

Binomial Probability Distributions
Section 5-3 Binomial Probability Distributions

Key Concept This section presents a basic definition of a binomial distribution along with notation, and it presents methods for finding probability values. Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two relevant categories such as acceptable/defective or survived/died.

Definitions A binomial probability distribution results from a procedure that meets all the following requirements: 1. The procedure has a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.) 3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure). Binomial probability distributions are important because they allow us to deal with circumstances in which the outcomes belong to TWO categories, such as pass/fall, acceptable/defective, etc. page 214 of Elementary Statistics, 10th Edition 4. The probability of a success remains the same in all trials.

Notation for Binomial Probability Distributions
S and F (success and failure) denote two possible categories of all outcomes; p and q will denote the probabilities of S and F, respectively, so P(S) = p (p = probability of success) The word success is arbitrary and does not necessarily represent something GOOD. If you are trying to find the probability of deaths from hang-gliding, the ‘success’ probability is that of dying from hang-gliding. Remind students to carefully look at the probability (percentage or rate) provided and whether it matches the probability desired in the question. It might be necessary to determine the complement of the given probability to establish the ‘p’ value of the problem. Page 214 of Elementary Statistics, 10th Edition. P(F) = 1 – p = q (q = probability of failure)

Notation (cont) n denotes the number of fixed trials.
x denotes a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive. p denotes the probability of success in one of the n trials. q denotes the probability of failure in one of the n trials. Page 214 of Elementary Statistics, 10th Edition. P(x) denotes the probability of getting exactly x successes among the n trials.

Important Hints Be sure that x and p both refer to the same category being called a success. When sampling without replacement, consider events to be independent if n < 0.05N. Page of Elementary Statistics, 10th Edition.

Methods for Finding Probabilities
We will now discuss three methods for finding the probabilities corresponding to the random variable x in a binomial distribution. Page 216 of Elementary Statistics, 10th Edition.

Method 1: Using the Binomial
Probability Formula P(x) = • px • qn-x (n – x )!x! n ! for x = 0, 1, 2, . . ., n where n = number of trials x = number of successes among n trials p = probability of success in any one trial q = probability of failure in any one trial (q = 1 – p) page 216 of Elementary Statistics, 10th Edition

Method 2: Using Table A-1 in Appendix A
Part of Table A-1 is shown below. With n = 12 and p = 0.80 in the binomial distribution, the probabilities of 4, 5, 6, and 7 successes are 0.001, 0.003, 0.016, and respectively. Page 217 of Elementary Statistics, 10th Edition.

Method 3: Using Technology
STATDISK, Minitab, Excel and the TI-83 Plus calculator can all be used to find binomial probabilities. STATDISK Minitab Page 218 of Elementary Statistics, 10th Edition.

Method 3: Using Technology
STATDISK, Minitab, Excel and the TI-83 Plus calculator can all be used to find binomial probabilities. Excel TI-83 Plus calculator Page 218 of Elementary Statistics, 10th Edition.

Strategy for Finding Binomial Probabilities
Use computer software or a TI-83 Plus calculator if available. If neither software nor the TI-83 Plus calculator is available, use Table A-1, if possible. If neither software nor the TI-83 Plus calculator is available and the probabilities can’t be found using Table A-1, use the binomial probability formula. Page 219 of Elementary Statistics, 10th Edition

Rationale for the Binomial Probability Formula
P(x) = • px • qn-x n ! (n – x )!x! The number of outcomes with exactly x successes among n trials The ‘counting’ factor of the formula counts the number of ways the x successes and (n-x) failures can be arranged - i.e.. the number of arrangements (Review section 4-7, page 181). Discussion is on page 219 of Elementary Statistics, 10th Edition.

Binomial Probability Formula
P(x) = • px • qn-x n ! (n – x )!x! Number of outcomes with exactly x successes among n trials The probability of x successes among n trials for any one particular order The remaining two factors of the formula will compute the probability of any one arrangement of successes and failures. This probability will be the same no matter what the arrangement is. The three factors multiplied together give the correct probability of ‘x’ successes.

The definition of the binomial probability distribution. Notation. Important hints. Three computational methods. Rationale for the formula.

Mean, Variance, and Standard Deviation for the Binomial Distribution
Section 5-4 Mean, Variance, and Standard Deviation for the Binomial Distribution

Key Concept In this section we consider important characteristics of a binomial distribution including center, variation and distribution. That is, we will present methods for finding its mean, variance and standard deviation. As before, the objective is not to simply find those values, but to interpret them and understand them.

For Any Discrete Probability Distribution: Formulas
Mean µ = [x • P(x)] Variance 2= [ x2 • P(x) ] – µ2 Std. Dev  = [ x2 • P(x) ] – µ2 These formulas were introduced in section 5-2, pages These formulas will produce the mean, standard deviation, and variance for any probability distribution. The difficult part of these formulas is that one must have all the P(x) values for the random variables in the distribution.

Binomial Distribution: Formulas
Std. Dev. = n • p • q Mean µ = n • p Variance 2 = n • p • q Where n = number of fixed trials p = probability of success in one of the n trials q = probability of failure in one of the n trials The obvious advantage to the formulas for the mean, standard deviation, and variance of a binomial distribution is that you do not need the values in the distribution table. You need only the n, p, and q values. A common error students tend to make is to forget to take the square root of (n)(p)(q) to find the standard deviation, especially if they used L1 and L2 lists in a graphical calculator to find the standard deviation with non-binomial distributions. The square root is built into the programming of the calculator and students do not have to remember it. These formulas do require the student to remember to take the square root of the three factors. Page 225 of Elementary Statistics, 10th Edition

Interpretation of Results
It is especially important to interpret results. The range rule of thumb suggests that values are unusual if they lie outside of these limits: Maximum usual values = µ + 2  Minimum usual values = µ – 2  Page 225 of Elementary Statistics, 10th Edition.

Mean,variance and standard deviation formulas for the any discrete probability distribution. Mean,variance and standard deviation formulas for the binomial probability distribution. Interpreting results.

The Poisson Distribution
Section 5-5 The Poisson Distribution

Key Concept The Poisson distribution is important because it is often used for describing the behavior of rare events (with small probabilities).

P(x) = where e  2.71828 Definition x! Formula µ x • e -µ
The Poisson distribution is a discrete probability distribution that applies to occurrences of some event over a specified interval. The random variable x is the number of occurrences of the event in an interval. The interval can be time, distance, area, volume, or some similar unit. P(x) = where e  µ x • e -µ x! Formula Some discussion of the what the value of ‘e’ represents might be necessary. A minimal scientific calculator with factorial key is required for this formula. Most instructors will want students to use the calculator representation of ‘e’, rather than the rounded version of ‘e’, which might provide a very slightly different final answer. Page 230 of Elementary Statistics, 10th Edition.

Poisson Distribution Requirements
The random variable x is the number of occurrences of an event over some interval. The occurrences must be random. The occurrences must be independent of each other. The occurrences must be uniformly distributed over the interval being used. Parameters The mean is µ. Page 231 of Elementary Statistics, 10th Edition. The standard deviation is  = µ .

Binomial Distribution
Difference from a Binomial Distribution The Poisson distribution differs from the binomial distribution in these fundamental ways: The binomial distribution is affected by the sample size n and the probability p, whereas the Poisson distribution is affected only by the mean μ. In a binomial distribution the possible values of the random variable x are 0, 1, n, but a Poisson distribution has possible x values of 0, 1, , with no upper limit. Page 231 of Elementary Statistics, 10th Edition.

Poisson as Approximation to Binomial
The Poisson distribution is sometimes used to approximate the binomial distribution when n is large and p is small. Rule of Thumb n  100 np  10 If n is large (greater than or equal to 100) and np is less than or equal to 10, then the Poisson distribution can be used to approximate the binomial distribution. Page 232 of Elementary Statistics, 10th Edition.

Poisson as Approximation to Binomial - μ
np  10 Value for μ = n • p Page 232 of Elementary Statistics, 10th Edition.

Definition of the Poisson distribution. Requirements for the Poisson distribution. Difference between a Poisson distribution and a binomial distribution. Poisson approximation to the binomial.

Chapter 6 Normal Probability Distributions
6-1 Overview 6-2 The Standard Normal Distribution 6-3 Applications of Normal Distributions 6-4 Sampling Distributions and Estimators 6-5 The Central Limit Theorem 6-6 Normal as Approximation to Binomial 6-7 Assessing Normality

Overview f(x) = Chapter focus is on: ( )2  2 p
Continuous random variables Normal distributions ( x- )2 -1  e2 f(x) =  2 p Formula 6-1 Figure 6-1

The Standard Normal Distribution
Section 6-2 The Standard Normal Distribution

Key Concept This section presents the standard normal distribution which has three properties: It is bell-shaped. It has a mean equal to 0. It has a standard deviation equal to 1. It is extremely important to develop the skill to find areas (or probabilities or relative frequencies) corresponding to various regions under the graph of the standard normal distribution.

Definition A continuous random variable has a uniform distribution if its values spread evenly over the range of probabilities. The graph of a uniform distribution results in a rectangular shape.

Definition A density curve is the graph of a continuous probability distribution. It must satisfy the following properties: 1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. (That is, the curve cannot fall below the x-axis.)

Area and Probability Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.

Using Area to Find Probability
Figure 6-3

Definition The standard normal distribution is a probability distribution with mean equal to 0 and standard deviation equal to 1, and the total area under its density curve is equal to 1.

Finding Probabilities - Table A-2
Inside back cover of textbook Formulas and Tables card Appendix

Finding Probabilities – Other Methods
STATDISK Minitab Excel TI-83/84

Table A-2 - Example

Using Table A-2 z Score Distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table A-2. Area Region under the curve; refer to the values in the body of Table A-2.

Example - Thermometers
If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is less than 1.58 degrees.

Example - Cont P(z < 1.58) = Figure 6-6

Look at Table A-2

Example - cont P (z < 1.58) = Figure 6-6

Example - cont P (z < 1.58) = 0.9429
The probability that the chosen thermometer will measure freezing water less than 1.58 degrees is

94.29% of the thermometers have readings less than 1.58 degrees.
Example - cont P (z < 1.58) = 94.29% of the thermometers have readings less than 1.58 degrees.

The probability that the chosen thermometer with a reading above
Example - cont If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads (at the freezing point of water) above –1.23 degrees. P (z > –1.23) = The probability that the chosen thermometer with a reading above -1.23 degrees is

89.07% of the thermometers have readings above –1.23 degrees.
Example - cont P (z > –1.23) = 89.07% of the thermometers have readings above –1.23 degrees.

Example - cont A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. P (z < –2.00) = P (z < 1.50) = P (–2.00 < z < 1.50) = – = The probability that the chosen thermometer has a reading between – 2.00 and 1.50 degrees is

Example - Modified A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. P (z < –2.00) = P (z < 1.50) = P (–2.00 < z < 1.50) = – = If many thermometers are selected and tested at the freezing point of water, then 91.04% of them will read between –2.00 and 1.50 degrees.

denotes the probability that the z score is greater than a.
Notation P(a < z < b) denotes the probability that the z score is between a and b. P(z > a) denotes the probability that the z score is greater than a. P(z < a) denotes the probability that the z score is less than a.

Finding a z Score When Given a Probability Using Table A-2
1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not a cumulative region from the left, work instead with a known region that is a cumulative region from the left. 2. Using the cumulative area from the left, locate the closest probability in the body of Table A-2 and identify the corresponding z score.

When Given Probabilities Finding the 95th Percentile
Finding z Scores When Given Probabilities 5% or 0.05 (z score will be positive) Figure 6-10 Finding the 95th Percentile

When Given Probabilities - cont Finding the 95th Percentile
Finding z Scores When Given Probabilities - cont 5% or 0.05 (z score will be positive) 1.645 Figure 6-10 Finding the 95th Percentile

When Given Probabilities - cont Finding the Bottom 2.5% and Upper 2.5%
Finding z Scores When Given Probabilities - cont (One z score will be negative and the other positive) Figure 6-11 Finding the Bottom 2.5% and Upper 2.5%

Recap In this section we have discussed: Density curves.
Relationship between area and probability Standard normal distribution. Using Table A-2.

Applications of Normal Distributions
Section 6-3 Applications of Normal Distributions

Key Concept This section presents methods for working with normal distributions that are not standard. That is, the mean is not 0 or the standard deviation is not 1, or both. The key concept is that we can use a simple conversion that allows us to standardize any normal distribution so that the same methods of the previous section can be used.

x – µ z =  Conversion Formula Formula 6-2
Round z scores to 2 decimal places

Converting to a Standard Normal Distribution
x –   z = Figure 6-12

Example – Weights of Water Taxi Passengers
In the Chapter Problem, we noted that the safe load for a water taxi was found to be 3500 pounds. We also noted that the mean weight of a passenger was assumed to be 140 pounds. Assume the worst case that all passengers are men. Assume also that the weights of the men are normally distributed with a mean of 172 pounds and standard deviation of 29 pounds. If one man is randomly selected, what is the probability he weighs less than 174 pounds?

Example - cont z = 174 – 172 29 = 0.07  =29  = 172 Figure 6-13

Example - cont  = 172  =29 P ( x < 174 lb.) = P(z < 0.07)
 = 172 P ( x < 174 lb.) = P(z < 0.07) = Figure 6-13

Cautions to Keep in Mind
1. Don’t confuse z scores and areas. z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A-2 lists z scores in the left column and across the top row, but areas are found in the body of the table. 2. Choose the correct (right/left) side of the graph. 3. A z score must be negative whenever it is located in the left half of the normal distribution. 4. Areas (or probabilities) are positive or zero values, but they are never negative.

Procedure for Finding Values Using Table A-2 and Formula 6-2
1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the x value(s) being sought. 2. Use Table A-2 to find the z score corresponding to the cumulative left area bounded by x. Refer to the body of Table A-2 to find the closest area, then identify the corresponding z score. 3. Using Formula 6-2, enter the values for µ, , and the z score found in step 2, then solve for x. x = µ + (z • ) (Another form of Formula 6-2) (If z is located to the left of the mean, be sure that it is a negative number.) 4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem. page 263 of Elementary Statistics, 10th Edition

Example – Lightest and Heaviest
Use the data from the previous example to determine what weight separates the lightest 99.5% from the heaviest 0.5%?

Example – Lightest and Heaviest - cont
x =  + (z ● ) x = (2.575  29) x = (247 rounded)

Example – Lightest and Heaviest - cont
The weight of 247 pounds separates the lightest 99.5% from the heaviest 0.5%

Non-standard normal distribution. Converting to a standard normal distribution. Procedures for finding values using Table A-2 and Formula 6-2.

Sampling Distributions and Estimators
Section 6-4 Sampling Distributions and Estimators

Key Concept The main objective of this section is to understand the concept of a sampling distribution of a statistic, which is the distribution of all values of that statistic when all possible samples of the same size are taken from the same population. We will also see that some statistics are better than others for estimating population parameters.

Definition The sampling distribution of a statistic (such as the sample proportion or sample mean) is the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population.

Definition The sampling distribution of a proportion is the distribution of sample proportions, with all samples having the same sample size n taken from the same population.

Properties Sample proportions tend to target the value of the population proportion. (That is, all possible sample proportions have a mean equal to the population proportion.) Under certain conditions, the distribution of the sample proportion can be approximated by a normal distribution.

Definition The sampling distribution of the mean is the distribution of sample means, with all samples having the same sample size n taken from the same population. (The sampling distribution of the mean is typically represented as a probability distribution in the format of a table, probability histogram, or formula.)

Definition The value of a statistic, such as the sample mean x, depends on the particular values included in the sample, and generally varies from sample to sample. This variability of a statistic is called sampling variability.

Estimators Some statistics work much better than others as estimators of the population. The example that follows shows this.

Example - Sampling Distributions
A population consists of the values 1, 2, and 5. We randomly select samples of size 2 with replacement. There are 9 possible samples. a. For each sample, find the mean, median, range, variance, and standard deviation. b. For each statistic, find the mean from part (a)

Sampling Distributions
A population consists of the values 1, 2, and 5. We randomly select samples of size 2 with replacement. There are 9 possible samples. a. For each sample, find the mean, median, range, variance, and standard deviation. See Table 6-7 on the next slide.

A population consists of the values 1, 2, and 5. We randomly select samples of size 2 with replacement. There are 9 possible samples. b. For each statistic, find the mean from part (a) The means are found near the bottom of Table 6-7.

Interpretation of Sampling Distributions We can see that when using a sample statistic to estimate a population parameter, some statistics are good in the sense that they target the population parameter and are therefore likely to yield good results. Such statistics are called unbiased estimators. Statistics that target population parameters: mean, variance, proportion Statistics that do not target population parameters: median, range, standard deviation

Sampling distribution of a statistic. Sampling distribution of a proportion. Sampling distribution of the mean. Sampling variability. Estimators.

The Central Limit Theorem
Section 6-5 The Central Limit Theorem

Key Concept The procedures of this section form the foundation for estimating population parameters and hypothesis testing – topics discussed at length in the following chapters.

Central Limit Theorem Given:
1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation . 2. Simple random samples all of size n are selected from the population. (The samples are selected so that all possible samples of the same size n have the same chance of being selected.)

Central Limit Theorem - cont
Conclusions: 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means is the population mean µ. 3. The standard deviation of all sample means is  n

Practical Rules Commonly Used
1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n (not just the values of n larger than 30).

x =  µx = µ  n Notation the mean of the sample means
the standard deviation of sample mean  (often called the standard error of the mean) µx = µ n x = 

Simulation With Random Digits
Generate 500,000 random digits, group into 5000 samples of 100 each. Find the mean of each sample. Even though the original 500,000 digits have a uniform distribution, the distribution of 5000 sample means is approximately a normal distribution!

Important Point As the sample size increases, the sampling distribution of sample means approaches a normal distribution. Result of example on p 282 of Elementary Statistics, 10th Edition

Example – Water Taxi Safety
Given the population of men has normally distributed weights with a mean of 172 lb and a standard deviation of 29 lb, a) if one man is randomly selected, find the probability that his weight is greater than lb. b) if 20 different men are randomly selected, find the probability that their mean weight is greater than 175 lb (so that their total weight exceeds the safe capacity of 3500 pounds).

Example – cont a) if one man is randomly selected, find the probability that his weight is greater than lb. z = 175 – 172 = 0.10 29

Example – cont b) if 20 different men are randomly selected, find the probability that their mean weight is greater than 172 lb. z = 175 – 172 = 0.46 29 20

Example - cont a) if one man is randomly selected, find the probability that his weight is greater than 175 lb. P(x > 175) = b) if 20 different men are randomly selected, their mean weight is greater than 175 lb. P(x > 175) = It is much easier for an individual to deviate from the mean than it is for a group of 20 to deviate from the mean.

Interpretation of Results
Given that the safe capacity of the water taxi is 3500 pounds, there is a fairly good chance (with probability ) that it will be overloaded with 20 randomly selected men.

Correction for a Finite Population
When sampling without replacement and the sample size n is greater than 5% of the finite population of size N, adjust the standard deviation of sample means by the following correction factor: N – n x =  n N – 1 finite population correction factor

Recap In this section we have discussed: Central limit theorem.
Practical rules. Effects of sample sizes. Correction for a finite population.

Normal as Approximation to Binomial
Section 6-6 Normal as Approximation to Binomial

Key Concept This section presents a method for using a normal distribution as an approximation to the binomial probability distribution. If the conditions of np ≥ 5 and nq ≥ 5 are both satisfied, then probabilities from a binomial probability distribution can be approximated well by using a normal distribution with mean μ = np and standard deviation σ = √npq

Review Binomial Probability Distribution
1. The procedure must have fixed number of trials. 2. The trials must be independent. 3. Each trial must have all outcomes classified into two categories. 4. The probability of success remains the same in all trials. Solve by binomial probability formula, Table A-1, or technology. A review of necessary requirements for a binomial probability distribution. page 214 of Elementary Statistics, 10th Edition

Approximation of a Binomial Distribution with a Normal Distribution
np  5 nq  5 then µ = np and  = npq and the random variable has distribution. (normal) a

Procedure for Using a Normal Distribution to Approximate a Binomial Distribution
1. Establish that the normal distribution is a suitable approximation to the binomial distribution by verifying np  5 and nq  5. 2. Find the values of the parameters µ and  by calculating µ = np and  = npq. 3. Identify the discrete value of x (the number of successes). Change the discrete value x by replacing it with the interval from x – 0.5 to x (See continuity corrections discussion later in this section.) Draw a normal curve and enter the values of µ , , and either x – 0.5 or x + 0.5, as appropriate.

Procedure for Using a Normal Distribution to Approximate a Binomial Distribution - cont
4. Change x by replacing it with x – 0.5 or x + 0.5, as appropriate. 5. Using x – 0.5 or x (as appropriate) in place of x, find the area corresponding to the desired probability by first finding the z score and finding the area to the left of the adjusted value of x.

Example – Number of Men Among Passengers
Finding the Probability of “At Least 122 Men” Among 213 Passengers Figure 6-21

(that is, adding and subtracting 0.5).
Definition When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a discrete whole number x in the binomial distribution by representing the single value x by the interval from x – 0.5 to x + 0.5 (that is, adding and subtracting 0.5). page 295 of Elementary Statistics, 10th Edition

Procedure for Continuity Corrections
1. When using the normal distribution as an approximation to the binomial distribution, always use the continuity correction. 2. In using the continuity correction, first identify the discrete whole number x that is relevant to the binomial probability problem. 3. Draw a normal distribution centered about µ, then draw a vertical strip area centered over x . Mark the left side of the strip with the number x – 0.5, and mark the right side with x For x = 122, draw a strip from to Consider the area of the strip to represent the probability of discrete whole number x.

Procedure for Continuity Corrections - cont
4. Now determine whether the value of x itself should be included in the probability you want. Next, determine whether you want the probability of at least x, at most x, more than x, fewer than x, or exactly x. Shade the area to the right or left of the strip, as appropriate; also shade the interior of the strip itself if and only if x itself is to be included. The total shaded region corresponds to the probability being sought.

x = at least 122 x = more than 122 x = at most 122 x = fewer than 122
Figure 6-22 x = at least 122 (includes 122 and above) x = more than 122 (doesn’t include 122) x = at most 122 (includes 122 and below) x = fewer than 122 (doesn’t include 122) x = exactly 122

Approximating a binomial distribution with a normal distribution. Procedures for using a normal distribution to approximate a binomial distribution. Continuity corrections.

Section 6-7 Assessing Normality

Key Concept This section provides criteria for determining whether the requirement of a normal distribution is satisfied. The criteria involve visual inspection of a histogram to see if it is roughly bell shaped, identifying any outliers, and constructing a new graph called a normal quantile plot.

Definition A normal quantile plot (or normal probability plot) is a graph of points (x,y), where each x value is from the original set of sample data, and each y value is the corresponding z score that is a quantile value expected from the standard normal distribution. page 302 of Elementary Statistics, 10th Edition

Methods for Determining Whether Data Have a Normal Distribution
1. Histogram: Construct a histogram. Reject normality if the histogram departs dramatically from a bell shape. 2. Outliers: Identify outliers. Reject normality if there is more than one outlier present. 3. Normal Quantile Plot: If the histogram is basically symmetric and there is at most one outlier, construct a normal quantile plot as follows:

Procedure for Determining Whether Data Have a Normal Distribution - cont
3. Normal Quantile Plot a. Sort the data by arranging the values from lowest to highest. b. With a sample size n, each value represents a proportion of 1/n of the sample. Using the known sample size n, identify the areas of 1/2n, 3/2n, 5/2n, 7/2n, and so on. These are the cumulative areas to the left of the corresponding sample values. c. Use the standard normal distribution (Table A-2, software or calculator) to find the z scores corresponding to the cumulative left areas found in Step (b).

Procedure for Determining Whether Data Have a Normal Distribution - cont
d. Match the original sorted data values with their corresponding z scores found in Step (c), then plot the points (x, y), where each x is an original sample value and y is the corresponding z score. e. Examine the normal quantile plot using these criteria: If the points do not lie close to a straight line, or if the points exhibit some systematic pattern that is not a straight-line pattern, then the data appear to come from a population that does not have a normal distribution. If the pattern of the points is reasonably close to a straight line, then the data appear to come from a population that has a normal distribution.

Example Interpretation: Because the points lie reasonably close to a straight line and there does not appear to be a systematic pattern that is not a straight-line pattern, we conclude that the sample appears to be a normally distributed population.

Recap In this section we have discussed: Normal quantile plot.
Procedure to determine if data have a normal distribution.

Chapter 7 Estimates and Sample Sizes
7-1 Overview 7-2 Estimating a Population Proportion 7-3 Estimating a Population Mean: σ Known 7-4 Estimating a Population Mean: σ Not Known 7-5 Estimating a Population Variance

Overview This chapter presents the beginning of inferential statistics. The two major applications of inferential statistics involve the use of sample data to (1) estimate the value of a population parameter, and (2) test some claim (or hypothesis) about a population. We introduce methods for estimating values of these important population parameters: proportions, means, and variances. We also present methods for determining sample sizes necessary to estimate those parameters.

Section 7-2 Estimating a Population Proportion

Key Concept In this section we present important methods for using a sample proportion to estimate the value of a population proportion with a confidence interval. We also present methods for finding the size of the sample needed to estimate a population proportion.

Requirements for Estimating a Population Proportion
1. The sample is a simple random sample. 2. The conditions for the binomial distribution are satisfied. (See Section 5-3.) 3. There are at least 5 successes and 5 failures.

Notation for Proportions
population proportion p = ˆ x n sample proportion of x successes in a sample of size n (pronounced ‘p-hat’) q = 1 - p = sample proportion of failures in a sample size of n ˆ ˆ

Definition A point estimate is a single value (or point) used to approximate a population parameter.

Definition ˆ The sample proportion p is the best point estimate of the population proportion p.

Example: 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Using these survey results, find the best point estimate of the proportion of all adult Minnesotans opposed to photo-cop use. Because the sample proportion is the best point estimate of the population proportion, we conclude that the best point estimate of p is When using the survey results to estimate the percentage of all adult Minnesotans that are opposed to photo-cop use, our best estimate is 51%.

Definition A confidence interval (or interval estimate) is a range (or an interval) of values used to estimate the true value of a population parameter. A confidence interval is sometimes abbreviated as CI.

Definition Most common choices are 90%, 95%, or 99%.
A confidence level is the probability 1-  (often expressed as the equivalent percentage value) that is the proportion of times that the confidence interval actually does contain the population parameter, assuming that the estimation process is repeated a large number of times. (The confidence level is also called degree of confidence, or the confidence coefficient.) Most common choices are 90%, 95%, or 99%. ( = 10%), ( = 5%), ( = 1%)

Example: 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Using these survey results, find the 95% confidence interval of the proportion of all adult Minnesotans opposed to photo-cop use. “We are 95% confident that the interval from to does contain the true value of p.” This means if we were to select many different samples of size 829 and construct the corresponding confidence intervals, 95% of them would actually contain the value of the population proportion p.

Using Confidence Intervals for Comparisons
Do not use the overlapping of confidence intervals as the basis for making formal and final conclusions about the equality of proportions.

Critical Values 1. We know from Section 6-6 that under certain conditions, the sampling distribution of sample proportions can be approximated by a normal distribution, as in Figure 7-2, following. 2. Sample proportions have a relatively small chance (with probability denoted by ) of falling in one of the red tails of Figure 7-2, following. 3. Denoting the area of each shaded tail by /2, we see that there is a total probability of  that a sample proportion will fall in either of the two red tails.

Critical Values 4. By the rule of complements (from Chapter 4), there is a probability of 1- that a sample proportion will fall within the inner region of Figure 7-2, following. 5. The z score separating the right-tail is commonly denoted by z /2 and is referred to as a critical value because it is on the borderline separating sample proportions that are likely to occur from those that are unlikely to occur.

z2 The Critical Value Figure 7-2

Notation for Critical Value
The critical value z/2 is the positive z value that is at the vertical boundary separating an area of /2 in the right tail of the standard normal distribution. (The value of –z/2 is at the vertical boundary for the area of /2 in the left tail.) The subscript /2 is simply a reminder that the z score separates an area of /2 in the right tail of the standard normal distribution.

Definition A critical value is the number on the borderline separating sample statistics that are likely to occur from those that are unlikely to occur. The number z/2 is a critical value that is a z score with the property that it separates an area of /2 in the right tail of the standard normal distribution. (See Figure 7-2).

Finding z2 for a 95% Confidence Level
 = 5%  2 = 2.5% = .025 -z2 z2 Critical Values

Finding z2 for a 95% Confidence Level - cont
 = 0.05 Use Table A-2 to find a z score of 1.96 z2 =  

Definition When data from a simple random sample are used to estimate a population proportion p, the margin of error, denoted by E, is the maximum likely (with probability 1 – ) difference between the observed proportion p and the true value of the population proportion p. The margin of error E is also called the maximum error of the estimate and can be found by multiplying the critical value and the standard deviation of the sample proportions, as shown in Formula 7-1, following. ˆ

Margin of Error of the Estimate of p
Formula 7-1 z a / 2 E = n ˆ p q

Confidence Interval for Population Proportion
p – E < < E ˆ p where z a / 2 E = n ˆ p q

Confidence Interval for Population Proportion - cont
p – E < < E p + E p ˆ ˆ (p – E, p + E) ˆ

Round-Off Rule for Confidence Interval Estimates of p
Round the confidence interval limits for p to three significant digits.

Procedure for Constructing a Confidence Interval for p
Verify that the required assumptions are satisfied. (The sample is a simple random sample, the conditions for the binomial distribution are satisfied, and the normal distribution can be used to approximate the distribution of sample proportions because np  5, and nq  5 are both satisfied.) 2. Refer to Table A-2 and find the critical value z/2 that corresponds to the desired confidence level. 3. Evaluate the margin of error E = ˆ n p q

Procedure for Constructing a Confidence Interval for p - cont
ˆ 4. Using the value of the calculated margin of error, E and the value of the sample proportion, p, find the values of p – E and p + E. Substitute those values in the general format for the confidence interval: ˆ ˆ ˆ ˆ p – E < p < p + E 5. Round the resulting confidence interval limits to three significant digits.

Example: 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Use these survey results. a) Find the margin of error E that corresponds to a 95% confidence level. b) Find the 95% confidence interval estimate of the population proportion p. c) Based on the results, can we safely conclude that the majority of adult Minnesotans oppose use the the photo-cop?

Example: 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Use these survey results. a) Find the margin of error E that corresponds to a 95% confidence level. ˆ ˆ First, we check for assumptions. We note that np =  5, and nq =  5. ˆ ˆ Next, we calculate the margin of error. We have found that p = 0.51, q = 1 – 0.51 = 0.49, z2 = 1.96, and n = 829. E = 1.96 (0.51)(0.49) 829 E =

Example: 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Use these survey results. b) Find the 95% confidence interval for the population proportion p. We substitute our values from Part a to obtain: 0.51 – < p < , 0.476 < p < 0.544

Example: 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Use these survey results. c) Based on the results, can we safely conclude that the majority of adult Minnesotans oppose use of the photo-cop? Based on the survey results, we are 95% confident that the limits of 47.6% and 54.4% contain the true percentage of adult Minnesotans opposed to the photo-cop. The percentage of opposed adult Minnesotans is likely to be any value between 47.6% and 54.4%. However, a majority requires a percentage greater than 50%, so we cannot safely conclude that the majority is opposed (because the entire confidence interval is not greater than 50%).

Sample Size Suppose we want to collect sample data with the objective of estimating some population. The question is how many sample items must be obtained?

Determining Sample Size
p q ˆ n (solve for n by algebra) ( )2 ˆ p q a / 2 Z n = E 2

Sample Size for Estimating Proportion p
When an estimate of p is known: ˆ Formula 7-2 ˆ ( )2 p q n = E 2 a / 2 z When no estimate of p is known: Formula 7-3 ( )2 0.25 n = E 2 a / 2 z ˆ

Example: Suppose a sociologist wants to determine the current percentage of U.S. households using . How many households must be surveyed in order to be 95% confident that the sample percentage is in error by no more than four percentage points? a) Use this result from an earlier study: In 1997, 16.9% of U.S. households used (based on data from The World Almanac and Book of Facts). b) Assume that we have no prior information suggesting a possible value of p. ˆ

Example: Suppose a sociologist wants to determine the current percentage of U.S. households using . How many households must be surveyed in order to be 95% confident that the sample percentage is in error by no more than four percentage points? a) Use this result from an earlier study: In 1997, 16.9% of U.S. households used (based on data from The World Almanac and Book of Facts). n = [za/2 ]2 p q E2 ˆ 0.042 = [1.96]2 (0.169)(0.831) = = 338 households To be 95% confident that our sample percentage is within four percentage points of the true percentage for all households, we should randomly select and survey 338 households.

Example: Suppose a sociologist wants to determine the current percentage of U.S. households using . How many households must be surveyed in order to be 95% confident that the sample percentage is in error by no more than four percentage points? ˆ b) Assume that we have no prior information suggesting a possible value of p. n = [za/2 ]2 • 0.25 E2 With no prior information, we need a larger sample to achieve the same results with 95% confidence and an error of no more than 4%. = (1.96)2 (0.25) 0.042 = = 601 households

Finding the Point Estimate and E from a Confidence Interval
Point estimate of p: p = (upper confidence limit) + (lower confidence limit) 2 ˆ ˆ Margin of Error: E = (upper confidence limit) — (lower confidence limit) 2

Recap In this section we have discussed: Point estimates.
Confidence intervals. Confidence levels. Critical values. Margin of error. Determining sample sizes.

Section 7-3 Estimating a Population Mean:  Known

Key Concept This section presents methods for using sample data to find a point estimate and confidence interval estimate of a population mean. A key requirement in this section is that we know the standard deviation of the population.

Requirements 1. The sample is a simple random sample. (All samples of the same size have an equal chance of being selected.) 2. The value of the population standard deviation  is known. 3. Either or both of these conditions is satisfied: The population is normally distributed or n > 30. page 338 of Elementary Statistics, 10th Edition

Point Estimate of the Population Mean
The sample mean x is the best point estimate of the population mean µ.

Sample Mean For all populations, the sample mean x is an unbiased estimator of the population mean , meaning that the distribution of sample means tends to center about the value of the population mean . 2. For many populations, the distribution of sample means x tends to be more consistent (with less variation) than the distributions of other sample statistics. page 339 of Elementary Statistics, 10th Edition

Example: A study found the body temperatures of 106 healthy adults
Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the point estimate of the population mean  of all body temperatures. Because the sample mean x is the best point estimate of the population mean , we conclude that the best point estimate of the population mean  of all body temperatures is 98.20o F.

Definition The margin of error is the maximum likely difference observed between sample mean x and population mean µ, and is denoted by E.

E = z/2 • Formula 7-4 Formula Margin of Error  n
Margin of error for mean (based on known σ)

Confidence Interval estimate of the Population Mean µ (with  Known)
x – E < µ < x + E or x + E or (x – E, x + E)

Definition The two values x – E and x + E are called confidence interval limits.

Procedure for Constructing a Confidence Interval for µ (with Known )
1. Verify that the requirements are satisfied. 2. Refer to Table A-2 and find the critical value z2 that corresponds to the desired degree of confidence. 3. Evaluate the margin of error E = z2 • / n . 4. Find the values of x – E and x + E. Substitute those values in the general format of the confidence interval: page 340 of Elementary Statistics, 10th Edition x – E < µ < x + E 5. Round using the confidence intervals round-off rules.

Round-Off Rule for Confidence Intervals Used to Estimate µ
When using the original set of data, round the confidence interval limits to one more decimal place than used in original set of data. When the original set of data is unknown and only the summary statistics (n, x, s) are used, round the confidence interval limits to the same number of decimal places used for the sample mean.

Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the margin of error E and the 95% confidence interval for µ. n = 106 x = 98.20o s = 0.62o  = 0.05 /2 = 0.025 z / 2 = 1.96 E = z / 2 •  = • = 0.12 n 106 x – E < μ < x + E 98.08o <  < o 98.20o – <  < 98.20o

Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the margin of error E and the 95% confidence interval for µ. n = 106 x = 98.20o s = 0.62o  = 0.05 /2 = 0.025 z / 2 = 1.96 E = z / 2 •  = • = 0.12 n 106 x – E <  < x + E 98.08o <  < o Based on the sample provided, the confidence interval for the population mean is o <  < 98.32o. If we were to select many different samples of the same size, 95% of the confidence intervals would actually contain the population mean .

Sample Size for Estimating Mean 
Formula 7-5 (z/2)   n = E 2 Where zα/2 = critical z score based on the desired confidence level E = desired margin of error σ = population standard deviation

Round-Off Rule for Sample Size n
When finding the sample size n, if the use of Formula 7-5 does not result in a whole number, always increase the value of n to the next larger whole number. page 343 of Elementary Statistics, 10th Edition

Finding the Sample Size n When  is Unknown
1. Use the range rule of thumb (see Section 3-3) to estimate the standard deviation as follows:   range/4. 2. Conduct a pilot study by starting the sampling process. Start the sample collection process and, using the first several values, calculate the sample standard deviation s and use it in place of . page 344 of Elementary Statistics, 10th Edition 3. Estimate the value of  by using the results of some other study that was done earlier.

Example: Assume that we want to estimate the mean IQ score for the population of statistics professors. How many statistics professors must be randomly selected for IQ tests if we want 95% confidence that the sample mean is within 2 IQ points of the population mean? Assume that  = 15, as is found in the general population.  = 0.05 /2 = 0.025 z / 2 = 1.96 E = 2  = 15 n = • = = 217 2 Example on page 344 of Elementary Statistics, 10th Edition With a simple random sample of only 217 statistics professors, we will be 95% confident that the sample mean will be within 2 IQ points of the true population mean .

Recap In this section we have discussed: Margin of error.
Confidence interval estimate of the population mean with σ known. Round off rules. Sample size for estimating the mean μ.

Section 7-4 Estimating a Population Mean:  Not Known

Key Concept This section presents methods for finding a confidence interval estimate of a population mean when the population standard deviation is not known. With σ unknown, we will use the Student t distribution assuming that certain requirements are satisfied.

Requirements with σ Unknown
1) The sample is a simple random sample. 2) Either the sample is from a normally distributed population, or n > 30. Use Student t distribution Smaller samples will have means that are likely to vary more. The greater variation is accounted for by the t distribution. Students usually like hearing about the history of how the Student t distribution got its name. page 350 of text See Note to Instructor in margin for other ideas.

Student t Distribution
If the distribution of a population is essentially normal, then the distribution of t = x - µ s n Student t distribution is usually referred to as the t distribution. is a Student t Distribution for all samples of size n. It is often referred to a a t distribution and is used to find critical values denoted by t/2.

Definition degrees of freedom = n – 1 in this section.
The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values. degrees of freedom = n – 1 in this section. Explain that other statistical procedures may have different formulas for degrees of freedom.

Margin of Error E for Estimate of  (With σ Not Known)
Formula 7-6 where t/2 has n – 1 degrees of freedom. n s E = t/ 2 Table A-3 lists values for tα/2

Confidence Interval for the Estimate of μ (With σ Not Known)
x – E < µ < x + E where E = t/2 n s t/2 found in Table A-3

Procedure for Constructing a Confidence Interval for µ (With σ Unknown)
1. Verify that the requirements are satisfied. 2. Using n - 1 degrees of freedom, refer to Table A-3 and find the critical value t2 that corresponds to the desired confidence level. 3. Evaluate the margin of error E = t2 • s / n . 4. Find the values of x - E and x + E. Substitute those values in the general format for the confidence interval: x – E < µ < x + E 5. Round the resulting confidence interval limits.

Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the margin of error E and the 95% confidence interval for µ.

Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the margin of error E and the 95% confidence interval for µ. n = 106 x = 98.20o s = 0.62o  = 0.05 /2 = 0.025 t / 2 = 1.96 E = t / 2 • s = • = n 106 x – E <  < x + E 98.08o <  < o

Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the margin of error E and the 95% confidence interval for µ. n = 106 x = 98.20o s = 0.62o  = 0.05 /2 = 0.025 t / 2 = 1.96 E = t / 2 • s = • = n 106 x – E <  < x + E 98.08o <  < o Based on the sample provided, the confidence interval for the population mean is o <  < 98.32o. The interval is the same here as in Section 7-2, but in some other cases, the difference would be much greater.

Important Properties of the Student t Distribution
1. The Student t distribution is different for different sample sizes (see Figure 7-5, following, for the cases n = 3 and n = 12). 2. The Student t distribution has the same general symmetric bell shape as the standard normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples. 3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean of z = 0). 4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1 (unlike the standard normal distribution, which has a  = 1). 5. As the sample size n gets larger, the Student t distribution gets closer to the normal distribution.

Student t Distributions for n = 3 and n = 12
Student t distributions have the same general shape and symmetry as the standard normal distribution, but reflect a greater variability that is expected with small samples. Figure 7-5

Choosing the Appropriate Distribution
Figure 7-6

Example: Flesch ease of reading scores for 12 different pages randomly selected from J.K. Rowling’s Harry Potter and the Sorcerer’s Stone. Find the 95% interval estimate of , the mean Flesch ease of reading score. (The 12 pages’ distribution appears to be bell-shaped with x = and s = 4.68.)

E = t2 s = (2.201)(4.68) = 2.97355 x – E < µ < x + E
Example: Flesch ease of reading scores for 12 different pages randomly selected from J.K. Rowling’s Harry Potter and the Sorcerer’s Stone. Find the 95% interval estimate of , the mean Flesch ease of reading score. (The 12 pages’ distribution appears to be bell-shaped with x = and s = 4.68.) E = t s = (2.201)(4.68) = n 12 x = 80.75 s =  = 0.05 /2 = 0.025 t/2 = 2.201 x – E < µ < x + E 80.75 – < µ < <  < 77.78 <  < 83.72 We are 95% confident that this interval contains the mean Flesch ease of reading score for all pages.

Finding the Point Estimate and E from a Confidence Interval
Point estimate of µ: x = (upper confidence limit) + (lower confidence limit) 2 Margin of Error: E = (upper confidence limit) – (lower confidence limit) 2

Confidence Intervals for Comparing Data
As before in Sections 7-2 and 7-3, do not use the overlapping of confidence intervals as the basis for making final conclusions about the equality of means.

Recap In this section we have discussed: Student t distribution.
Degrees of freedom. Margin of error. Confidence intervals for μ with σ unknown. Choosing the appropriate distribution. Point estimates. Using confidence intervals to compare data.

Section 7-5 Estimating a Population Variance

Key Concept This section presents methods for (1) finding a confidence interval estimate of a population standard deviation or variance and (2) determining the sample size required to estimate a population standard deviation or variance. We also introduce the chi-square distribution, which is used for finding a confidence interval estimate for σ or σ 2.

Requirements 1. The sample is a simple random sample.
2. The population must have normally distributed values (even if the sample is large). The requirement of a normal distribution is critical with this test. Use of non-normal distributions can lead to gross errors. Use the methods discussed in Section 6-7, Assessing Normality, to verify a normal distribution before proceeding with this procedure.

Chi-Square Distribution
 2 = 2 (n – 1) s2 Formula 7-7 where n = sample size s 2 = sample variance 2 = population variance One place where students tend to make errors with this formula is switching the values for s and sigma. Careful reading of the problem should eliminate this difficulty. page 364 of Elementary Statistics, 10th Edition

Properties of the Distribution of the Chi-Square Statistic
1. The chi-square distribution is not symmetric, unlike the normal and Student t distributions. Figure 7-9 Chi-Square Distribution for df = 10 and df = 20 As the number of degrees of freedom increases, the distribution becomes more symmetric. Figure 7-8 Chi-Square Distribution Different degrees of freedom produce different shaped chi-square distributions.

Properties of the Distribution of the Chi-Square Statistic - cont
2. The values of chi-square can be zero or positive, but they cannot be negative. 3. The chi-square distribution is different for each number of degrees of freedom, which is df = n – 1 in this section. As the number increases, the chi-square distribution approaches a normal distribution. In Table A-4, each critical value of 2 corresponds to an area given in the top row of the table, and that area represents the cumulative area located to the right of the critical value.

Example: Find the critical values of 2 that determine critical regions containing an area of in each tail. Assume that the relevant sample size is 10 so that the number of degrees of freedom is 10 – 1, or 9.  = 0.05 /2 = 0.025 1 /2 = 0.975

Critical Values of the Chi-Square Distribution
Areas to the right of each tail Figure 7-10

Estimators of 2 The sample variance s is the best point estimate of the population variance 2 . 2

Confidence Interval (or Interval Estimate) for the Population Variance  2
(n – 1)s 2 (n – 1)s 2 2    2 2 Right-tail CV R L Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula.

Confidence Interval (or Interval Estimate) for the Population Variance  2 - cont
(n – 1)s 2 (n – 1)s 2 2    2 2 Right-tail CV L Left-tail CV R Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula.

Confidence Interval (or Interval Estimate) for the Population Variance 2 - cont
2  (n – 1)s 2  2 Right-tail CV Left-tail CV 2 R L Confidence Interval for the Population Standard Deviation  Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula. (n – 1)s 2  2  L R

Procedure for Constructing a Confidence Interval for  or  2
1. Verify that the required assumptions are satisfied. 2. Using n – 1 degrees of freedom, refer to Table A-4 and find the critical values 2R and 2Lthat correspond to the desired confidence level. 3. Evaluate the upper and lower confidence interval limits using this format of the confidence interval: 2  (n – 1)s 2  2 R L

Procedure for Constructing a Confidence Interval for  or  2 - cont
4. If a confidence interval estimate of  is desired, take the square root of the upper and lower confidence interval limits and change  2 to . 5. Round the resulting confidence level limits. If using the original set of data to construct a confidence interval, round the confidence interval limits to one more decimal place than is used for the original set of data. If using the sample standard deviation or variance, round the confidence interval limits to the same number of decimals places.

Confidence Intervals for Comparing Data
As in previous sections, do not use the overlapping of confidence intervals as the basis for making final conclusions about the equality of variances or standard deviations.

Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the 95% confidence interval for . n = 106 x = 98.2o s = 0.62o  = 0.05 /2 = 0.025 1 – /2 = 0.975  2R = ,  2L = (106 – 1)(0.62)2 < 2 < (106 – 1)(0.62)2 0.31 < 2 < 0.54 0.56 <  < 0.74 We are 95% confident that the limits of 0.56°F and 0.74°F contain the true value of . We are 95% confident that the standard deviation of body temperatures of all healthy people is between 0.56°F and 0.74°F.

Determining Sample Sizes
Instead of presenting the much more complex procedures for determining sample size necessary to estimate the population variance, this table will be used.

Example: We want to estimate , the standard deviation off all body temperatures. We want to be 95% confident that our estimate is within 10% of the true value of . How large should the sample be? Assume that the population is normally distributed. From Table 7-2, we can see that 95% confidence and an error of 10% for  correspond to a sample of size 191.

Recap In this section we have discussed: The chi-square distribution.
Using Table A-4. Confidence intervals for the population variance and standard deviation. Determining sample sizes.

Chapter 8 Hypothesis Testing
8-1 Overview 8-2 Basics of Hypothesis Testing 8-3 Testing a Claim about a Proportion 8-4 Testing a Claim About a Mean: σ Known 8-5 Testing a Claim About a Mean: σ Not Known 8-6 Testing a Claim About a Standard Deviation or Variance

Definitions In statistics, a hypothesis is a claim or statement about a property of a population. A hypothesis test (or test of significance) is a standard procedure for testing a claim about a property of a population. page 386 of Elementary Statistics, 10th Edition Various examples are provided below definition box

Rare Event Rule for Inferential Statistics
If, under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption is probably not correct. Example on page 386 of Elementary Statistics, 10th Edition Introduce the word ‘significant’ in regard to hypothesis testing.

Example: ProCare Industries, Ltd
Example: ProCare Industries, Ltd., once provided a product called “Gender Choice,” which, according to advertising claims, allowed couples to “increase your chances of having a boy up to 85%, a girl up to 80%.” Gender Choice was available in blue packages for couples wanting a baby boy and (you guessed it) pink packages for couples wanting a baby girl. Suppose we conduct an experiment with 100 couples who want to have baby girls, and they all follow the Gender Choice “easy-to-use in-home system” described in the pink package. For the purpose of testing the claim of an increased likelihood for girls, we will assume that Gender Choice has no effect. Using common sense and no formal statistical methods, what should we conclude about the assumption of no effect from Gender Choice if 100 couples using Gender Choice have 100 babies consisting of a) 52 girls?; b) 97 girls?

Example: ProCare Industries, Ltd.: Part a)
a) We normally expect around 50 girls in 100 births. The result of 52 girls is close to 50, so we should not conclude that the Gender Choice product is effective. If the 100 couples used no special method of gender selection, the result of 52 girls could easily occur by chance. The assumption of no effect from Gender Choice appears to be correct. There isn’t sufficient evidence to say that Gender Choice is effective.

Example: ProCare Industries, Ltd.: Part b)
b) The result of 97 girls in 100 births is extremely unlikely to occur by chance. We could explain the occurrence of 97 girls in one of two ways: Either an extremely rare event has occurred by chance, or Gender Choice is effective. The extremely low probability of getting 97 girls is strong evidence against the assumption that Gender Choice has no effect. It does appear to be effective.

Basics of Hypothesis Testing
Section 8-2 Basics of Hypothesis Testing Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Tom Wegleitner, Centreville, VA

Key Concept This section presents individual components of a hypothesis test, and the following sections use those components in comprehensive procedures. The role of the following should be understood: null hypothesis alternative hypothesis test statistic critical region significance level critical value P-value Type I and II error

Section 8-2 Objectives Given a claim, identify the null hypothesis and the alternative hypothesis, and express them both in symbolic form. Given a claim and sample data, calculate the value of the test statistic. Given a significance level, identify the critical value(s). Given a value of the test statistic, identify the P-value. State the conclusion of a hypothesis test in simple, non-technical terms. Example on page 387 of Elementary Statistics, 10th Edition

Example: Let’s again refer to the Gender Choice product that was once distributed by ProCare Industries. ProCare Industries claimed that couples using the pink packages of Gender Choice would have girls at a rate that is greater than 50% or Let’s again consider an experiment whereby 100 couples use Gender Choice in an attempt to have a baby girl; let’s assume that the 100 babies include exactly 52 girls, and let’s formalize some of the analysis. Under normal circumstances the proportion of girls is 0.5, so a claim that Gender Choice is effective can be expressed as p > Using a normal distribution as an approximation to the binomial distribution, we find P(52 or more girls in 100 births) =

Example: Let’s again refer to the Gender Choice product that was once distributed by ProCare Industries. ProCare Industries claimed that couples using the pink packages of Gender Choice would have girls at a rate that is greater than 50% or Let’s again consider an experiment whereby 100 couples use Gender Choice in an attempt to have a baby girl; let’s assume that the 100 babies include exactly 52 girls, and let’s formalize some of the analysis Figure 8-1, following, shows that with a probability of 0.5, the outcome of 52 girls in 100 births is not unusual.

Figure 8-1 We do not reject random chance as a reasonable explanation. We conclude that the proportion of girls born to couples using Gender Choice is not significantly greater than the number that we would expect by random chance.

Observations Claim: For couples using Gender Choice, the proportion of girls is p > 0.5. Working assumption: The proportion of girls is p = 0.5 (with no effect from Gender Choice). ˆ The sample resulted in 52 girls among 100 births, so the sample proportion is p = 52/100 = 0.52. Assuming that p = 0.5, we use a normal distribution as an approximation to the binomial distribution to find that P (at least 52 girls in 100 births) = There are two possible explanations for the result of 52 girls in 100 births: Either a random chance event (with probability ) has occurred, or the proportion of girls born to couples using Gender Choice is greater than 0.5. There isn’t sufficient evidence to support Gender Choice’s claim.

Components of a Formal Hypothesis Test

Null Hypothesis: H0 The null hypothesis (denoted by H0) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value. We test the null hypothesis directly. Either reject H0 or fail to reject H0. Emphasize “equal to”.

Alternative Hypothesis: H1
The alternative hypothesis (denoted by H1 or Ha or HA) is the statement that the parameter has a value that somehow differs from the null hypothesis. The symbolic form of the alternative hypothesis must use one of these symbols: , <, >. Give examples of different ways to word “not equal to,” < and >, such as ‘is different from’, ‘fewer than’, ‘more than’, etc.

Note about Forming Your Own Claims (Hypotheses)
If you are conducting a study and want to use a hypothesis test to support your claim, the claim must be worded so that it becomes the alternative hypothesis. By examining the flowchart for the Wording of the Final Conclusion, Figure 8-7, page 397 of Elementary Statistics, 10th Edition, this requirement for support of a statement becomes clear.

Note about Identifying H0 and H1
Figure 8-2

Example: Identify the Null and Alternative Hypothesis
Example: Identify the Null and Alternative Hypothesis. Refer to Figure 8-2 and use the given claims to express the corresponding null and alternative hypotheses in symbolic form. The proportion of drivers who admit to running red lights is greater than 0.5. b) The mean height of professional basketball players is at most 7 ft. c) The standard deviation of IQ scores of actors is equal to 15.

Example: Identify the Null and Alternative Hypothesis. Refer to Figure 8-2 and use the given claims to express the corresponding null and alternative hypotheses in symbolic form. a) The proportion of drivers who admit to running red lights is greater than 0.5. In Step 1 of Figure 8-2, we express the given claim as p > In Step 2, we see that if p > 0.5 is false, then p  0.5 must be true. In Step 3, we see that the expression p > 0.5 does not contain equality, so we let the alternative hypothesis H1 be p > 0.5, and we let H0 be p = 0.5.

Example: Identify the Null and Alternative Hypothesis. Refer to Figure 8-2 and use the given claims to express the corresponding null and alternative hypotheses in symbolic form. b) The mean height of professional basketball players is at most 7 ft. In Step 1 of Figure 8-2, we express “a mean of at most 7 ft” in symbols as   7. In Step 2, we see that if   7 is false, then µ > 7 must be true. In Step 3, we see that the expression µ > 7 does not contain equality, so we let the alternative hypothesis H1 be µ > 0.5, and we let H0 be µ = 7.

Example: Identify the Null and Alternative Hypothesis. Refer to Figure 8-2 and use the given claims to express the corresponding null and alternative hypotheses in symbolic form. c) The standard deviation of IQ scores of actors is equal to 15. In Step 1 of Figure 8-2, we express the given claim as  = 15. In Step 2, we see that if  = 15 is false, then   15 must be true. In Step 3, we let the alternative hypothesis H1 be   15, and we let H0 be  = 15.

Test Statistic The test statistic is a value used in making a decision about the null hypothesis, and is found by converting the sample statistic to a score with the assumption that the null hypothesis is true.

Test Statistic - Formulas Test statistic for proportions
z = p - p  pq n  Test statistic for proportions z = x - µx  n Test statistic for mean Test statistic for standard deviation 2 = (n – 1)s2  2

Example: A survey of n = 880 randomly selected adult drivers showed that 56% (or p = 0.56) of those respondents admitted to running red lights. Find the value of the test statistic for the claim that the majority of all adult drivers admit to running red lights (In Section 8-3 we will see that there are assumptions that must be verified. For this example, assume that the required assumptions are satisfied and focus on finding the indicated test statistic.)

Solution: The preceding example showed that the given claim results in the following null and alternative hypotheses: H0: p = 0.5 and H1: p > Because we work under the assumption that the null hypothesis is true with p = 0.5, we get the following test statistic: n pq  z = p – p  = (0.5)(0.5) 880 = 3.56

Interpretation: We know from previous chapters that a z score of 3
Interpretation: We know from previous chapters that a z score of 3.56 is exceptionally large. It appears that in addition to being “more than half,” the sample result of 56% is significantly more than 50%. See figure following.

Critical Region, Critical Value, Test Statistic

Critical Region The critical region (or rejection region) is the set of all values of the test statistic that cause us to reject the null hypothesis. For example, see the red-shaded region in the previous figure.

Significance Level The significance level (denoted by ) is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true. This is the same  introduced in Section Common choices for  are 0.05, 0.01, and 0.10.

Critical Value A critical value is any value that separates the critical region (where we reject the null hypothesis) from the values of the test statistic that do not lead to rejection of the null hypothesis. The critical values depend on the nature of the null hypothesis, the sampling distribution that applies, and the significance level . See the previous figure where the critical value of z = corresponds to a significance level of  = 0.05.

Two-tailed, Right-tailed, Left-tailed Tests
The tails in a distribution are the extreme regions bounded by critical values. page 394 of Elementary Statistics, 10th Edition

 is divided equally between the two tails of the critical
Two-tailed Test H0: = H1:   is divided equally between the two tails of the critical region Means less than or greater than

Right-tailed Test H0: = H1: > Points Right

Left-tailed Test H0: = H1: < Points Left

P-Value The P-value (or p-value or probability value) is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. The null hypothesis is rejected if the P-value is very small, such as 0.05 or less.

Conclusions in Hypothesis Testing
We always test the null hypothesis. The initial conclusion will always be one of the following: 1. Reject the null hypothesis. 2. Fail to reject the null hypothesis. page 394 of Elementary Statistics, 10th Edition

Decision Criterion Traditional method:
Reject H0 if the test statistic falls within the critical region. Fail to reject H0 if the test statistic does not fall within the critical region.

Decision Criterion - cont
P-value method: Reject H0 if the P-value   (where  is the significance level, such as 0.05). Fail to reject H0 if the P-value > .

Another option: Instead of using a significance level such as 0.05, simply identify the P-value and leave the decision to the reader.

Confidence Intervals: Because a confidence interval estimate of a population parameter contains the likely values of that parameter, reject a claim that the population parameter has a value that is not included in the confidence interval.

Procedure for Finding P-Values
Figure 8-6

Example: Finding P-values
Example: Finding P-values. First determine whether the given conditions result in a right-tailed test, a left-tailed test, or a two-tailed test, then find the P-values and state a conclusion about the null hypothesis. a) A significance level of  = 0.05 is used in testing the claim that p > 0.25, and the sample data result in a test statistic of z = 1.18. b) A significance level of  = 0.05 is used in testing the claim that p  0.25, and the sample data result in a test statistic of z = 2.34.

Example: Finding P-values. First determine whether the given conditions result in a right-tailed test, a left-tailed test, or a two-tailed test, then find the P-values and state a conclusion about the null hypothesis. a) With a claim of p > 0.25, the test is right-tailed. Because the test is right-tailed, Figure 8-6 shows that the P-value is the area to the right of the test statistic z = We refer to Table A-2 and find that the area to the right of z = 1.18 is The P-value of is greater than the significance level  = 0.05, so we fail to reject the null hypothesis. The P-value of is relatively large, indicating that the sample results could easily occur by chance.

Example: Finding P-values. First determine whether the given conditions result in a right-tailed test, a left-tailed test, or a two-tailed test, then find the P-values and state a conclusion about the null hypothesis. b) With a claim of p  0.25, the test is two-tailed. Because the test is two-tailed, and because the test statistic of z = 2.34 is to the right of the center, Figure 8-6 shows that the P-value is twice the area to the right of z = We refer to Table A-2 and find that the area to the right of z = 2.34 is , so P-value = x = The P-value of is less than or equal to the significance level, so we reject the null hypothesis. The small P-value o shows that the sample results are not likely to occur by chance.

Wording of Final Conclusion
Figure 8-7

Accept Versus Fail to Reject
Some texts use “accept the null hypothesis.” We are not proving the null hypothesis. The sample evidence is not strong enough to warrant rejection (such as not enough evidence to convict a suspect). page 396 of Elementary Statistics, 10th Edition The term ‘accept’ is somewhat misleading, implying incorrectly that the null has been proven. The phrase ‘fail to reject’ represents the result more correctly.

Type I Error A Type I error is the mistake of rejecting the null hypothesis when it is true. The symbol (alpha) is used to represent the probability of a type I error. page 398 of Elementary Statistics, 10th Edition

Type II Error A Type II error is the mistake of failing to reject the null hypothesis when it is false. The symbol  (beta) is used to represent the probability of a type II error. Page 398 of Elementary Statistics, 10th Edition

Example: Assume that we a conducting a hypothesis test of the claim p > Here are the null and alternative hypotheses: H0: p = 0.5, and H1: p > 0.5. a) Identify a type I error. b) Identify a type II error.

Example: Assume that we a conducting a hypothesis test of the claim p > Here are the null and alternative hypotheses: H0: p = 0.5, and H1: p > 0.5. a) A type I error is the mistake of rejecting a true null hypothesis, so this is a type I error: Conclude that there is sufficient evidence to support p > 0.5, when in reality p = 0.5.

Example: Assume that we a conducting a hypothesis test of the claim p > Here are the null and alternative hypotheses: H0: p = 0.5, and H1: p > 0.5. b) A type II error is the mistake of failing to reject the null hypothesis when it is false, so this is a type II error: Fail to reject p = 0.5 (and therefore fail to support p > 0.5) when in reality p > 0.5.

Type I and Type II Errors

Controlling Type I and Type II Errors
For any fixed , an increase in the sample size n will cause a decrease in  For any fixed sample size n, a decrease in  will cause an increase in . Conversely, an increase in  will cause a decrease in . To decrease both  and , increase the sample size. page 399 of Elementary Statistics, 10th Edition

Definition The power of a hypothesis test is the probability (1 - ) of rejecting a false null hypothesis, which is computed by using a particular significance level and a particular value of the population parameter that is an alternative to the value assumed true in the null hypothesis. That is, the power of the hypothesis test is the probability of supporting an alternative hypothesis that is true.

Comprehensive Hypothesis Test – P-Value Method

Comprehensive Hypothesis Test – Traditional Method

Comprehensive Hypothesis Test - cont
A confidence interval estimate of a population parameter contains the likely values of that parameter. We should therefore reject a claim that the population parameter has a value that is not included in the confidence interval.

Comprehensive Hypothesis Test - cont
Caution: In some cases, a conclusion based on a confidence interval may be different from a conclusion based on a hypothesis test. See the comments in the individual sections.

Null and alternative hypotheses. Test statistics. Significance levels. P-values. Decision criteria. Type I and II errors. Power of a hypothesis test.

Testing a Claim About a Proportion
Section 8-3 Testing a Claim About a Proportion

Key Concept This section presents complete procedures for testing a hypothesis (or claim) made about a population proportion. This section uses the components introduced in the previous section for the P-value method, the traditional method or the use of confidence intervals.

Requirements for Testing Claims About a Population Proportion p
1) The sample observations are a simple random sample. 2) The conditions for a binomial distribution are satisfied (Section 5-3). 3) The conditions np  5 and nq  5 are satisfied, so the binomial distribution of sample proportions can be approximated by a normal distribution with µ = np and  = npq .

p = x (sample proportion) n
Notation n = number of trials  p = x (sample proportion) n p = population proportion (used in the null hypothesis) q = 1 – p

Test Statistic for Testing a Claim About a Proportion
p – p pq n z = 

P-Value Method Use the same method as described in Section 8-2 and in Figure 8-8. Use the standard normal distribution (Table A-2).

Traditional Method Use the same method as described in Section 8-2 and in Figure 8-9.

Confidence Interval Method
Use the same method as described in Section 8-2 and in Table 8-2.

Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > The sample data are n = 880, and p = 0.56.  np = (880)(0.5) = 440  5 nq = (880)(0.5) = 440  5

Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > The sample data are n = 880, and p = We will use the P-value Method.  pq n p – p z =  0.56 – 0.5 (0.5)(0.5) 880 = = 3.56 H0: p = 0.5 H1: p > 0.5  = 0.05 Referring to Table A-2, we see that for values of z = 3.50 and higher, we use for the cumulative area to the left of the test statistic. The P-value is 1 – =

Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > The sample data are n = 880, and p = We will use the P-value Method.  pq n p – p z =  0.56 – 0.5 (0.5)(0.5) 880 = = 3.56 H0: p = 0.5 H1: p > 0.5  = 0.05 Since the P-value of is less than the significance level of  = 0.05, we reject the null hypothesis. There is sufficient evidence to support the claim.

Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > The sample data are n = 880, and p = We will use the P-value Method.  z = 3.56 H0: p = 0.5 H1: p > 0.5  = 0.05

Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > The sample data are n = 880, and p = We will use the Traditional Method.  pq n p – p z =  0.56 – 0.5 (0.5)(0.5) 880 = = 3.56 H0: p = 0.5 H1: p > 0.5  = 0.05 This is a right-tailed test, so the critical region is an area of We find that z = is the critical value of the critical region. We reject the null hypothesis. There is sufficient evidence to support the claim.

Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > The sample data are n = 880, and p = We will use the confidence interval method.  For a one-tailed hypothesis test with significance level , we will construct a confidence interval with a confidence level of 1 – 2. We construct a 90% confidence interval. We obtain < p < We are 90% confident that the true value of p is contained within the limits of and Thus we support the claim that p > 0.5.

CAUTION When testing claims about a population proportion, the traditional method and the P-value method are equivalent and will yield the same result since they use the same standard deviation based on the claimed proportion p. However, the confidence interval uses an estimated standard deviation based upon the sample proportion p. Consequently, it is possible that the traditional and P-value methods may yield a different conclusion than the confidence interval method. A good strategy is to use a confidence interval to estimate a population proportion, but use the P-value or traditional method for testing a hypothesis. 

(determining the sample proportion of households with cable TV)
 Obtaining P  p sometimes is given directly “10% of the observed sports cars are red” is expressed as p = 0.10  and 54 do not” is calculated using p sometimes must be calculated “96 surveyed households have cable TV  p = = = 0.64 x n 96 (96+54)  (determining the sample proportion of households with cable TV)

Example: When Gregory Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 428 peas with green pods and 152 peas with yellow pods. According to Mendel’s theory, 1/4 of the offspring peas should have yellow pods. Use a 0.05 significance level with the P-value method to test the claim that the proportion of peas with yellow pods is equal to 1/4. We note that n = = 580, so p = 0.262, and p = 0.25. 

Example: When Gregory Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 428 peas with green pods and 152 peas with yellow pods. According to Mendel’s theory, 1/4 of the offspring peas should have yellow pods. Use a 0.05 significance level with the P-value method to test the claim that the proportion of peas with yellow pods is equal to 1/4. z = p – p pq n  H0: p = 0.25 H1: p  0.25 n = 580  = 0.05 p = 0.262 0.262 – 0.25 (0.25)(0.75) 580 = = 0.67  Since this is a two-tailed test, the P-value is twice the area to the right of the test statistic. Using Table A-2, z = 0.67 is 1 – =

Example: When Gregory Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 428 peas with green pods and 152 peas with yellow pods. According to Mendel’s theory, 1/4 of the offspring peas should have yellow pods. Use a 0.05 significance level with the P-value method to test the claim that the proportion of peas with yellow pods is equal to 1/4. z = p – p pq n  H0: p = 0.25 H1: p  0.25 n = 580  = 0.05 p = 0.262 0.262 – 0.25 (0.25)(0.75) 580 = = 0.67  The P-value is 2(0.2514) = We fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that 1/4 of the peas have yellow pods.

Test statistics for claims about a proportion. P-value method. Confidence interval method. Obtaining p.

Testing a Claim About a Mean:  Known
Section 8-4 Testing a Claim About a Mean:  Known

Key Concept This section presents methods for testing a claim about a population mean, given that the population standard deviation is a known value. This section uses the normal distribution with the same components of hypothesis tests that were introduced in Section 8-2.

Requirements for Testing Claims About a Population Mean (with  Known)
1) The sample is a simple random sample. 2) The value of the population standard deviation  is known. 3) Either or both of these conditions is satisfied: The population is normally distributed or n > 30.

Test Statistic for Testing a Claim About a Mean (with  Known)
x – µx z =  n

Example: We have a sample of 106 body temperatures having a mean of 98
Example: We have a sample of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation  is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the P-value method. H0:  = 98.6 H1:   98.6  = 0.05 x = 98.2  = 0.62 = z = x – µx  n 98.2 – 98.6 = − 6.64 0.62 106 This is a two-tailed test and the test statistic is to the left of the center, so the P-value is twice the area to the left of z = – We refer to Table A-2 to find the area to the left of z = –6.64 is , so the P-value is 2(0.0001) =

Example: We have a sample of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation  is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the P-value method. z = –6.64 H0:  = 98.6 H1:   98.6  = 0.05 x = 98.2  = 0.62

Example: We have a sample of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation  is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the P-value method. H0:  = 98.6 H1:   98.6  = 0.05 x = 98.2  = 0.62 z = –6.64 Because the P-value of is less than the significance level of  = 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that the mean body temperature of healthy adults differs from 98.6°F.

Example: We have a sample of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation  is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the traditional method. z = –6.64 H0:  = 98.6 H1:   98.6  = 0.05 x = 98.2  = 0.62 We now find the critical values to be z = –1.96 and z = We would reject the null hypothesis, since the test statistic of z = –6.64 would fall in the critical region. There is sufficient evidence to conclude that the mean body temperature of healthy adults differs from 98.6°F.

Example: We have a sample of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation  is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F. Use the confidence interval method. H0:  = 98.6 H1:   98.6  = 0.05 x = 98.2  = 0.62 For a two-tailed hypothesis test with a 0.05 significance level, we construct a 95% confidence interval. Use the methods of Section 7-2 to construct a 95% confidence interval: 98.08 <  < 98.32 We are 95% confident that the limits of and contain the true value of , so it appears that 98.6 cannot be the true value of .

Underlying Rationale of Hypothesis Testing
If, under a given assumption, there is an extremely small probability of getting sample results at least as extreme as the results that were obtained, we conclude that the assumption is probably not correct. When testing a claim, we make an assumption (null hypothesis) of equality. We then compare the assumption and the sample results and we form one of the following conclusions:

Underlying Rationale of Hypotheses Testing - cont
If the sample results (or more extreme results) can easily occur when the assumption (null hypothesis) is true, we attribute the relatively small discrepancy between the assumption and the sample results to chance. If the sample results cannot easily occur when that assumption (null hypothesis) is true, we explain the relatively large discrepancy between the assumption and the sample results by concluding that the assumption is not true, so we reject the assumption.

Requirements for testing claims about population means, σ known. P-value method. Traditional method. Confidence interval method. Rationale for hypothesis testing.

Testing a Claim About a Mean:  Not Known
Section 8-5 Testing a Claim About a Mean:  Not Known

Key Concept This section presents methods for testing a claim about a population mean when we do not know the value of σ. The methods of this section use the Student t distribution introduced earlier.

Requirements for Testing Claims About a Population Mean (with  Not Known)
1) The sample is a simple random sample. 2) The value of the population standard deviation  is not known. 3) Either or both of these conditions is satisfied: The population is normally distributed or n > 30.

Test Statistic for Testing a Claim About a Mean (with  Not Known)
x – µx t = s n P-values and Critical Values Found in Table A-3 Degrees of freedom (df) = n – 1

Important Properties of the Student t Distribution
1. The Student t distribution is different for different sample sizes (see Figure 7-5 in Section 7-4). 2. The Student t distribution has the same general bell shape as the normal distribution; its wider shape reflects the greater variability that is expected when s is used to estimate . 3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean of z = 0). 4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1 (unlike the standard normal distribution, which has = 1). 5. As the sample size n gets larger, the Student t distribution gets closer to the standard normal distribution.

Choosing between the Normal and Student t Distributions when Testing a Claim about a Population Mean µ Use the Student t distribution when  is not known and either or both of these conditions is satisfied: The population is normally distributed or n > 30.

Example: Data Set 13 in Appendix B of the text includes weights of 13 red M&M candies randomly selected from a bag containing 465 M&Ms. The weights (in grams) have a mean x = and a standard deviation s = g. The bag states that the net weight of the contents is g, so the M&Ms must have a mean weight that is 396.9/465 = g in order to provide the amount claimed. Use the sample data with a 0.05 significance level to test the claim of a production manager that the M&Ms have a mean that is actually greater than g. Use the traditional method. The sample is a simple random sample and we are not using a known value of σ. The sample size is n = 13 and a normal quartile plot suggests the weights are normally distributed.

Example: Data Set 13 in Appendix B of the text includes weights of 13 red M&M candies randomly selected from a bag containing 465 M&Ms. The weights (in grams) have a mean x = and a standard deviation s = g. The bag states that the net weight of the contents is g, so the M&Ms must have a mean weight that is 396.9/465 = g in order to provide the amount claimed. Use the sample data with a 0.05 significance level to test the claim of a production manager that the M&Ms have a mean that is actually greater than g. Use the traditional method. H0:  = H1:  >  = 0.05 x = s = n = 13 x – µx – 0.0576 13 = 0.626 = t = s n The critical value, from Table A-3, is t = 1.782

Example: Data Set 13 in Appendix B of the text includes weights of 13 red M&M candies randomly selected from a bag containing 465 M&Ms. The weights (in grams) have a mean x = and a standard deviation s = g. The bag states that the net weight of the contents is g, so the M&Ms must have a mean weight that is 396.9/465 = g in order to provide the amount claimed. Use the sample data with a 0.05 significance level to test the claim of a production manager that the M&Ms have a mean that is actually greater than g. Use the traditional method. H0:  = H1:  >  = 0.05 x = s = n = 13 t = 0.626 Critical Value t = 1.782 Because the test statistic of t = does not fall in the critical region, we fail to reject H0. There is not sufficient evidence to support the claim that the mean weight of the M&Ms is greater than g.

Normal Distribution Versus Student t Distribution
The critical value in the preceding example was t = 1.782, but if the normal distribution were being used, the critical value would have been z = The Student t critical value is larger (farther to the right), showing that with the Student t distribution, the sample evidence must be more extreme before we can consider it to be significant.

P-Value Method Use software or a TI-83/84 Plus calculator.
If technology is not available, use Table A-3 to identify a range of P-values.

Example: Assuming that neither software nor a TI-83 Plus calculator is available, use Table A-3 to find a range of values for the P-value corresponding to the given results. a) In a left-tailed hypothesis test, the sample size is n = 12, and the test statistic is t = –2.007. b) In a right-tailed hypothesis test, the sample size is n = 12, and the test statistic is t = c) In a two-tailed hypothesis test, the sample size is n = 12, and the test statistic is t = –3.456.

Example: Assuming that neither software nor a TI-83 Plus calculator is available, use Table A-3 to find a range of values for the P-value corresponding to the given results.

Example: Assuming that neither software nor a TI-83 Plus calculator is available, use Table A-3 to find a range of values for the P-value corresponding to the given results. a) The test is a left-tailed test with test statistic t = –2.007, so the P-value is the area to the left of – Because of the symmetry of the t distribution, that is the same as the area to the right of Any test statistic between and has a right-tailed P- value that is between and We conclude that < P-value < 0.05.

Example: Assuming that neither software nor a TI-83 Plus calculator is available, use Table A-3 to find a range of values for the P-value corresponding to the given results. b) The test is a right-tailed test with test statistic t = 1.222, so the P-value is the area to the right of Any test statistic less than has a right-tailed P-value that is greater than We conclude that P-value > 0.10.

Example: Assuming that neither software nor a TI-83 Plus calculator is available, use Table A-3 to find a range of values for the P-value corresponding to the given results. c) The test is a two-tailed test with test statistic t = – The P-value is twice the area to the right of – Any test statistic greater than has a two-tailed P- value that is less than We conclude that P- value < 0.01.

Assumptions for testing claims about population means, σ unknown. Student t distribution. P-value method.

Testing a Claim About a Standard Deviation or Variance
Section 8-6 Testing a Claim About a Standard Deviation or Variance

Key Concept This section introduces methods for testing a claim made about a population standard deviation σ or population variance σ 2. The methods of this section use the chi-square distribution that was first introduced in Section 7-5.

Requirements for Testing Claims About  or  2
1. The sample is a simple random sample. 2. The population has a normal distribution. (This is a much stricter requirement than the requirement of a normal distribution when testing claims about means.)

Chi-Square Distribution
Test Statistic  2 = (n – 1) s 2 2 n = sample size s 2 = sample variance 2 = population variance (given in null hypothesis)

P-Values and Critical Values for Chi-Square Distribution
Use Table A-4. The degrees of freedom = n –1.

Properties of Chi-Square Distribution
All values of  2 are nonnegative, and the distribution is not symmetric (see Figure 8-13, following). There is a different distribution for each number of degrees of freedom (see Figure 8-14, following). The critical values are found in Table A-4 using n – 1 degrees of freedom.

Properties of Chi-Square Distribution - cont
Figure 8-13 Properties of the Chi-Square Distribution There is a different distribution for each number of degrees of freedom. Chi-Square Distribution for 10 and 20 Degrees of Freedom Figure 8-14

Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that  = 15. H0:  = 15 H1:   15  = 0.05 n = 13 s = 7.2 = 2.765  2 = (n – 1)s2 2 (13 – 1)(7.2)2 152 =

Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that  = 15. H0:  = 15 H1:   15  = 0.05 n = 13 s = 7.2  2 = 2.765

Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that  = 15.  2 = 2.765 H0:  = 15 H1:   15  = 0.05 n = 13 s = 7.2 The critical values of and are found in Table A-4, in the 12th row (degrees of freedom = n – 1) in the column corresponding to and

Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that  = 15.  2 = 2.765 H0:  = 15 H1:   15  = 0.05 n = 13 s = 7.2 Because the test statistic is in the critical region, we reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the standard deviation is equal to 15.

Tests for claims about standard deviation and variance. Test statistic. Chi-square distribution. Critical values.

Chapter 9 Inferences from Two Samples
9-1 Overview 9-2 Inferences About Two Proportions 9-3 Inferences About Two Means: Independent Samples 9-4 Inferences from Matched Pairs 9-5 Comparing Variation in Two Samples

Overview There are many important and meaningful situations in which it becomes necessary to compare two sets of sample data. This chapter extends the same methods introduced in Chapters 7 and 8 to situations involving two samples instead of only one. page 456 of Elementary Statistics, 10th Edition Examples in the discussion

Inferences About Two Proportions
Section 9-2 Inferences About Two Proportions

Key Concept This section presents methods for using two sample proportions for constructing a confidence interval estimate of the difference between the corresponding population proportions, or testing a claim made about the two population proportions.

Requirements 1. We have proportions from two independent simple random samples. 2. For each of the two samples, the number of successes is at least 5 and the number of failures is at least 5. page 457 of Elementary Statistics, 10th Edition

Notation for Two Proportions
For population 1, we let: p1 = population proportion n1 = size of the sample x1 = number of successes in the sample ^ p1 = x1 (the sample proportion) q1 = 1 – p1 n1 The corresponding meanings are attached to p2, n2 , x2 , p2. and q2 , which come from population 2. ^

Pooled Sample Proportion
The pooled sample proportion is denoted by p and is given by: = p n1 + n2 x1 + x2 We denote the complement of p by q, so q = 1 – p

Test Statistic for Two Proportions
For H0: p1 = p2 H1: p1  p2 , H1: p1 < p2 , H1: p 1> p2 + z = ( p1 – p2 ) – ( p1 – p2 ) ^ n1 pq n2 Example given at the bottom of page of Elementary Statistics, 10th Edition.

Test Statistic for Two Proportions - cont
For H0: p1 = p2 H1: p1  p2 , H1: p1 < p2 , H1: p 1> p2 where p1 – p 2 = 0 (assumed in the null hypothesis) = p1 ^ x1 n1 p2 x2 n2 and and q = 1 – p n1 + n2 p = x1 + x2

Test Statistic for Two Proportions - cont
P-value: Use Table A-2. (Use the computed value of the test statistic z and find its P-value by following the procedure summarized by Figure 8-6 in the text.) Critical values: Use Table A-2. (Based on the significance level α, find critical values by using the procedures introduced in Section in the text.)

Example: For the sample data listed in the Table below, use a 0
Example: For the sample data listed in the Table below, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.

Example: For the sample data listed in the previous Table, use a 0
Example: For the sample data listed in the previous Table, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400 H0: p1 = p2, H1: p1 > p2 p = x1 + x2 = = n1 + n q = 1 – =

Example: For the sample data listed in the previous Table, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. (0.120 – 0.105) – 0 ( )( ) + ( )( ) z = 0.64 z = 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400

Example: For the sample data listed in the previous Table, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400 z = 0.64 This is a right-tailed test, so the P-value is the area to the right of the test statistic z = The P-value is Because the P-value of is greater than the significance level of  = 0.05, we fail to reject the null hypothesis.

Example: For the sample data listed in the previous Table, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. z = 0.64 Because we fail to reject the null hypothesis, we conclude that there is not sufficient evidence to support the claim that the proportion of black drivers stopped by police is greater than that for white drivers. This does not mean that racial profiling has been disproved. The evidence might be strong enough with more data. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400

Example: For the sample data listed in the previous Table, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400

Confidence Interval Estimate of p1 - p2
p1 q1 p2 q2 + ^ where E = z ( p1 – p2 ) – E < ( p1 – p2 ) < ( p1 – p2 ) + E Example at the bottom of page 461 of Elementary Statistics, 10th Edition Rationale for the procedures of this section on page of Elementary Statistics, 10th Edition

Example: For the sample data listed in the previous Table, find a 90% confidence interval estimate of the difference between the two population proportions. n1 n2 + p1 q1 p2 q2 ^ E = z E = 1.645 200 1400 (.12)(.88)+ (0.105)(0.895) E = 0.040 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 n1 200 ^ n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 n2 1400

Example: For the sample data listed in the previous table, use a 0
Example: For the sample data listed in the previous table, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 n1 200 ^ n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 n2 1400 (0.120 – 0.105) – < ( p1– p2) < (0.120 – 0.105) –0.025 < ( p1– p2) < 0.055

Why Do the Procedures of This Section Work?
The text contains a detailed explanation of how and why the test statistic given for hypothesis tests is justified. Be sure to study it carefully.

Requirements for inferences about two proportions. Notation. Pooled sample proportion. Hypothesis tests.

Inferences About Two Means: Independent Samples
Section 9-3 Inferences About Two Means: Independent Samples

Key Concept This section presents methods for using sample data from two independent samples to test hypotheses made about two population means or to construct confidence interval estimates of the difference between two population means.

Part 1: Independent Samples with σ1 and σ2 Unknown and Not Assumed Equal

Definitions Two samples are independent if the sample values selected from one population are not related to or somehow paired or matched with the sample values selected from the other population. Two samples are dependent (or consist of matched pairs) if the members of one sample can be used to determine the members of the other sample. Text will use the wording ‘matched pairs’. Example at bottom of page of Elementary Statistics, 10th Edition

Requirements 1. σ1 an σ2 are unknown and no assumption is made about the equality of σ1 and σ2 . 2. The two samples are independent. 3. Both samples are simple random samples. 4. Either or both of these conditions are satisfied: The two sample sizes are both large (with n1 > 30 and n2 > 30) or both samples come from populations having normal distributions. page 470 of Elementary Statistics, 10th Edition

Hypothesis Test for Two Means: Independent Samples
(x1 – x2) – (µ1 – µ2) t = n1 n2 + s1. s2 2

Test Statistic for Two Means: Independent Samples
Hypothesis Test - cont Test Statistic for Two Means: Independent Samples Degrees of freedom: In this book we use this simple and conservative estimate: df = smaller of n1 – 1 and n2 – 1. P-values: Refer to Table A-3. Use the procedure summarized in Figure 8-6. Critical values: Refer to Table A-3.

McGwire Versus Bonds Sample statistics are shown for the distances of the home runs hit in record-setting seasons by Mark McGwire and Barry Bonds. Use a 0.05 significance level to test the claim that the distances come from populations with different means. McGwire Bonds n x s

McGwire Versus Bonds - cont
Below is a Statdisk plot of the data

Claim: 1  2 Ho : 1 = 2 H1 : 1  2  = 0.05 n1 – 1 = 69 n2 – 1 = 72 df = 69 t.025 = 1.994

McGwire Versus Bonds - cont Test Statistic for Two Means:
(x1 – x2) – (µ1 – µ2) t = s1 s2 2 2 + n1 n2

McGwire Versus Bonds - cont Test Statistic for Two Means:
(418.5 – 403.7) – 0 45.52 30.62 + 70 73 = 2.273

Claim: 1  2 Ho : 1 = 2 H1 : 1  2  = 0.05

Claim: 1  2 Ho : 1 = 2 H1 : 1  2  = 0.05 There is significant evidence to support the claim that there is a difference between the mean home run distances of Mark McGwire and Barry Bonds. Reject the Null Hypothesis

(x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E
Confidence Intervals (x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E + n1 n2 s1 s2 where E = z 2

McGwire Versus Bonds Confidence Interval Method
Using the data given in the preceding example, construct a 95% confidence interval estimate of the difference between the mean home run distances of Mark McGwire and Barry Bonds. n1 n2 + s1 s2 E = t 2 E = 1.994 70 73 45.5 30.6 E = 13.0

McGwire Versus Bonds Confidence Interval Method - cont
Using the data given in the preceding example, construct a 95% confidence interval estimate of the difference between the mean home run distances of Mark McGwire and Barry Bonds. (418.5 – 403.7) – 13.0 < (1 – 2) < (418.5 – 403.7) 1.8 < (1 – 2) < 27.8 We are 95% confident that the limits of 1.8 ft and 27.8 ft actually do contain the difference between the two population means.

Part 2: Alternative Methods

Independent Samples with σ1 and σ2 Known.

Requirements 1. The two population standard deviations are both known.
2. The two samples are independent. 3. Both samples are simple random samples. 4. Either or both of these conditions are satisfied: The two sample sizes are both large (with n1 > 30 and n2 > 30) or both samples come from populations having normal distributions.

Hypothesis Test for Two Means: Independent Samples with σ1 and σ2 Both Known
(x1 – x2) – (µ1 – µ2) z = σ1 σ2 2 2 + n1 n2 P-values and critical values: Refer to Table A-2.

Confidence Interval: Independent Samples with σ1 and σ2 Both Known
(x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E + n1 n2 σ1 σ2 where E = z 2

Methods for Inferences About Two Independent Means
Figure 9-3

Assume that σ1 = σ2 and Pool the Sample Variances.

Requirements 1. The two population standard deviations are not known, but they are assumed to be equal. That is σ1 = σ2. 2. The two samples are independent. 3. Both samples are simple random samples. 4. Either or both of these conditions are satisfied: The two sample sizes are both large (with n1 > 30 and n2 > 30) or both samples come from populations having normal distributions.

Hypothesis Test Statistic for Two Means: Independent Samples and σ1 = σ2
= (n1 – 1) (n2 -1) (x1 – x2) – (µ1 – µ2) t = n1 n2 + sp 2 sp. Where s1 s2 (n1 – 1) + (n2 – 1) and the number of degrees of freedom is df = n1 + n2 - 2

Confidence Interval Estimate of μ1 – μ2: Independent Samples with σ1 = σ2
(x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E + n1 n2 sp where E = t 2 and number of degrees of freedom is df = n1 + n2 - 2

Strategy Unless instructed otherwise, use the following strategy:
Assume that σ1 and σ2 are unknown, do not assume that σ1 = σ2, and use the test statistic and confidence interval given in Part 1 of this section. (See Figure 9-3.)

Independent samples with the standard deviations unknown and not assumed equal. Alternative method where standard deviations are known Alternative method where standard deviations are assumed equal and sample variances are pooled.

Inferences from Matched Pairs
Section 9-4 Inferences from Matched Pairs

Key Concept In this section we develop methods for testing claims about the mean difference of matched pairs. For each matched pair of sample values, we find the difference between the two values, then we use those sample differences to test claims about the population difference or to construct confidence interval estimates of the population difference.

Requirements 1. The sample data consist of matched pairs.
2. The samples are simple random samples. 3. Either or both of these conditions is satisfied: The number of matched pairs of sample data is large (n > 30) or the pairs of values have differences that are from a population having a distribution that is approximately normal. page 485 of Elementary Statistics, 10th Edition

Notation for Matched Pairs
d = individual difference between the two values of a single matched pair µd = mean value of the differences d for the population of paired data d = mean value of the differences d for the paired sample data (equal to the mean of the x – y values) sd = standard deviation of the differences d for the paired sample data n = number of pairs of data.

Hypothesis Test Statistic for Matched Pairs
d – µd sd n where degrees of freedom = n – 1 page 485 of Elementary Statistics, 10th Edition

P-values and Critical Values
Use Table A-3 (t-distribution). page 485 of Elementary Statistics, 10th Edition Hypothesis example given on this page

Confidence Intervals for Matched Pairs
d – E < µd < d + E where E = t/2 sd n Critical values of tα/2 : Use Table A-3 with n – 1 degrees of freedom.

Are Forecast Temperatures Accurate?
The following Table consists of five actual low temperatures and the corresponding low temperatures that were predicted five days earlier. The data consist of matched pairs, because each pair of values represents the same day. Use a 0.05 significance level to test the claim that there is a difference between the actual low temperatures and the low temperatures that were forecast five days earlier.

Are Forecast Temperatures Accurate? - cont

d = –13.2 s = 10.7 n = 5 t/2 = (found from Table A-3 with 4 degrees of freedom and 0.05 in two tails)

H0: d = 0 H1: d  0 t = d – µd n sd = –13.2 – 0 = –2.759 10.7 5

H0: d = 0 H1: d  0 t = d – µd n sd = –13.2 – 0 = –2.759 10.7 5 Because the test statistic does not fall in the critical region, we fail to reject the null hypothesis.

H0: d = 0 H1: d  0 t = d – µd n sd = –13.2 – 0 = –2.759 10.7 5 The sample data in the previous Table do not provide sufficient evidence to support the claim that actual and five-day forecast low temperatures are different.

Using the same sample matched pairs in the previous Table, construct a 95% confidence interval estimate of d , which is the mean of the differences between actual low temperatures and five-day forecasts.

E = t/2 sd n E = (2.776)( ) 10.7 5 = 13.3

d – E < d < d + E –13.2 – 13.3 < d < – –26.5 < d < 0.1

In the long run, 95% of such samples will lead to confidence intervals that actually do contain the true population mean of the differences.

Requirements for inferences from matched pairs. Notation. Hypothesis test. Confidence intervals.

Comparing Variation in Two Samples
Section 9-5 Comparing Variation in Two Samples

Key Concept This section presents the F test for using two samples to compare two population variances (or standard deviations). We introduce the F distribution that is used for the F test. Note that the F test is very sensitive to departures from normal distributions.

Measures of Variation s = standard deviation of sample
 = standard deviation of population s2 = variance of sample 2 = variance of population page 495 of Elementary Statistics, 10th Edition

1. The two populations are independent of each other.
Requirements 1. The two populations are independent of each other. 2. The two populations are each normally distributed. Students may need a reminder of the definition of independent from Section 9-2. The normal distribution requirement is quite strict with the methods discussed in this section. Use of the normal quantile plots (section 6-7) will be important to determine normality

s1 = larger of the two sample variances
Notation for Hypothesis Tests with Two Variances or Standard Deviations s1 = larger of the two sample variances n1 = size of the sample with the larger variance 1 = variance of the population from which the sample with the larger variance was drawn The symbols s2 , n2 , and 2 are used for the other sample and population. 2 2

Test Statistic for Hypothesis Tests with Two Variances
= s2 2 Where s12 is the larger of the two sample variances Critical Values: Using Table A-5, we obtain critical F values that are determined by the following three values: 1. The significance level  2. Numerator degrees of freedom = n1 – 1 3. Denominator degrees of freedom = n2 – 1

The F distribution is not symmetric.
Properties of the F Distribution The F distribution is not symmetric. Values of the F distribution cannot be negative. The exact shape of the F distribution depends on two different degrees of freedom. page 496 of Elementary Statistics, 10th Edition

Properties of the F Distribution - cont
If the two populations do have equal variances, then F = will be close to 1 because and are close in value. s1 2 s2 s1 2 s 2 page 497 of Elementary Statistics, 10th Edition Give some numerical examples of the computation of F close to 1.

Properties of the F Distribution - cont
If the two populations have radically different variances, then F will be a large number. Remember, the larger sample variance will be s1 . 2 Give some numerical examples of an F value that is not close to 1.

Conclusions from the F Distribution
Consequently, a value of F near 1 will be evidence in favor of the conclusion that 1 = 2 . 2 But a large value of F will be evidence against the conclusion of equality of the population variances.

Coke Versus Pepsi Data Set 12 in Appendix B includes the weights (in pounds) of samples of regular Coke and regular Pepsi. Sample statistics are shown. Use the 0.05 significance level to test the claim that the weights of regular Coke and the weights of regular Pepsi have the same standard deviation. Regular Coke Regular Pepsi n x s

Coke Versus Pepsi Claim: 1 = 2 Ho : 1 = 2 H1 : 1  2  = 0.05 s1
Value of F = s1 s2 2 = = We should recognize that the F test is extremely sensitive to distributions that are not normally distributed, so this conclusion might make it appear that there is no significant difference between the population variances when there really is a difference that was hidden by non-normal distributions.

Coke Versus Pepsi Claim: 1 = 2 Ho : 1 = 2 H1 : 1  2  = 0.05
We should recognize that the F test is extremely sensitive to distributions that are not normally distributed, so this conclusion might make it appear that there is no significant difference between the population variances when there really is a difference that was hidden by non-normal distributions. There is not sufficient evidence to warrant rejection of the claim that the two variances are equal.

Requirements for comparing variation in two samples Notation. Hypothesis test. Confidence intervals. F test and distribution.

Chapter 10 Correlation and Regression
10-1 Overview 10-2 Correlation 10-3 Regression 10-4 Variation and Prediction Intervals 10-5 Multiple Regression 10-6 Modeling

Overview This chapter introduces important methods for making inferences about a correlation (or relationship) between two variables, and describing such a relationship with an equation that can be used for predicting the value of one variable given the value of the other variable. We consider sample data that come in pairs. page 517 of Elementary Statistics, 10th Edition

Section 10-2 Correlation Created by Erin Hodgess, Houston, Texas
Revised to accompany 10th Edition, Tom Wegleitner, Centreville, VA

Key Concept This section introduces the linear correlation coefficient r, which is a numerical measure of the strength of the relationship between two variables representing quantitative data. Because technology can be used to find the value of r, it is important to focus on the concepts in this section, without becoming overly involved with tedious arithmetic calculations.

Part 1: Basic Concepts of Correlation

Definition A correlation exists between two variables when one of them is related to the other in some way.

Definition The linear correlation coefficient r measures the strength of the linear relationship between paired x- and y- quantitative values in a sample. page 518 of Elementary Statistics, 10th Edition

Exploring the Data We can often see a relationship between two variables by constructing a scatterplot. Figure 10-2 following shows scatterplots with different characteristics. Relate a scatter plot to the algebraic plotting of number pairs (x,y).

Scatterplots of Paired Data
Page 519 of Elementary Statistics, 10th Edition Figure 10-2

Requirements 1. The sample of paired (x, y) data is a random sample of independent quantitative data. 2. Visual examination of the scatterplot must confirm that the points approximate a straight-line pattern. 3. The outliers must be removed if they are known to be errors. The effects of any other outliers should be considered by calculating r with and without the outliers included. page 520 of Elementary Statistics, 10th Edition Explain to students the difference between the ‘paired’ data of this chapter and the investigation of two groups of data in Chapter 9.

Notation for the Linear Correlation Coefficient
n represents the number of pairs of data present.  denotes the addition of the items indicated. x denotes the sum of all x-values. x2 indicates that each x-value should be squared and then those squares added. (x)2 indicates that the x-values should be added and the total then squared. xy indicates that each x-value should be first multiplied by its corresponding y-value. After obtaining all such products, find their sum. r represents linear correlation coefficient for a sample.  represents linear correlation coefficient for a population.

Formula The linear correlation coefficient r measures the strength of a linear relationship between the paired values in a sample. nxy – (x)(y) n(x2) – (x) n(y2) – (y)2 r = Formula 10-1 Calculators can compute r

Interpreting r Using Table A-6: If the absolute value of the computed value of r exceeds the value in Table A-6, conclude that there is a linear correlation. Otherwise, there is not sufficient evidence to support the conclusion of a linear correlation. Using Software: If the computed P-value is less than or equal to the significance level, conclude that there is a linear correlation. Otherwise, there is not sufficient evidence to support the conclusion of a linear correlation.

Correlation Coefficient r
Rounding the Linear Correlation Coefficient r Round to three decimal places so that it can be compared to critical values in Table A-6. Use calculator or computer if possible. Page 521 of Elementary Statistics, 10th Edition

Example: Calculating r
Using the simple random sample of data below, find the value of r. 3 5 1 8 6 4 Data x y page 521 of Elementary Statistics, 10th Edition.

Example: Calculating r - cont
page 522 of Elementary Statistics, 10th Edition.

3 5 1 8 6 4 Data x y nxy – (x)(y) n(x2) – (x) n(y2) – (y)2 r = 4(61) – (12)(23) 4(44) – (12) (141) – (23)2 -32 33.466 = Page 522 of Elementary Statistics, 10th Edition

Given r = , if we use a 0.05 significance level we conclude that there is a linear correlation between x and y since the absolute value of r exceeds the critical value of However, if we use a 0.01 significance level, we do not conclude that there is a linear correlation because the absolute value of r does not exceed the critical value of

Example: Old Faithful Given the sample data in Table 10-1, find the value of the linear correlation coefficient r, then refer to Table A-6 to determine whether there is a significant linear correlation between the duration and interval of the eruption times. Using the same procedure previously illustrated, we find that r = Referring to Table A-6, we locate the row for which n = 8. Using the critical value for  = 0.05, we have Because r = 0.926, its absolute value exceeds 0.707, so we conclude that there is a linear correlation between the duration and interval after the eruption times. Page 523 of Elementary Statistics, 10th Edition.

Properties of the Linear Correlation Coefficient r
2. The value of r does not change if all values of either variable are converted to a different scale. 3. The value of r is not affected by the choice of x and y. Interchange all x- and y-values and the value of r will not change. 4. r measures strength of a linear relationship. page 524 of Elementary Statistics, 10th Edition If using a graphics calculator for demonstration, it will be an easy exercise to switch the x and y values to show that the value of r will not change.

Interpreting r : Explained Variation
The value of r2 is the proportion of the variation in y that is explained by the linear relationship between x and y. page 524 of Elementary Statistics, 10th Edition If using a graphics calculator for demonstration, it will be an easy exercise to switch the x and y values to show that the value of r will not change.

Example: Old Faithful Using the duration/interval data in Table 10-1, we have found that the value of the linear correlation coefficient r = What proportion of the variation of the interval after eruption times can be explained by the variation in the duration times? With r = 0.926, we get r2 = We conclude that (or about 86%) of the variation in interval after eruption times can be explained by the linear relationship between the duration times and the interval after eruption times. This implies that 14% of the variation in interval after eruption times cannot be explained by the duration times. page 524 of Elementary Statistics, 10th Edition

Common Errors Involving Correlation
1. Causation: It is wrong to conclude that correlation implies causality. 2. Averages: Averages suppress individual variation and may inflate the correlation coefficient. 3. Linearity: There may be some relationship between x and y even when there is no linear correlation. page 525 of Elementary Statistics, 10th Edition

Part 2: Formal Hypothesis Test

Formal Hypothesis Test
We wish to determine whether there is a significant linear correlation between two variables. We present two methods. In both methods let H0: = (no significant linear correlation) H1:  (significant linear correlation) page 527 of Elementary Statistics, 10th Edition

Hypothesis Test for a Linear Correlation
page 526 of Elementary Statistics, 10th Edition Figure 10-3

Method 1: Test Statistic is t
1 – r 2 n – 2 r t = Critical values: Use Table A-3 with degrees of freedom = n – 2 This is the first example in the text where the degrees of freedom for Table A-3 is different from n Special note should be made of this.

Method 1 - cont P-value: Conclusion: Use Table A-3 with
degrees of freedom = n – 2 Conclusion: If the absolute value of t is > critical value from Table A-3, reject H0 and conclude that there is a linear correlation. If the absolute value of t ≤ critical value, fail to reject H0; there is not sufficient evidence to conclude that there is a linear correlation.

Method 2: Test Statistic is r Test statistic: r
Critical values: Refer to Table A-6 Conclusion: If the absolute value of r is > critical value from Table A-6, reject H0 and conclude that there is a linear correlation. If the absolute value of r ≤ critical value, fail to reject H0; there is not sufficient evidence to conclude that there is a linear correlation. This method is preferred by some instructors because the calculations are easier.

Example: Old Faithful t = t = r
Using the duration/interval data in Table 10-1, test the claim that there is a linear correlation between the duration of an eruption and the time interval after that eruption. Use Method 1. 1 – r 2 n – 2 r t = page 527 of Elementary Statistics, 10th Edition 1 – 8 – 2 0.926 t = = 6.008

Method 1: Test Statistic is t (follows format of earlier chapters)
This is the drawing used to verify the position of the sample data t value in regard to the critical t values for the example which begins on page 527 of Elementary Statistics, 10th Edition. Drawing is at the top of page 528. Figure 10-4

Example: Old Faithful Using the duration/interval data in Table 10-1, test the claim that there is a linear correlation between the duration of an eruption and the time interval after that eruption. Use Method 2. The test statistic is r = The critical values of r = 0.707 are found in Table A-6 with n = 8 and  = 0.05. Page 528 of Elementary Statistics, 10th Edition

Method 2: Test Statistic is r
Test statistic: r Critical values: Refer to Table A-6 (8 degrees of freedom) This is the drawing used to verify the position of the sample data r value in regard to the critical r values for the example which begins on page 527 of Elementary Statistics, 10th Edition. Drawing is at the bottom page 528. Figure 10-5

Example: Old Faithful Using the duration/interval data in Table 10-1, test the claim that there is a linear correlation between the duration of an eruption and the time interval after that eruption. Using either of the two methods, we find that the absolute value of the test statistic does exceed the critical value (Method 1: > Method 2: > 0.707); that is, the test statistic falls in the critical region. This example begins on page 527 of Elementary Statistics, 10th Edition. We therefore reject the null hypothesis. There is sufficient evidence to support the claim of a linear correlation between the duration times of eruptions and the time intervals after the eruptions.

Justification for r Formula
The text presents a detailed rationale for the use of Formula The student should study it carefully. For one point (7,23) the differences from the centroid are found. These values would be used in the r formula.

Recap In this section, we have discussed: Correlation.
The linear correlation coefficient r. Requirements, notation and formula for r. Interpreting r. Formal hypothesis testing (two methods). Rationale for calculating r.

Section 10-3 Regression

Key Concept The key concept of this section is to describe the relationship between two variables by finding the graph and the equation of the straight line that best represents the relationship. The straight line is called a regression line and its equation is called the regression equation.

Part 1: Basic Concepts of Regression

Regression The regression equation expresses a relationship between x (called the independent variable, predictor variable or explanatory variable), and y (called the dependent variable or response variable). ^ The typical equation of a straight line y = mx + b is expressed in the form y = b0 + b1x, where b0 is the y-intercept and b1 is the slope. ^

Requirements 1. The sample of paired (x, y) data is a random sample of quantitative data. 2. Visual examination of the scatterplot shows that the points approximate a straight-line pattern. 3. Any outliers must be removed if they are known to be errors. Consider the effects of any outliers that are not known errors.

Definitions Regression Equation y = b0 + b1x Regression Line
Given a collection of paired data, the regression equation y = b0 + b1x ^ algebraically describes the relationship between the two variables. Regression Line The graph of the regression equation is called the regression line (or line of best fit, or least squares line).

Notation for Regression Equation
Population Parameter Sample Statistic ^ y-intercept of regression equation  b0 Slope of regression equation  b1 Equation of the regression line y = 0 + 1 x y = b0 + b1 x

calculators or computers can compute these values
Formulas for b0 and b1 Formula 10-2 n(xy) – (x) (y) b1 = (slope) n(x2) – (x)2 b0 = y – b1 x (y-intercept) Formula 10-3 calculators or computers can compute these values

The regression line fits the sample points best.
Special Property The regression line fits the sample points best.

Rounding the y-intercept b0 and the Slope b1
Round to three significant digits. If you use the formulas 10-2 and 10-3, try not to round intermediate values. page 543 of Elementary Statistics, 10th Edition

Calculating the Regression Equation
3 5 1 8 6 4 Data x y In Section 10-2, we used these values to find that the linear correlation coefficient of r = – Use this sample to find the regression equation. page 543 of Elementary Statistics, 10th Edition.

Regression Equation - cont
Calculating the Regression Equation - cont 3 5 1 8 6 4 Data x y n(xy) – (x) (y) n(x2) –(x)2 b1 = 4(61) – (12) (23) 4(44) – (12)2 -32 32 = –1 n = 4 x = 12 y = 23 x2 = 44 y2 = 141 xy = 61 page 543 of Elementary Statistics, 10th Edition.

Calculating the Regression Equation - cont 3 5 1 8 6 4 Data x y n = 4 x = 12 y = 23 x2 = 44 y2 = 141 xy = 61 b0 = y – b1 x 5.75 – (–1)(3) = 8.75 Page 543 of Elementary Statistics, 10th Edition.

Calculating the Regression Equation - cont 3 5 1 8 6 4 Data x y The estimated equation of the regression line is: n = 4 x = 12 y = 23 x2 = 44 y2 = 141 xy = 61 y = 8.75 – 1x ^ Page 543 of Elementary Statistics, 10th Edition.

Example: Old Faithful y = 34.8 + 0.234x
Given the sample data in Table 10-1, find the regression equation. Using the same procedure as in the previous example, we find that b1 = and b0 = Hence, the estimated regression equation is: Page 544 of Elementary Statistics, 10th Edition. y = x ^

Example: Old Faithful - cont
Given the sample data in Table 10-1, find the regression equation. page 544 of Elementary Statistics, 10th Edition.

Part 2: Beyond the Basics of Regression

Predictions In predicting a value of y based on some given value of x ... 1. If there is not a linear correlation, the best predicted y-value is y. 2. If there is a linear correlation, the best predicted y-value is found by substituting the x-value into the regression equation.

Procedure for Predicting

Example: Old Faithful Given the sample data in Table 10-1, we found that the regression equation is y = x. Assuming that the current eruption has a duration of x = 180 sec, find the best predicted value of y, the time interval after this eruption. ^ page 546 of Elementary Statistics, 10th Edition.

Example: Old Faithful Given the sample data in Table 10-1, we found that the regression equation is y = x. Assuming that the current eruption has a duration of x = 180 sec, find the best predicted value of y, the time interval after this eruption. ^ We must consider whether there is a linear correlation that justifies the use of that equation. We do have a significant linear correlation (with r = 0.926). page 546 of Elementary Statistics, 10th Edition.

Given the sample data in Table 10-1, we found that the regression equation is y = x. Assuming that the current eruption has a duration of x = 180 sec, find the best predicted value of y, the time interval after this eruption. ^ y = x (180) = 76.9 min ^ page 546 of Elementary Statistics, 10th Edition. The predicted time interval is 76.9 min.

Guidelines for Using The
Regression Equation 1. If there is no linear correlation, don’t use the regression equation to make predictions. 2. When using the regression equation for predictions, stay within the scope of the available sample data. 3. A regression equation based on old data is not necessarily valid now. 4. Don’t make predictions about a population that is different from the population from which the sample data were drawn. page 547 of Elementary Statistics, 10th Edition

Definitions Marginal Change Outlier
The marginal change is the amount that a variable changes when the other variable changes by exactly one unit. Outlier An outlier is a point lying far away from the other data points. Influential Point An influential point strongly affects the graph of the regression line. page 547 – 548 of Elementary Statistics, 10th Edition The slope b1 in the regression equation represents the marginal change in y that occurs when x changes by one unit.

Definition Residual The residual for a sample of paired (x, y) data, is the difference (y - y) between an observed sample y-value and the value of y, which is the value of y that is predicted by using the regression equation. ^ ^ residual = observed y – predicted y = y - y page 549 of Elementary Statistics, 10th Edition

Definitions Least-Squares Property Residual Plot ^
A straight line has the least-squares property if the sum of the squares of the residuals is the smallest sum possible. Residual Plot A scatterplot of the (x, y) values after each of the y-coordinate values have been replaced by the residual value y – y. That is, a residual plot is a graph of the points (x, y – y) ^ page of Elementary Statistics, 10th Edition

Residual Plot Analysis
If a residual plot does not reveal any pattern, the regression equation is a good representation of the association between the two variables. If a residual plot reveals some systematic pattern, the regression equation is not a good representation of the association between the two variables.

Residual Plots

The basic concepts of regression. Rounding rules. Using the regression equation for predictions. Interpreting he regression equation. Outliers Residuals and least-squares. Residual plots.

Variation and Prediction Intervals
Section 10-4 Variation and Prediction Intervals

Key Concept In this section we proceed to consider a method for constructing a prediction interval, which is an interval estimate of a predicted value of y.

Definition Total Deviation
The total deviation of (x, y) is the vertical distance y – y, which is the distance between the point (x, y) and the horizontal line passing through the sample mean y.

Definition Explained Deviation
The explained deviation is the vertical distance y - y, which is the distance between the predicted y-value and the horizontal line passing through the sample mean y. ^

Definition Unexplained Deviation
The unexplained deviation is the vertical distance y - y, which is the vertical distance between the point (x, y) and the regression line. (The distance y - y is also called a residual, as defined in Section 10-3.) ^

Unexplained, Explained, and Total Deviation
Figure 10-9

Relationships (y - y) = (y - y) + (y - y)
(total deviation) = (explained deviation) + (unexplained deviation) (y - y) = (y - y) (y - y) ^ (total variation) = (explained variation) + (unexplained variation) (y - y) 2 =  (y - y)  (y - y) 2 ^ Formula 10-4

r2 = Definition Coefficient of determination
is the amount of the variation in y that is explained by the regression line. r2 = explained variation. total variation page 559 of Elementary Statistics, 10th Edition The value of r2 is the proportion of the variation in y that is explained by the linear relationship between x and y.

Definition ^ Standard Error of Estimate
The standard error of estimate, denoted by se is a measure of the differences (or distances) between the observed sample y-values and the predicted values y that are obtained using the regression equation. ^

Standard Error of Estimate
 (y – y)2 n – 2 ^ se = or se =  y2 – b0  y – b1  xy n – 2 Formula 10-5 page 560 of Elementary Statistics, 10th Edition. Example found on page 561.

Example: Old Faithful Given the sample data in Table 10-1, find the standard error of estimate se for the duration/interval data. n = 8 y2 = 60,204 y = 688 xy = 154,378 b0 = b1 = se = n - 2  y2 - b0  y - b1  xy 8 – 2 60,204 – ( )(688) – ( )(154,378) page 561 of Elementary Statistics, 10th Edition.

Given the sample data in Table 10-1, find the standard error of estimate se for the duration/interval data. n = 8 y2 = 60,204 y = 688 xy = 154,378 b0 = b1 = se = = 4.97 (rounded) page 561 of Elementary Statistics, 10th Edition.

Prediction Interval for an Individual y
y - E < y < y + E ^ where E = t2 se n(x2) – (x)2 n(x0 – x)2 1 n x0 represents the given value of x t2 has n – 2 degrees of freedom

Example: Old Faithful For the paired duration/interval after eruption times in Table 10-1, we have found that for a duration of 180 sec, the best predicted time interval after the eruption is 76.9 min. Construct a 95% prediction interval for the time interval after the eruption, given that the duration of the eruption is 180 sec (so that x = 180). E = t2 se + n(x2) – (x)2 n(x0 – x)2 1 + 1 n page 562 of Elementary Statistics, 10th Edition. E = (2.447)( ) 8(399,451) – (1751)2 8(180 – )2 1 + 1 8 E = 13.4 (rounded) +

For the paired duration/interval after eruption times in Table 10-1, we have found that for a duration of 180 sec, the best predicted time interval after the eruption is 76.9 min. Construct a 95% prediction interval for the time interval after the eruption, given that the duration of the eruption is 180 sec (so that x = 180). y – E < y < y + E 76.9 – 13.4 < y < 63.5 < y < 90.3 ^ page 562 of Elementary Statistics, 10th Edition.

Explained and unexplained variation. Coefficient of determination. Standard error estimate. Prediction intervals.

Section 10-5 Multiple Regression

Key Concept This section presents a method for analyzing a linear relationship involving more than two variables. We focus on three key elements: The multiple regression equation. The values of the adjusted R2. The P-value.

Definition y = b0 + b1x1 + b2x2 + . . . + bkxk.
Multiple Regression Equation A linear relationship between a response variable y and two or more predictor variables (x1, x2, x , xk) The general form of the multiple regression equation is page 566 of Elementary Statistics, 10th Edition y = b0 + b1x1 + b2x bkxk. ^

Notation ^ y = b0 + b1 x1+ b2 x2+ b3 x bk xk (General form of the estimated multiple regression equation) n = sample size k = number of predictor variables y = predicted value of y x1, x2, x , xk are the predictor variables ^ page 567 of Elementary Statistics, 10th Edition

Notation - cont ß0 = the y-intercept, or the value of y when all of the predictor variables are 0 b0 = estimate of ß0 based on the sample data ß1, ß2, ß , ßk are the coefficients of the independent variables x1, x2, x , xk b1, b2, b , bk are the sample estimates of the coefficients ß1, ß2, ß , ßk page 567 of Elementary Statistics, 10th Edition

Technology Use a statistical software package such as
STATDISK Minitab Excel The computations of this section will require a statistical software package. The TI-83/84 Plus calculator does not have multiple regression features, but a supplement may be available as an application that can be stored. page 567 of Elementary Statistics, 10th Edition

Example: Old Faithful Using the sample data in Table 10-1 included with the Chapter Problem, find the multiple regression in which the response (y) variable is the time interval after an eruption and the predictor (x) variables are the duration time of the eruption and height of the eruption. page 567 of Elementary Statistics, 10th Edition.

Example: Old Faithful - cont The Minitab results are shown here:
page 567 of Elementary Statistics, 10th Edition

The Minitab results give: Interval After = Duration – Height We can write this equation as: y = x1 – 0.098x2 page 567 of Elementary Statistics, 10th Edition.

Definition Multiple coefficient of determination
The multiple coefficient of determination R2 is a measure of how well the multiple regression equation fits the sample data. Adjusted coefficient of determination The adjusted coefficient of determination is the multiple coefficient of determination R2 modified to account for the number of variables and the sample size.

(1– R2) [n – (k + 1)] Adjusted R2 (n – 1) Adjusted R2 = 1 –
Formula 10-6 where n = sample size k = number of independent (x) variables

Finding the Best Multiple Regression Equation
1. Use common sense and practical considerations to include or exclude variables. 2. Consider the P-value. Select an equation having overall significance, as determined by the P-value found in the computer display. 3. Consider equations with high values of adjusted R2 and try to include only a few variables. If an additional predictor variable is included, the value of adjusted R2 does not increase by a substantial amount. For a given number of predictor (x) variables, select the equation with the largest value of adjusted R2. In weeding out predictor (x) variables that don’t have much of an effect on the response (y) variable, it might be helpful to find the linear correlation coefficient r for each of the paired variables being considered. Page 570 of Elementary Statistics, 10th Edition

Dummy Variables Many applications involve a dichotomous variable which has only two possible discrete values (such as male/female, dead/alive, etc.). A common procedure is to represent the two possible discrete values by 0 and 1, where 0 represents “failure” and 1 represents success. A dichotomous variable with the two values 0 and 1 is called a dummy variable.

Logistic Regression We can use the methods of this section if the dummy variable is the predictor variable. If the dummy variable is the response variable we need to use a method known as logistic regression. A discussion of logistic regression is given in the text.

The multiple regression equation. Adjusted R2. Finding the best multiple regression equation. Dummy variables and logistic regression.

Section 10-6 Modeling

Key Concept This section introduces some basic concepts of developing a mathematical model, which is a function that “fits” or describes real-world data. Unlike Section 10-3, we will not be restricted to a model that must be linear.

TI-83/84 Plus Generic Models
Linear: y = a + bx Quadratic: y = ax2 + bx + c Logarithmic: y = a + b ln x Exponential: y = abx Power: y = axb

The slides that follow illustrate the graphs of some common models displayed on a TI-83/84 Plus Calculator

page 577 of Elementary Statistics, 10th Edition

Development of a Good Mathematical Model
Look for a Pattern in the Graph: Examine the graph of the plotted points and compare the basic pattern to the known generic graphs of a linear function. Find and Compare Values of R2: Select functions that result in larger values of R2, because such larger values correspond to functions that better fit the observed points. Think: Use common sense. Don’t use a model that leads to predicted values known to be totally unrealistic. Pages page 577 of Elementary Statistics, 10th Edition

The concept of mathematical modeling. Graphs from a TI-83/84 Plus calculator. Rules for developing a good mathematical model.

Elementary Statistics Tenth Edition and the Triola Statistics Series
by Mario F. Triola

Chapter 11 Multinomial Experiments and Contingency Tables
11-1 Overview 11-2 Multinomial Experiments: Goodness-of-fit 11-3 Contingency Tables: Independence and Homogeneity 11-4 McNemar’s Test for Matched Pairs

Overview and Multinomial Experiments:
Section 11-1 & 11-2 Overview and Multinomial Experiments: Goodness of Fit

Overview We focus on analysis of categorical (qualitative or attribute) data that can be separated into different categories (often called cells). Use the 2 (chi-square) test statistic (Table A- 4). The goodness-of-fit test uses a one-way frequency table (single row or column). The contingency table uses a two-way frequency table (two or more rows and columns). page 590 of Elementary Statistics, 10th Edition

Key Concept Given data separated into different categories, we will test the hypothesis that the distribution of the data agrees with or “fits” some claimed distribution. The hypothesis test will use the chi-square distribution with the observed frequency counts and the frequency counts that we would expect with the claimed distribution. page 591 of Elementary Statistics, 10th Edition

Multinomial Experiment
Definition Multinomial Experiment This is an experiment that meets the following conditions: 1. The number of trials is fixed. 2. The trials are independent. 3. All outcomes of each trial must be classified into exactly one of several different categories. 4. The probabilities for the different categories remain constant for each trial. Note that these conditions are similar to those of a binomial experiment. A binomial experiment has only two categories, whereas a multinomial experiment has more than two categories.

Example: Last Digits of Weights
When asked, people often provide weights that are somewhat lower than their actual weights. So how can researchers verify that weights were obtained through actual measurements instead of asking subjects? page 591 of Elementary Statistics, 10th Edition The analysis of the last digits of data is often used to detect data that have been reported instead of measured.

Test the claim that the digits in Table 11-2 do not occur with the same frequency. Table 11-2 summarizes the last digit of weights of 80 randomly selected students. page 592 of Elementary Statistics, 10th Edition

Verify that the four conditions of a multinomial experiment are satisfied. 1. The number of trials (last digits) is the fixed number 80. 2. The trials are independent, because the last digit of any individual’s weight does not affect the last digit of any other weight. 3. Each outcome (last digit) is classified into exactly 1 of 10 different categories. The categories are 0, 1, … , 9. 4. Finally, in testing the claim that the 10 digits are equally likely, each possible digit has a probability of 1/10, and by assumption, that probability remains constant for each subject. page 591 of Elementary Statistics, 10th Edition

Definition Goodness-of-fit Test
A goodness-of-fit test is used to test the hypothesis that an observed frequency distribution fits (or conforms to) some claimed distribution. page 592 of Elementary Statistics, 10th Edition

Goodness-of-Fit Test Notation
O represents the observed frequency of an outcome. E represents the expected frequency of an outcome. k represents the number of different categories or outcomes. n represents the total number of trials. page 592 of Elementary Statistics, 10th Edition

If all expected frequencies are equal:
the sum of all observed frequencies divided by the number of categories n E = k page 593 of Elementary Statistics, 10th Edition

If expected frequencies are not all equal:
Each expected frequency is found by multiplying the sum of all observed frequencies by the probability for the category. E = n p page 593 of Elementary Statistics, 10th Edition

Goodness-of-Fit Test in Multinomial Experiments
Requirements The data have been randomly selected. The sample data consist of frequency counts for each of the different categories. For each category, the expected frequency is at least 5. (The expected frequency for a category is the frequency that would occur if the data actually have the distribution that is being claimed. There is no requirement that the observed frequency for each category must be at least 5.) page 593 of Elementary Statistics, 10th Edition

Test Statistics 2 =  (O – E)2 E Critical Values 1. Found in Table A- 4 using k – 1 degrees of freedom, where k = number of categories. 2. Goodness-of-fit hypothesis tests are always right-tailed.

A close agreement between observed and expected values will lead to a small value of 2 and a large P-value. A large disagreement between observed and expected values will lead to a large value of 2 and a small P-value. A significantly large value of 2 will cause a rejection of the null hypothesis of no difference between the observed and the expected.

Relationships Among the 2 Test Statistic, P-Value, and Goodness-of-Fit
Figure 11-3 page 594 of Elementary Statistics, 10th Edition

Example: Last Digit Analysis
Test the claim that the digits in Table 11-2 do not occur with the same frequency. H0: p0 = p1 =  = p9 H1: At least one of the probabilities is different from the others.  = 0.05 k – 1 = 9 2.05, 9 = page 595 of Elementary Statistics, 10th Edition

Test the claim that the digits in Table 11-2 do not occur with the same frequency. Because the 80 digits would be uniformly distributed through the 10 categories, each expected frequency should be 8.

Test the claim that the digits in Table 11-2 do not occur with the same frequency. page 596 of Elementary Statistics, 10th Edition

Test the claim that the digits in Table 11-2 do not occur with the same frequency. From Table 11-3, the test statistic is 2 = Since the critical value is , we reject the null hypothesis of equal probabilities. There is sufficient evidence to support the claim that the last digits do not occur with the same relative frequency.

Example: Detecting Fraud Unequal Expected Frequencies
In the Chapter Problem, it was noted that statistics can be used to detect fraud. Table 11-1 lists the percentages for leading digits from Benford’s Law. page 597 of Elementary Statistics, 10th Edition

Example: Detecting Fraud
Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks. Observed Frequencies and Frequencies Expected with Benford’s Law

Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks. H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, p4 = 0.097, p5 = 0.079, p6 = 0.067, p7 = 0.058, p8 = and p9 = 0.046 H1: At least one of the proportions is different from the claimed values.  = 0.01 k – 1 = 8 2.01,8 = page 597 of Elementary Statistics, 10th Edition

Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks. The test statistic is 2 = Since the critical value is , we reject the null hypothesis. There is sufficient evidence to reject the null hypothesis. At least one of the proportions is different than expected.

Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks. Figure 11-5 page 596 of Elementary Statistics, 10th Edition

page 599 of Elementary Statistics, 10th Edition Figure 11-6 Comparison of Observed Frequencies and Frequencies Expected with Benford’s Law

Multinomial Experiments: Goodness-of-Fit - Equal Expected Frequencies - Unequal Expected Frequencies Test the hypothesis that an observed frequency distribution fits (or conforms to) some claimed distribution.

Contingency Tables: Independence and Homogeneity
Section 11-3 Contingency Tables: Independence and Homogeneity Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

Key Concept In this section we consider contingency tables (or two-way frequency tables), which include frequency counts for categorical data arranged in a table with a least two rows and at least two columns. We present a method for testing the claim that the row and column variables are independent of each other. We will use the same method for a test of homogeneity, whereby we test the claim that different populations have the same proportion of some characteristics. page 606 of Elementary Statistics, 10th Edition

(or two-way frequency table)
Definition Contingency Table (or two-way frequency table) A contingency table is a table in which frequencies correspond to two variables. (One variable is used to categorize rows, and a second variable is used to categorize columns.) page 606 of Elementary Statistics, 10th Edition Contingency tables have at least two rows and at least two columns.

Case-Control Study of Motorcycle Drivers
Is the color of the motorcycle helmet somehow related to the risk of crash related injuries? 491 213 704 377 112 489 31 8 39 899 333 1232 Black White Yellow/Orange Row Totals Controls (not injured) Cases (injured or killed) Column Totals page 606 of Elementary Statistics, 10th Edition

Definition Test of Independence
A test of independence tests the null hypothesis that there is no association between the row variable and the column variable in a contingency table. (For the null hypothesis, we will use the statement that “the row and column variables are independent.”) page 607 of Elementary Statistics, 10th Edition

Requirements The sample data are randomly selected and are represented as frequency counts in a two-way table. The null hypothesis H0 is the statement that the row and column variables are independent; the alternative hypothesis H1 is the statement that the row and column variables are dependent. For every cell in the contingency table, the expected frequency E is at least 5. (There is no requirement that every observed frequency must be at least 5. Also there is no requirement that the population must have a normal distribution or any other specific distribution.) page 607 of Elementary Statistics, 10th Edition

Test of Independence Test Statistic
2 =  (O – E)2 E Critical Values 1. Found in Table A-4 using degrees of freedom = (r – 1)(c – 1) r is the number of rows and c is the number of columns 2. Tests of Independence are always right-tailed. page 607 of Elementary Statistics, 10th Edition Same chi-square formula as for multinomial tables.

Total number of all observed frequencies in the table
Expected Frequency (row total) (column total) (grand total) E = Total number of all observed frequencies in the table page 607 of Elementary Statistics, 10th Edition

Test of Independence This procedure cannot be used to establish a direct cause-and-effect link between variables in question. Dependence means only there is a relationship between the two variables.

Expected Frequency for Contingency Tables
grand total row total column total (probability of a cell) n • p page 609 of Elementary Statistics, 10th Edition E = (row total) (column total) (grand total)

491 213 704 377 112 489 31 8 39 899 333 1232 Black White Yellow/Orange Row Totals Controls (not injured) Cases (injured or killed) Column Totals 899 1232 704 For the upper left hand cell: (row total) (column total) E = (grand total) = E = (899)(704) 1232

Row Totals Black White Yellow/Orange Controls (not injured) Expected 491 213 704 377 112 489 31 8 39 899 333 1232 Cases (injured or killed) Expected Column Totals (row total) (column total) E = (grand total) = E = (899)(704) 1232

491 213 704 377 112 489 31 8 39 899 333 1232 Black White Yellow/Orange Row Totals Controls (not injured) Expected Cases (injured or killed) Column Totals 28.459 10.541 Expected Calculate expected for all cells. To interpret this result for the upper left hand cell, we can say that although 491 riders with black helmets were not injured, we would have expected the number to be if crash related injuries are independent of helmet color.

Using a 0.05 significance level, test the claim that group (control or case) is independent of the helmet color. H0: Whether a subject is in the control group or case group is independent of the helmet color. (Injuries are independent of helmet color.) H1: The group and helmet color are dependent.

Row Totals Black White Yellow/Orange Controls (not injured) Expected 491 213 704 377 112 489 28.459 10.541 31 8 39 899 333 1232 Cases (injured or killed) Expected Column Totals

H0: Row and column variables are independent. H1: Row and column variables are dependent. The test statistic is 2 = 8.775  = 0.05 The number of degrees of freedom are (r–1)(c–1) = (2–1)(3–1) = 2. The critical value (from Table A-4) is 2.05,2 = The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

Figure 11-4 page 610 of Elementary Statistics, 10th Edition We reject the null hypothesis. It appears there is an association between helmet color and motorcycle safety.

Relationships Among Key Components in Test of Independence
Figure 11-8 page 611 of Elementary Statistics, 10th Edition

Definition Test of Homogeneity
In a test of homogeneity, we test the claim that different populations have the same proportions of some characteristics. page 611 of Elementary Statistics, 10th Edition

How to Distinguish Between a Test of Homogeneity and a Test for Independence:
Were predetermined sample sizes used for different populations (test of homogeneity), or was one big sample drawn so both row and column totals were determined randomly (test of independence)? The key to identifying it is a test of homogeneity is the predetermined sample sizes.

Example: Influence of Gender
Using Table 11-6 with a 0.05 significance level, test the effect of pollster gender on survey responses by men. page 612 of Elementary Statistics, 10th Edition

Using Table 11-6 with a 0.05 significance level, test the effect of pollster gender on survey responses by men. H0: The proportions of agree/disagree responses are the same for the subjects interviewed by men and the subjects interviewed by women. H1: The proportions are different. page 612 of Elementary Statistics, 10th Edition

Using Table 11-6 with a 0.05 significance level, test the effect of pollster gender on survey responses by men. Minitab page 613 of Elementary Statistics, 10th Edition

Contingency tables where categorical data is arranged in a table with a least two rows and at least two columns. * Test of Independence tests the claim that the row and column variables are independent of each other. * Test of Homogeneity tests the claim that different populations have the same proportion of some characteristics.

McNemar’s Test for Matched Pairs
Section 11-4 McNemar’s Test for Matched Pairs This section is optional; it can easily be omitted. Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

Key Concept The Contingency table procedures in Section 11-3 are based on independent data. For 2 x 2 tables consisting of frequency counts that result from matched pairs, we do not have independence, and for such cases, we can use McNemar’s test for matched pairs. page 621 of Elementary Statistics, 10th Edition

Table 11-9 is a general table summarizing the frequency counts that result from matched pairs.
The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

Definition McNemar’s Test uses frequency counts from matched pairs of nominal data from two categories to test the null hypothesis that for a table such as Table 11-9, the frequencies b and c occur in the same proportion. Note that these conditions are similar to those of a binomial experiment. A binomial experiment has only two categories, whereas a multinomial experiment has more than two categories.

Requirements The sample data have been randomly selected.
The sample data consist of matched pairs of frequency counts. The data are at the nominal level of measurement, and each observation can be classified two ways: (1) According to the category distinguishing values with each matched pair, and (2) according to another category with two possible values. For tables such as Table 11-9, the frequencies are such that b + c ≥ 10. The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

Requirements Test Statistic (for testing the null hypothesis that
for tables such as Table 11-9, the frequencies b and c occur in the same proportion): Where the frequencies of b and c are obtained from the 2 x 2 table with a format similar to Table 11-9. Critical values 1. The critical region is located in the right tail only. 2. The critical values are found in Table A- 4 by using degrees of freedom = 1. The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

Example: Comparing Treatments
Table summarizes the frequency counts that resulted from the matched pairs using Pedacream on one foot and Fungacream on the other foot. The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

H0: The following two proportions are the same: The proportion of subjects with no cure on the Pedacream-treated foot and a cure on the Fungacream-treated foot. The proportion of subjects with a cure on the Pedacream-treated and no cure on the Based on the results, does there appear to be a difference? The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

Note that the requirements have been met: The data consist of matched pairs of frequency counts from randomly selected subjects. Each observation can be categorized according to two variables. (One variable has values of “Pedacream” and “Fungacream,” and the other variable has values of “cured” and “not cured.”) b = 8 and c = 40, so b + c ≥ 10. The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

Test statistic Critical value from Table A-4 The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4. Reject the null hypothesis. It appears the creams produce different results.

Note that the test did not use the categories where both feet were cured or where neither foot was cured. Only results from categories that are different were used. Definition . Discordant pairs of results come from pairs of categories in which the two categories are different (as in cure/no cure or no cure/cure). The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

McNemar’s test for matched pairs. Data are place in a 2 x 2 table where each observation is classified in two ways. The test only compares categories that are different (discordant pairs).

by Mario F. Triola

Chapter 12 Analysis of Variance
12-1 Overview 12-2 One-Way ANOVA 12-3 Two-Way ANOVA

Overview and One-Way ANOVA
Section 12-1 & 12-2 Overview and One-Way ANOVA

Key Concept This section introduces the method of one-way analysis of variance, which is used for tests of hypotheses that three or more population means are all equal. page 637 of Elementary Statistics, 10th Edition

Overview Analysis of variance (ANOVA) is a method for testing the hypothesis that three or more population means are equal. For example: H0: µ1 = µ2 = µ3 = µk H1: At least one mean is different This method to test if more than two population means are equal is similar to the way we tested two population means equal in Chapter 9. Discussion of why one cannot just test two samples at a time on page 637 of Elementary Statistics, 10th Edition .

ANOVA Methods Require the F-Distribution
1. The F- distribution is not symmetric; it is skewed to the right. 2. The values of F can be 0 or positive; they cannot be negative. 3. There is a different F-distribution for each pair of degrees of freedom for the numerator and denominator. Critical values of F are given in Table A-5 page 636 of Elementary Statistics, 10th Edition Because the procedures used in this chapter require complicated calculations, the use and interpretation of computer software or a TI-83 Plus calculator is strongly encouraged.

F - distribution Figure 12-1

An Approach to Understanding ANOVA
One-Way ANOVA An Approach to Understanding ANOVA Understand that a small P-value (such as 0.05 or less) leads to rejection of the null hypothesis of equal means. With a large P-value (such as greater than 0.05), fail to reject the null hypothesis of equal means. Develop an understanding of the underlying rationale by studying the examples in this section. page 637 of Elementary Statistics, 10th Edition The method used is called one-way analysis of variance because we use a single property, or characteristic, for categorizing the populations.

An Approach to Understanding ANOVA
One-Way ANOVA An Approach to Understanding ANOVA Become acquainted with the nature of the SS (sum of squares) and MS (mean square) values and their role in determining the F test statistic, but use statistical software packages or a calculator for finding those values. page 637 of Elementary Statistics, 10th Edition The method used is called one-way analysis of variance because we use a single property, or characteristic, for categorizing the populations.

Definition Treatment (or factor)
A treatment (or factor) is a property or characteristic that allows us to distinguish the different populations from one another. Use computer software or TI-83/84 for ANOVA calculations if possible. page 638 of Elementary Statistics, 10th Edition

One-Way ANOVA Requirements
1. The populations have approximately normal distributions. 2. The populations have the same variance 2 (or standard deviation  ). 3. The samples are simple random samples. 4. The samples are independent of each other. 5. The different samples are from populations that are categorized in only one way. page 638 of Elementary Statistics, 10th Edition The requirements of normality and equal variances are somewhat relaxed, because the methods in this section work reasonably well unless a population has a distribution that is very nonnormal or the population variances differ by large amounts.

Procedure for testing Ho: µ1 = µ2 = µ3 = . . .
1. Use STATDISK, Minitab, Excel, or a TI-83/84 Calculator to obtain results. 2. Identify the P-value from the display. 3. Form a conclusion based on these criteria: If P-value  , reject the null hypothesis of equal means. If P-value > , fail to reject the null hypothesis of equal means. Page 638 of Elementary Statistics, 10th Edition

Procedure for testing Ho: µ1 = µ2 = µ3 = . . .
Caution when interpreting ANOVA results: When we conclude that there is sufficient evidence to reject the claim of equal population means, we cannot conclude from ANOVA that any particular mean is different from the others. There are several other tests that can be used to identify the specific means that are different, and those procedures are called multiple comparison procedures, and they are discussed later in this section. Page 638 of Elementary Statistics, 10th Edition

Example: Weights of Poplar Trees
Do the samples come from populations with different means? Page 635 of Elementary Statistics, 10th Edition, in opening example

Do the samples come from populations with different means? H0: 1 = 2 = 3 = 4 H1: At least one of the means is different from the others. For a significance level of  = 0.05, use STATDISK, Minitab, Excel, or a TI-83/84 calculator to test the claim that the four samples come from populations with means that are not all the same. Page 638 of Elementary Statistics, 10th Edition

Do the samples come from populations with different means? STATDISK Minitab Page 639 of Elementary Statistics, 10th Edition

Do the samples come from populations with different means? Excel TI-83/84 PLUS Page 639 of Elementary Statistics, 10th Edition

Do the samples come from populations with different means? H0: 1 = 2 = 3 = 4 H1: At least one of the means is different from the others. The displays all show a P-value of approximately Because the P-value is less than the significance level of  = 0.05, we reject the null hypothesis of equal means. Page 640 of Elementary Statistics, 10th Edition There is sufficient evidence to support the claim that the four population means are not all the same. We conclude that those weights come from populations having means that are not all the same.

ANOVA Fundamental Concepts
Estimate the common value of 2: 1. The variance between samples (also called variation due to treatment) is an estimate of the common population variance  2 that is based on the variability among the sample means. 2. The variance within samples (also called variation due to error) is an estimate of the common population variance 2 based on the sample variances. Even with heavy use of technology with this topic, it is important to have a basic understanding of underlying principles. page 640 of Elementary Statistics, 10th Edition

ANOVA Fundamental Concepts
Test Statistic for One-Way ANOVA F = variance between samples variance within samples Page 640 of Elementary Statistics, 10th Edition An excessively large F test statistic is evidence against equal population means.

Relationships Between the F Test Statistic and P-Value
Figure 12-2 Page 641 of Elementary Statistics, 10th Edition

Calculations with Equal Sample Sizes
Variance between samples = n where = variance of sample means sx 2 Variance within samples = sp 2 Page 641 of Elementary Statistics, 10th Edition where sp = pooled variance (or the mean of the sample variances) 2

Example: Sample Calculations

Use Table 12-2 to calculate the variance between samples, variance within samples, and the F test statistic. 1. Find the variance between samples = For the means 5.5, 6.0 & 6.0, the sample variance is = = 4 X = 2. Estimate the variance within samples by calculating the mean of the sample variances page 642 of Elementary Statistics, 10th Edition 3 = =

Use Table 12-2 to calculate the variance between samples, variance within samples, and the F test statistic. 3. Evaluate the F test statistic F = F = = variance between samples variance within samples page 642 of Elementary Statistics, 10th Edition 0.3332 2.3333

Critical Value of F Right-tailed test
Degree of freedom with k samples of the same size n numerator df = k – 1 denominator df = k(n – 1)

Calculations with Unequal Sample Sizes
F = = variance between samples variance within samples (ni – 1)si (ni – 1) 2 ni(xi – x)2 k –1 where x = mean of all sample scores combined k = number of population means being compared ni = number of values in the ith sample xi = mean of values in the ith sample si = variance of values in the ith sample 2 page 643 of Elementary Statistics, 10th Edition

Key Components of the ANOVA Method
SS(total), or total sum of squares, is a measure of the total variation (around x) in all the sample data combined. Formula 12-1 SS(total) = (x – x)2 page 644 of Elementary Statistics, 10th Edition

SS(treatment), also referred to as SS(factor) or SS(between groups) or SS(between samples), is a measure of the variation between the sample means. Formula 12-2 SS(treatment) = n1(x1 – x)2 + n2(x2 – x) nk(xk – x)2 = ni(xi - x)2 page 644 of Elementary Statistics, 10th Edition

SS(error), (also referred to as SS(within groups) or SS(within samples), is a sum of squares representing the variability that is assumed to be common to all the populations being considered. SS(error) = (n1 –1)s1 + (n2 –1)s2 + (n3 –1)s nk(xk –1)si = (ni – 1)si Formula 12-3 page 644 of Elementary Statistics, 10th Edition 2

Given the previous expressions for SS(total), SS(treatment), and SS(error), the following relationship will always hold. Formula 12-4 SS(total) = SS(treatment) + SS(error) page 644 of Elementary Statistics, 10th Edition

Mean Squares (MS) MS(treatment) =
MS(treatment) is a mean square for treatment, obtained as follows: Formula 12-5 MS(treatment) = SS (treatment) k – 1 MS(error) is a mean square for error, obtained as follows: page 645 of Elementary Statistics, 10th Edition Formula 12-6 MS(error) = SS (error) N – k

Mean Squares (MS) MS(total) =
MS(total) is a mean square for the total variation, obtained as follows: Formula 12-7 MS(total) = SS(total) N – 1 page 645 of Elementary Statistics, 10th Edition

Test Statistic for ANOVA with Unequal Sample Sizes
MS (treatment) MS (error) Formula 12-8 Numerator df = k – 1 Denominator df = N – k page 645 of Elementary Statistics, 10th Edition

Table 12-3 has a format often used in computer displays. page 646 of Elementary Statistics, 10th Edition

Identifying Means That Are Different
After conducting an analysis of variance test, we might conclude that there is sufficient evidence to reject a claim of equal population means, but we cannot conclude from ANOVA that any particular mean is different from the others. page 646 of Elementary Statistics, 10th Edition

Identifying Means That Are Different
Informal procedures to identify the specific means that are different Use the same scale for constructing boxplots of the data sets to see if one or more of the data sets are very different from the others. Construct confidence interval estimates of the means from the data sets, then compare those confidence intervals to see if one or more of them do not overlap with the others. page 646 of Elementary Statistics, 10th Edition

Bonferroni Multiple Comparison Test
Step 1. Do a separate t test for each pair of samples, but make the adjustments described in the following steps. Step 2. For an estimate of the variance σ2 that is common to all of the involved populations, use the value of MS(error). page 647 of Elementary Statistics, 10th Edition

Step 2 (cont.) Using the value of MS(error), calculate the value of the test statistic, as shown below. (This example shows the comparison for Sample 1 and Sample 4.) page 647 of Elementary Statistics, 10th Edition

Step 2 (cont.) Change the subscripts and use another pair of samples until all of the different possible pairs of samples have been tested. Step 3. After calculating the value of the test statistic t for a particular pair of samples, find either the critical t value or the P-value, but make the following adjustment: page 647 of Elementary Statistics, 10th Edition

Step 3 (cont.) P-value: Use the test statistic t with df = N-k, where N is the total number of sample values and k is the number of samples. Find the P-value the usual way, but adjust the P-value by multiplying it by the number of different possible pairings of two samples. (For example, with four samples, there are six different possible pairings, so adjust the P-value by multiplying it by 6.) page 647 of Elementary Statistics, 10th Edition

Step 3 (cont.) Critical value: When finding the critical value, adjust the significance level α by dividing it by the number of different possible pairings of two samples. (For example, with four samples, there are six different possible pairings, so adjust the P-value by dividing it by 6.) page 647 of Elementary Statistics, 10th Edition

Example: Weights of Poplar Trees Using the Bonferroni Test
Using the data in Table 12-1, we concluded that there is sufficient evidence to warrant rejection of the claim of equal means. The Bonferroni test requires a separate t test for each different possible pair of samples. page 648 of Elementary Statistics, 10th Edition

We begin with testing H0: μ1 = μ4 . From Table 12-1: x1 = 0.184, n1 = 5, x4 = 1.334, n4 = 5 page 648 of Elementary Statistics, 10th Edition

Test statistic = – 3.484 df = N – k = 20 – 4 = 16 Two-tailed P-value is , but adjust it by multiplying by 6 (the number of different possible pairs of samples) to get P-value = Because the adjusted P-value is less than α = 0.05, reject the null hypothesis. It appears that Samples 1 and 4 have significantly different means. page 648 of Elementary Statistics, 10th Edition

SPSS Bonferroni Results page 649 of Elementary Statistics, 10th Edition

One-way analysis of variance (ANOVA) Calculations with Equal Sample Sizes Calculations with Unequal Sample Sizes Identifying Means That Are Different Bonferroni Multiple Comparison Test

Section 12-3 Two-Way ANOVA

Key Concepts The analysis of variance procedure introduced in Section 12-2 is referred to as one-way analysis of variance because the data are categorized into groups according to a single factor (or treatment). In this section we introduce the method of two-way analysis of variance, which is used with data partitioned into categories according to two factors. page 655 of Elementary Statistics, 10th Edition

Two-Way Analysis of Variance
Two-Way ANOVA involves two factors. The data are partitioned into subcategories called cells. page 655 of Elementary Statistics, 10th Edition

Example: Poplar Tree Weights

Definition There is an interaction between two factors if the effect of one of the factors changes for different categories of the other factor. page 655 of Elementary Statistics, 10th Edition It is important to consider the interaction rather than just conducting another one-way ANOVA test on another factor.

Exploring Data Calculate the mean for each cell. page 656 of Elementary Statistics, 10th Edition

Exploring Data Display the means on a graph. If a graph such as Figure 12-3 results in line segments that are approximately parallel, we have evidence that there is not an interaction between the row and column variables. page 656 of Elementary Statistics, 10th Edition

Minitab page 657 of Elementary Statistics, 10th Edition

Requirements 1. For each cell, the sample values come
from a population with a distribution that is approximately normal. 2. The populations have the same variance 2. 3. The samples are simple random samples. 4. The samples are independent of each other. 5. The sample values are categorized two ways. 6. All of the cells have the same number of sample values. page 657 of Elementary Statistics, 10th Edition

Minitab, Excel, TI-83/4 or STATDISK can be used.
Two-Way ANOVA calculations are quite involved, so we will assume that a software package is being used. Minitab, Excel, TI-83/4 or STATDISK can be used.

Figure 12- 4 Procedure for Two-Way ANOVAS

Using the Minitab display: Step 1: Test for interaction between the effects: H0: There are no interaction effects. H1: There are interaction effects. F = = = 0.17 MS(interaction) MS(error) page 657 of Elementary Statistics, 10th Edition The P-value is shown as 0.915, so we fail to reject the null hypothesis of no interaction between the two factors.

Using the Minitab display: Step 2: Check for Row/Column Effects: H0: There are no effects from the row factors. H1: There are row effects. H0: There are no effects from the column factors. H1: There are column effects. F = = = 0.81 MS(site) MS(error) page 658 of Elementary Statistics, 10th Edition The P-value is shown as 0.374, so we fail to reject the null hypothesis of no effects from site.

Special Case: One Observation per Cell and No Interaction
If our sample data consist of only one observation per cell, we lose MS(interaction), SS(interaction), and df(interaction). If it seems reasonable to assume that there is no interaction between the two factors, make that assumption and then proceed as before to test the following two hypotheses separately: page 659 of Elementary Statistics, 10th Edition H0: There are no effects from the row factors. H0: There are no effects from the column factors.

Minitab page 659 & 660 of Elementary Statistics, 10th Edition

Row factor: F = = = 0.28 MS(site) MS(error) The P-value in the Minitab display is Fail to reject the null hypothesis. page 660 of Elementary Statistics, 10th Edition

Column factor: F = = = 0.73 MS(site) MS(error) The P-value in the Minitab display is Fail to reject the null hypothesis. page 660 of Elementary Statistics, 10th Edition

Two- way analysis of variance (ANOVA) Special case: One observation per cell and no interaction

by Mario F. Triola

Chapter 13 Nonparametric Statistics
13-1 Overview 13-2 Sign Test 13-3 Wilcoxon Signed-Ranks Test for Matched Pairs Wilcoxon Rank-Sum Test for Two Independent Samples Kruskal-Wallis Test 13-6 Rank Correlation 13-7 Runs Test for Randomness

Overview Definitions Parametric tests have requirements about the nature or shape of the populations involved. Nonparametric tests do not require that samples come from populations with normal distributions or have any other particular distributions. Consequently, nonparametric tests are called distribution-free tests. page 676 of Elementary Statistics, 10th Edition

Advantages of Nonparametric Methods
1. Nonparametric methods can be applied to a wide variety of situations because they do not have the more rigid requirements of the corresponding parametric methods. In particular, nonparametric methods do not require normally distributed populations. 2. Unlike parametric methods, nonparametric methods can often be applied to categorical data, such as the genders of survey respondents. 3. Nonparametric methods usually involve simpler computations than the corresponding parametric methods and are therefore easier to understand and apply. page 676 of Elementary Statistics, 10th Edition

Disadvantages of Nonparametric Methods
1. Nonparametric methods tend to waste information because exact numerical data are often reduced to a qualitative form. 2. Nonparametric tests are not as efficient as parametric tests, so with a nonparametric test we generally need stronger evidence (such as a larger sample or greater differences) before we reject a null hypothesis. page of Elementary Statistics, 10th Edition

Efficiency of Nonparametric Methods
Explanation of efficiency values are in the middle of page 677 of Elementary Statistics, 10th Edition.

Definitions Data are sorted when they are arranged according to some criterion, such as smallest to the largest or best to worst. A rank is a number assigned to an individual sample item according to its order in the sorted list. The first item is assigned a rank of 1, the second is assigned a rank of 2, and so on. page 677 of Elementary Statistics, 10th Edition

Handling Ties in Ranks Find the mean of the ranks involved and assign this mean rank to each of the tied items. Sorted Data 4 5 10 11 12 Preliminary Ranking 1 2 3 4 5 6 7 8 Rank 1 3 5 6 7.5 Mean is 3. page 678 of Elementary Statistics, 10th Edition. Exercise 19 in Section 13-2 gives other approaches for handling ties. Mean is 7.5.

Section 13-2 Sign Test

Key Concept The main objective of this section is to understand the sign test procedure, which involves converting data values to plus and minus signs, then testing for disproportionately more of either sign. page 678 of Elementary Statistics, 10th Edition

Definition Sign Test 1) Claims involving matched pairs of sample data;
The sign test is a nonparametric (distribution free) test that uses plus and minus signs to test different claims, including: 1) Claims involving matched pairs of sample data; 2) Claims involving nominal data; 3) Claims about the median of a single population. page 678 of Elementary Statistics, 10th Edition The sign test allows a decision of significance rather than just guessing.

Basic Concept of the Sign Test
The basic idea underlying the sign test is to analyze the frequencies of the plus and minus signs to determine whether they are significantly different. page 678 of Elementary Statistics, 10th Edition The sign test allows a decision of significance rather than just guessing.

Figure 13-1 Sign Test Procedure
Flowchart on page 679 of Elementary Statistics, 10th Edition

Requirements 1. The sample data have been randomly selected.
2. There is no requirement that the sample data come from a population with a particular distribution, such as a normal distribution. page 680 of Elementary Statistics, 10th Edition

Notation for Sign Test x = the number of times n = the total number of
the less frequent sign occurs n = the total number of positive and negative signs combined page 680 of Elementary Statistics, 10th Edition

Test Statistic Critical values
For n  25: x (the number of times the less frequent sign occurs) z = For n > 25: n (x + 0.5) – 2 Critical values For n  25, critical x values are in Table A-7. For n > 25, critical z values are in Table A-2. page 680 of Elementary Statistics, 10th Edition

Claims Involving Matched Pairs
When using the sign test with data that are matched pairs, we convert the raw data to plus and minus signs as follows: Subtract each value of the second variable from the corresponding value of the first variable. Record only the sign of the difference found in step 1. Exclude ties: that is, any matched pairs in which both values are equal. page of Elementary Statistics, 10th Edition

Key Concept Underlying This Use of the Sign Test
If the two sets of data have equal medians, the number of positive signs should be approximately equal to the number of negative signs. page 681 of Elementary Statistics, 10th Edition Example for a claim involving nominal data on page 682.

Example: Yields of Corn
from Different Seeds Use the data in Table 13-3 with a 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. page 681 of Elementary Statistics, 10th Edition

from Different Seeds Use the data in Table 13-3 with a 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. H0: The median of the differences is equal to 0. H1: The median of the differences is not equal to 0.  = 0.05 x = minimum(7, 4) = 4 (From Table 13-3, there are 7 negative signs and 4 positive signs.) Critical value = 1 (From Table A-7 where n = 11 and  = 0.05) page 681 of Elementary Statistics, 10th Edition

from Different Seeds Use the data in Table 13-3 with a 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. H0: The median of the differences is equal to 0. H1: The median of the differences is not equal to 0. With a test statistic of x = 4 and a critical value of 1, we fail to reject the null hypothesis of no difference. There is not sufficient evidence to warrant rejection of the claim that the median of the differences is equal to 0. page of Elementary Statistics, 10th Edition

Claims Involving Nominal Data
The nature of nominal data limits the calculations that are possible, but we can identify the proportion of the sample data that belong to a particular category. Then we can test claims about the corresponding population proportion p. page 682 of Elementary Statistics, 10th Edition.

Example: Gender Selection
Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. The procedures are for cases in which n > 25. Note that the only requirement is that the sample data are randomly selected. page of Elementary Statistics, 10th Edition H0: p = 0.5 (the proportion of girls is 0.5) H1: p  0.5

Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. Denoting girls by the positive sign (+) and boys by the negative sign (–), we have 295 positive signs and 30 negative signs. Test statistic x = minimum(295, 30) = 30 The test involves two tails. page 683 of Elementary Statistics, 10th Edition

Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. n (x + 0.5) – 2 z = n 2 page 683 of Elementary Statistics, 10th Edition ( ) – z = 325 2 = –14.64

Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. With  = 0.05 in a two-tailed test, the critical values are z =  1.96. The test statistic z = is less than We reject the null hypothesis that p = 0.5. There is sufficient evidence to warrant rejection of the claim that the method of gender selection has no effect. page 683 of Elementary Statistics, 10th Edition

Of the 325 babies born to parents using the XSORT method of gender selection, 295 were girls. Use the sign test and a 0.05 significance level to test the claim that this method of gender selection has no effect. Figure 13.2 page 683 of Elementary Statistics, 10th Edition.

Claims About the Median of a Single Population
The negative and positive signs are based on the claimed value of the median. page 684 of Elementary Statistics, 10th Edition

Example: Body Temperature
Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6°F. There are 68 subjects with temperatures below 98.6°F, 23 subjects with temperatures above 98.6°F, and 15 subjects with temperatures equal to 98.6°F. H0: Median is equal to 98.6°F. H1: Median is less than 98.6°F. page 684 of Elementary Statistics, 10th Edition Since the claim is that the median is less than 98.6°F. the test involves only the left tail.

Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6°F. Discard the 15 zeros. Use ( – ) to denote the 68 temperatures below 98.6°F, and use ( + ) to denote the 23 temperatures above 98.6°F. So n = 91 and x = 23 page 684 of Elementary Statistics, 10th Edition

Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6°F. (x + 0.5) – z = n 2 ( ) – z = 91 2 = – 4.61 page 684 of Elementary Statistics, 10th Edition

Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6°F. We use Table A-2 to get the critical z value of –1.645. The test statistic of z = –4.61 falls into the critical region. We reject the null hypothesis. We support the claim that the median body temperature of healthy adults is less than 98.6°F. page of Elementary Statistics, 10th Edition

Use the temperatures for 12:00 A.M. on Day 2 in Data Set 2 in Appendix B. Use the sign test to test the claim that the median is less than 98.6°F. Figure 13.3 page 685 of Elementary Statistics, 10th Edition

Sign tests where data are assigned plus or minus signs and then tested to see if the number of plus and minus signs is equal. Sign tests can be performed on claims involving: Matched pairs Nominal data The median of a single population

Wilcoxon Signed-Ranks Test for Matched Pairs
Section 13-3 Wilcoxon Signed-Ranks Test for Matched Pairs

Key Concept The Wilcoxon signed-ranks test uses ranks of sample data consisting of matched pairs. This test is used with a null hypothesis that the population of differences from the matched pairs has a median equal to zero. page 689 of Elementary Statistics, 10th Edition

Definition The Wilcoxon signed-ranks test is a nonparametric test that uses ranks of sample data consisting of matched pairs. It is used to test the null hypothesis that the population of differences has a median of zero. H0: The matched pairs have differences that come from a population with a median equal to zero. H1: The matched pairs have differences that come from a population with a nonzero median. The Wilcoxon signed-ranks test can also be used to test the claim that a sample comes from a population with a specified median. See Exercise 13 on page 695.

Wilcoxon Signed-Ranks Test Requirements
1. The data consist of matched pairs that have been randomly selected. 2. The population of differences (found from the pairs of data) has a distribution that is approximately symmetric, meaning that the left half of its histogram is roughly a mirror image of its right half. (There is no requirement that the data have a normal distribution.) page 690 of Elementary Statistics, 10th Edition

Notation T = the smaller of the following two sums:
1. The sum of the absolute values of the negative ranks of the nonzero differences d 2. The sum of the positive ranks of the nonzero differences d page 690 of Elementary Statistics, 10th Edition

Test Statistic for the Wilcoxon Signed-Ranks Test for Matched Pairs
For n  30, the test statistic is T. z = For n > 30, the test statistic is 4 T – n(n + 1) n(n +1) (2n +1) 24 page 690 of Elementary Statistics, 10th Edition

Critical Values for the Wilcoxon Signed-Ranks Test for Matched Pairs
For n  30, the critical T value is found in Table A-8. For n > 30, the critical z values are found in Table A-2. page 690 of Elementary Statistics, 10th Edition

Procedure for Finding the Value of the Test Statistic
Step 1: For each pair of data, find the difference d by subtracting the second value from the first. Keep the signs, but discard any pairs for which d = 0. Step 2: Ignore the signs of the differences, then sort the differences from lowest to highest and replace the differences by the corresponding rank value. When differences have the same numerical value, assign to them the mean of the ranks involved in the tie. Step 3: Attach to each rank the sign difference from which it came. That is, insert those signs that were ignored in step 2. Step 4: Find the sum of the absolute values of the negative ranks. Also find the sum of the positive ranks. first of 2 slides

Step 5: Let T be the smaller of the two sums found in Step 4. Either sum could be used, but for a simplified procedure we arbitrarily select the smaller of the two sums. Step 6: Let n be the number of pairs of data for which the difference d is not 0. Step 7: Determine the test statistic and critical values based on the sample size, as shown above. Step 8: When forming the conclusion, reject the null hypothesis if the sample data lead to a test statistic that is in the critical region - that is, the test statistic is less than or equal to the critical value(s). Otherwise, fail to reject the null hypothesis. second of 2 slides

Example: Does the Type of Seed
Affect Corn Growth? Use the data in Table 13-4 with the Wilcoxon signed-ranks test and 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. page 691 of Elementary Statistics, 10th Edition

Affect Corn Growth? Use the data in Table 13-4 with the Wilcoxon signed-ranks test and 0.05 significance level to test the claim that there is no difference between the yields from the regular and kiln-dried seed. H0: There is no difference between the times of the first and second trials. H1: There is a difference between the times of the first and second trials. page 692 of Elementary Statistics, 10th Edition

Affect Corn Growth? The ranks of differences in row four of the table are found by ranking the absolute differences, handling ties by assigning the mean of the ranks. The signed ranks in row five of the table are found by attaching the sign of the differences to the ranks. page 692 of Elementary Statistics, 10th Edition Stem and leaf plots can be useful for finding ranks The differences in row three of the table are found by computing the first time – second time.

Affect Corn Growth? Calculate the Test Statistic Step 1: In Table 13- 4, the row of differences is obtained by computing this difference for each pair of data: d = yield from regular seed – yield from kiln- dried seed Step 2: Ignoring their signs, we rank the absolute differences from lowest to highest. Step 3: The bottom row of Table is created by attaching to each rank the sign of the corresponding differences. page 692 of Elementary Statistics, 10th Edition first of 4 slides

Affect Corn Growth? Calculate the Test Statistic Step 3 (cont.): Step 3 (cont.): If there really is no difference between the yields from the two types of seed (as in the null hypothesis), we expect the sum of the positive ranks to be approximately equal to the sum of the absolute values of the negative ranks. Step 4: We now find the sum of the absolute values of the negative ranks, and we also find the sum of the positive ranks. second of 4 slides

Affect Corn Growth? Calculate the Test Statistic Step 4 (cont.): Sum of absolute values of negative ranks: 51 (from ) Sum of positive ranks: 15 (from ) Step 5: Letting T be the smaller of the two sums found in Step 4, we find that T = 15. Step 6: Letting n be the number of pairs of data for which the difference d is not 0, we have n = 11. third of 4 slides.

Affect Corn Growth? Calculate the Test Statistic Step 7: Because n = 11, we have n ≤ 30, so we use a test statistic of T = 15. From Table A- 8, the critical T = 11 (using n = 11 and a = 0.05 in two tails). Step 8: The test statistic T = 15 is not less than or equal to the critical value of 11, so we fail to reject the null hypothesis. Step 7: fourth of 4 slides It appears that there is no difference between yields from regular seed and kiln-dried seed.

The Wilcoxon signed-ranks test which uses matched pairs. The hypothesis is that the matched pairs have differences that come from a population with a median equal to zero.

Wilcoxon Rank-Sum Test for Two Independent Samples
Section 13- 4 Wilcoxon Rank-Sum Test for Two Independent Samples

Key Concept The Wilcoxon signed-ranks test (Section 13-3) involves matched pairs of data. The Wilcoxon rank-sum test of this section involves two independent samples that are not related or somehow matched or paired. page 695 of Elementary Statistics, 10th Edition

Definition The Wilcoxon rank-sum test is a nonparametric test that uses ranks of sample data from two independent populations. It is used to test the null hypothesis that the two independent samples come from populations with equal medians. H0: The two samples come from populations with equal medians. H1: The two samples come from populations with different medians. page 695 of Elementary Statistics, 10th Edition Remind students that samples are independent if the sample values selected from one population are not related or somehow matched or paired with the sample values from the other population.

Basic Concept If two samples are drawn from identical populations and the individual values are all ranked as one combined collection of values, then the high and low ranks should fall evenly between the two samples. This test is equivalent to the Mann-Whitney U Test (see Exercise 13).

Requirements 1. There are two independent samples of randomly selected data. 2. Each of the two samples has more than values. 3. There is no requirement that the two populations have a normal distribution or any other particular distribution. page 696 of Elementary Statistics, 10th Edition

Notation for the Wilcoxon Rank-Sum Test
n1 = size of Sample 1 n2 = size of Sample 2 R1 = sum of ranks for Sample 1 R2 = sum of ranks for Sample 2 R = same as R1 (sum of ranks for Sample 1) R = mean of the sample R values that is expected when the two populations have equal medians R = standard deviation of the sample R values that is expected with two populations having equal medians page 696 of Elementary Statistics, 10th Edition

Test Statistic for the Wilcoxon Rank-Sum Test
R – R z = R n1 (n1 + n2 + 1) 2 =  R where R = n1 n2 (n1 + n2 + 1) 12 page 696 of Elementary Statistics, 10th Edition n1 = size of the sample from which the rank sum R is found n2 = size of the other sample R = sum of ranks of the sample with size n1

Critical Values for the Wilcoxon Rank-Sum Test
Critical values can be found in Table A-2 (because the test statistic is based on the normal distribution). page 696 of Elementary Statistics, 10th Edition

1. Temporarily combine the two samples into one big sample, then replace each sample value with its rank. 2. Find the sum of the ranks for either one of the two samples. 3. Calculate the value of the z test statistic as shown in the previous slide, where either sample can be used as ‘Sample 1’. Unlike the corresponding hypothesis tests in Section 9-3, the Wilcoxon rank-sum test does not require normally distributed populations.

Example: BMI of Men and Women
The data in Table 13-5 are from Data Set 1 in Appendix B and use only the first 13 sample values for men and the first 12 sample values for women. The numbers in parentheses are their ranks beginning with a rank of 1 assigned to the lowest value of 17.7. R1 and R2 at the bottom denote the sum of ranks. page 697 of Elementary Statistics, 10th Edition

Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. The requirements of having two independent and random samples and each having more than 10 values are met. page 697 of Elementary Statistics, 10th Edition H0: Men and women have BMI values with equal medians H1: Men and women have BMI values with medians that are not equal

Example: BMI of Men and Women Procedures .
Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. Procedures 1. Rank all 25 BMI measurements combined. This is done in Table 13-5. 2. Find the sum of the ranks of either one of the samples. For men the sum of ranks is R = … = 187 first slide of 2

Example: BMI of Men and Women Procedures (cont.) .
3. Calculate the value of the z test statistic. second slide of 2

Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. A large positive value of z would indicate that the higher ranks are found disproportionately in Sample 1, and a large negative value of z would indicate that Sample 1 had a disproportionate share of lower ranks. page 698 of Elementary Statistics, 10th Edition

Use the data in Table 13-5 with the Wilcoxon rank-sum test and a 0.05 significance level to test the claim that the median BMI of men is equal to the median BMI of women. We have a two tailed test (with  = 0.05), so the critical values are 1.96 and –1.96. The test statistic of 0.98 does not fall within the critical region, so we fail to reject the null hypothesis that men and women have BMI values with equal medians. It appears that BMI values of men and women are basically the same. page 698 of Elementary Statistics, 10th Edition

The preceding example used only 13 of the 40 sample BMI values for men listed in Data Set 1 in Appendix B, and it used only 12 of the 40 BMI values for women. Do the results change if we use all 40 sample values for both men and women? The null and alternative hypotheses are the same. Example of large sample

In the Minitab display below ETA1 and ETA2 denote the medians of the first and second samples, respectively. The rank sum for men is W = The P-value is (or after adjustment for ties). Minitab page of Elementary Statistics, 10th Edition

Because the P-value is greater than α = 0.05, we fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have BMI values with equal medians. Minitab page of Elementary Statistics, 10th Edition

The Wilcoxon Rank-Sum Test for Two Independent Samples. It is used to test the null hypothesis that the two independent samples come from populations with equal medians.

Section 13-5 Kruskal-Wallis Test

Key Concept This section introduces the Kruskal-Wallis test, which uses ranks of data from three or more independent samples to test the null hypothesis that the samples come from populations with equal medians. This is equivalent to the one-way ANOVA that tests for equal means, but this test does not require that the populations have normal distributions.

Kruskal-Wallis Test Definition .
The Kruskal-Wallis test (also called the H test) is a nonparametric test that uses ranks of sample data from three or more independent populations. It is used to test the null hypothesis that the independent samples come from populations with the equal medians. page 702 of Elementary Statistics, 10th Edition The ANOVA test was used in Chapter 11 to test hypotheses that differences in means among several samples were significant, but that method requires all the involved populations have normal distributions. H0: The samples come from populations with equal medians. H1: The samples come from populations with medians that are not all equal.

Kruskal-Wallis Test We compute the test statistic H, which has a distribution that can be approximated by the chi-square (2 ) distribution as long as each sample has at least 5 observations. When we use the chi-square distribution in this context, the number of degrees of freedom is k – 1, where k is the number of samples. page 702 of Elementary Statistics, 10th Edition For a quick review of the key features of the chi-square distribution, see Section 7-5

Kruskal-Wallis Test Requirements
1. We have at least three independent samples, all of which are randomly selected. 2. Each sample has at least 5 observations. 3. There is no requirement that the populations have a normal distribution or any other particular distribution. page 702 of Elementary Statistics, 10th Edition

Kruskal-Wallis Test Notation n1 = number of observations in Sample 1
N = total number of observations in all observations combined k = number of samples R1 = sum of ranks for Sample 1 n1 = number of observations in Sample 1 For Sample 2, the sum of ranks is R2 and the number of observations is n2 , and similar notation is used for the other samples. page 702 of Elementary Statistics, 10th Edition

Kruskal-Wallis Test Test Statistic Critical Values
1. Test is right-tailed. 2. df = k – 1 (Because the test statistic H can be approximated by the 2 distribution, use Table A- 4). The Kruskal-Wallis test statistic H is the rank version of the test statistic F used in ANOVA.

Procedure for Finding the Value of the Test Statistic H
1 Temporarily combine all samples into one big sample and assign a rank to each sample value. 2. For each sample, find the sum of the ranks and find the sample size. 3. Calculate H by using the results of Step 2 and the notation and test statistic given on the preceding slide. page 703 of Elementary Statistics, 10th Edition

Procedure for Finding the Value of the Test Statistic H
The test statistic H is basically a measure of the variance of the rank sums R1 , R2 , … , R k. If the ranks are distributed evenly among the sample groups, then H should be a relatively small number. If the samples are very different, then the ranks will be excessively low in some groups and high in others, with the net effect that H will be large. page 703 of Elementary Statistics, 10th Edition

Example: Effects of Treatments on Poplar Tree Weights
Table 13-6 lists weights of poplar trees given different treatments. (Numbers in parentheses are ranks.) page of Elementary Statistics, 10th Edition

Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. Are requirements met? There are three or more independent and random samples. Each sample size is 5. (Requirement is at least 5.) page of Elementary Statistics, 10th Edition H0: The populations of poplar tree weights from the four treatments have equal medians. H1: The four population medians are not all equal.

Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. The following statistics come from Table 13-6: n1 = 5, n2 = 5, n3 = 5, n4 = 5 N = 20 R1 = 45, R2 = 37.5, R3 = 42.5, R4 = 85 page of Elementary Statistics, 10th Edition

Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. Evaluate the test statistic. . page 704 of Elementary Statistics, 10th Edition

Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. Find the critical value. . Because each sample has at least five observations, the distribution of H is approximately a chi-square distribution. The Kruskal-Wallis test statistic H is the rank version of the test statistic F used in ANOVA. df = k – 1 = 4 – 1 = α = 0.05 From Table A-4 the critical value =

Use the data in Table 13-6 with the Kruskal-Wallis test to test the claim that the four samples come from populations with equal medians. The test statistic is in the critical region, so we reject the null hypothesis of equal medians. At least one of the medians appears to be different from the others. page 704 of Elementary Statistics, 10th Edition

The Kruskal-Wallis Test is the non-parametric equivalent of ANOVA. It tests the hypothesis that three or more populations have equal means. The populations do not have to be normally distributed.

Section 13-6 Rank Correlation

Key Concept This section describes the nonparametric method of rank correlation, which uses paired data to test for an association between two variables. In Chapter 10 we used paired sample data to compute values for the linear correlation coefficient r, but in this section we use ranks as a the basis for computing the rank correlation coefficient rs . page 708 of Elementary Statistics, 10th Edition We use the notation rs for the rank correlation coefficient so that we don’t confuse it with the linear correlation coefficient r. The subscript s is used to honor Charles Spearman who originated the approach.

Rank Correlation Definition
The rank correlation test (or Spearman’s rank correlation test) is a non-parametric test that uses ranks of sample data consisting of matched pairs. It is used to test for an association between two variables, so the null and alternative hypotheses are as follows (where ρs denotes the rank correlation coefficient for the entire population): Ho: ρs = 0 (There is no correlation between the two variables.) H1: ρs  0 (There is a correlation between the two variables.) page 708 of Elementary Statistics, 10th Edition

Advantages Rank correlation has these advantages over the parametric methods discussed in Chapter 10: The nonparametric method of rank correlation can be used in a wider variety of circumstances than the parametric method of linear correlation. With rank correlation, we can analyze paired data that are ranks or can be converted to ranks. Rank correlation can be used to detect some (not all) relationships that are not linear. page 708 of Elementary Statistics, 10th Edition

Disadvantages A disadvantage of rank correlation is its efficiency rating of 0.91, as described in Section 13-1. This efficiency rating shows that with all other circumstances being equal, the nonparametric approach of rank correlation requires 100 pairs of sample data to achieve the same results as only 91 pairs of sample observations analyzed through parametric methods, assuming that the stricter requirements of the parametric approach are met. page 708 of Elementary Statistics, 10th Edition

Figure 13-4 Rank Correlation for Testing H0: s = 0

Requirements The sample paired data have been randomly selected.
Unlike the parametric methods of Section 10-2, there is no requirement that the sample pairs of data have a bivariate normal distribution. There is no requirement of a normal distribution for any population. page 710 of Elementary Statistics, 10th Edition

Notation rs = rank correlation coefficient for sample paired data (rs is a sample statistic) s = rank correlation coefficient for all the population data (s is a population parameter) n = number of pairs of data d = difference between ranks for the two values within a pair page 710 of Elementary Statistics, 10th Edition

Rank Correlation Test Statistic
No ties: After converting the data in each sample to ranks, if there are no ties among ranks for either variable, the exact value of the test statistic can be calculated using this formula: Ties: After converting the data in each sample to ranks, if either variable has ties among its ranks, the exact value of the test statistic rs can be found by using Formula 10-1 with the ranks: page 710 of Elementary Statistics, 10th Edition

Rank Correlation Critical values:
If n  30, critical values are found in Table A-9. If n > 30, use Formula 13-1. Formula 13-1 page 710 of Elementary Statistics, 10th Edition Table A-9 in this edition includes some values that have been changed from those in Table A-9 from earlier editions. where the value of z corresponds to the significance level. (For example, if a = 0.05, z – 1.96.)

Example: Rankings of Colleges
Use the data in Table 13-7 to determine if there is a correlation between the student rankings and the rankings of the magazine. page of Elementary Statistics, 10th Edition

Use the data in Table 13-7 to determine if there is a correlation between the student rankings and the rankings of the magazine. H0: s = 0 H1: s  0 Since neither variable has ties in the ranks: page 711 of Elementary Statistics, 10th Edition

Use the data in Table 13-7 to determine if there is a correlation between the student rankings and the rankings of the magazine. H0: s = 0 H1: s  0 From Table A-9 the critical values are 0.738. Because the test statistic of rs = does not exceed the critical value, we fail to reject the null hypothesis. There is not sufficient evidence to support a claim of a correlation between the rankings of the students and the magazine. page 711 of Elementary Statistics, 10th Edition

Large Sample Case Assume that the preceding example is expanded by including a total of 40 colleges and that the test statistic rs is found to be If the significance level of  = 0.05, what do you conclude about the correlation? Since n = 40 exceeds 30, we find the critical value from Formula 13-1 page 712 of Elementary Statistics, 10th Edition

Large Sample Case Assume that the preceding example is expanded by including a total of 40 colleges and that the test statistic rs is found to be If the significance level of  = 0.05, what do you conclude about the correlation? The test statistic of rs = does not exceed the critical value of 0.314, so we fail to reject the null hypothesis. There is not sufficient evidence to support the claim of a correlation between students and the magazine. page 712 of Elementary Statistics, 10th Edition

Example: Detecting a Nonlinear Pattern
The data in Table are the numbers of games played and the last scores (in millions) of a Raiders of the Lost Ark pinball game. We expect that there should be an association between the number of games played and the pinball score. H0: s = 0 H1: s  0 page 712 of Elementary Statistics, 10th Edition

There are no ties among ranks of either list. page 713 of Elementary Statistics, 10th Edition

Since n = 9 is less than 30, use Table A-9 Critical values are ± 0.700 The sample statistic exceeds 0.700, so we conclude that there is significant evidence to reject the null hypothesis of no correlation. There appears to be correlation between the number of games played and the score. page 713 of Elementary Statistics, 10th Edition

If the preceding example is done using the methods of Chapter 9, the linear correlation coefficient is r = This leads to the conclusion that there is not enough evidence to support the claim of a significant linear correlation, whereas the nonlinear test found that there was enough evidence. The Excel scatter diagram shows that there is a non-linear relationship that the parametric method would not have detected. Excel page 713 of Elementary Statistics, 10th Edition

Rank correlation which is the non-parametric equivalent of testing for correlation described in Chapter 10. It uses ranks of matched pairs to test for association. Sometimes rank correlation can detect non-linear correlation that the parametric test will not recognize.

Runs Test for Randomness
Section 13-7 Runs Test for Randomness page 717 of Elementary Statistics, 10th Edition

Key Concept This section introduces the runs test for randomness, which can be used to determine whether the sample data in a sequence are in a random order. This test is based on sample data that have two characteristics, and it analyzes runs of those characteristics to determine whether the runs appear to result from some random process, or whether the runs suggest that the order of the data is not random. page 717 of Elementary Statistics, 10th Edition Suggestion: use the seating arrangement of the class to test for randomness in the way that males and females are seated.

Definitions A run is a sequence of data having the same characteristic; the sequence is preceded and followed by data with a different characteristic or by no data at all. The runs test uses the number of runs in a sequence of sample data to test for randomness in the order of the data. page 717 of Elementary Statistics, 10th Edition This is not a test to determine whether there is a biased sample.

Fundamental Principles of the Run Test
Reject randomness if the number of runs is very low or very high. Example: The sequence of genders FFFFFMMMMM is not random because it has only 2 runs, so the number of runs is very low. Example: The sequence of genders FMFMFMFMFM is not random because there are 10 runs, which is very high. It is important to note that the runs test for randomness is based on the order in which the data occur; it is not based on the frequency of the data. page 717 of Elementary Statistics, 10th Edition

Figure 13-5 Procedure for Runs Test for Randomness

Requirements 1. The sample data are arranged according to some ordering scheme, such as the order in which the sample values were obtained. 2. Each data value can be categorized into one of two separate categories (such as male/female). page 719 of Elementary Statistics, 10th Edition

Notation n1 = number of elements in the sequence that have one particular characteristic (The characteristic chosen for n1 is arbitrary.) n2 = number of elements in the sequence that have the other characteristic G = number of runs page 719 of Elementary Statistics, 10th Edition

For Small Samples (n1 ≤ 20 and n2 ≤ 20) and a = 0.05: Test Statistic Test statistic is the number of runs G Critical Values Critical values are found in Table A-10. page 719 of Elementary Statistics, 10th Edition

For Small Samples (n1 ≤ 20 and n2 ≤ 20) and a = 0.05: Decision criteria Reject randomness if the number of runs G is: less than or equal to the smaller critical value found in Table A-10. or greater than or equal to the larger critical value found in Table A-10. page 719 of Elementary Statistics, 10th Edition

For Small Samples (n1 ≤ 20 and n2 ≤ 20) and a = 0.05: Test Statistic where page 719 of Elementary Statistics, 10th Edition and

For Large Samples (n1 > 20 or n2 > 20) or a ≠ 0.05: Critical Values Critical values of z: Use Table A-2. page 719 of Elementary Statistics, 10th Edition

Separate the runs as shown below.
Example: Small Sample Genders of Bears Listed below are the genders of the first 10 bears from Data Set 6 in Appendix B. Use a 0.05 significance level to test for randomness in the sequence of genders. M M M M F F M M F F Separate the runs as shown below. M M M M F F M M F F 2nd run 3rd run 4th run 1st run page of Elementary Statistics, 10th Edition

Example: Small Sample Genders of Bears
M M M M F F M M F F 2nd run 3rd run 4th run 1st run n1 = total number of males = 6 n2 = total number of females = 4 G = number of runs = 4 Because n1 ≤ 20 and n2 ≤ 20 and a = 0.05, the test statistic is G = 4 page 720 of Elementary Statistics, 10th Edition

Example: Small Sample Genders of Bears
M M M M F F M M F F 1st run 2nd run 3rd run 4th run From Table A-10, the critical values are 2 and 9. Because G = 4 is not less than or equal to 2, nor is it greater than or equal to 9, we do not reject randomness. It appears the sequence of genders is random. page 720 of Elementary Statistics, 10th Edition

Boston Rainfall on Mondays
Example: Large Sample Boston Rainfall on Mondays Refer to the rainfall amounts for Boston as listed in Data Set 10 in Appendix B. Is there sufficient evidence to support the claim that rain on Mondays is not random? D D D D R D R D D R D D R D D D R D D R R R D D D D R D R D R R R D R D D D R D D D R D R D D R D D D R H0: The sequence is random. H1: The sequence is not random. n1 = number of Ds = 33 n2 = number or Rs = 19 G = number of runs = 30 page 720 of Elementary Statistics, 10th Edition

Example: Large Sample Boston Rainfall on Mondays Since n1 > 20, we must calculate z using the formulas: page 721 of Elementary Statistics, 10th Edition

Example: Large Sample Boston Rainfall on Mondays The critical values are z = and 1.96. The test statistic of z = 1.48 does not fall within the critical region, so we fail to reject the null hypothesis of randomness. The given sequence does appear to be random. page 721 of Elementary Statistics, 10th Edition

The runs test for randomness which can be used to determine whether the sample data in a sequence are in a random order. We reject randomness if the number of runs is very low or very high.

Lecture Slides Elementary Statistics Tenth Edition

Statistical Process Control
Chapter 14 Statistical Process Control 14-1 Overview 14-2 Control Charts for Variation and Mean 14-3 Control Charts for Attributes

Overview and Control Charts for Variation and Mean
Section 14-1 and 14-2 Overview and Control Charts for Variation and Mean

Key Concept The main objective of this section is to construct run charts, R charts, and charts so that we can monitor important features of data over time. We will use such charts to determine whether some process is statistically stable (or within statistical control). page 734 of Elementary Statistics, 10th Edition

Control Charts for Variation and Mean
Definition . Process data are data arranged according to some time sequence. They are measurements of a characteristic of goods or services that result from some combination of equipment, people, materials, and conditions. Important characteristics of process data can change over time. page 734 of Elementary Statistics, 10th Edition

Definition . A run chart is a sequential plot of individual data values over time. One axis (usually vertical) is used for the data values, and the other axis (usually horizontal) is used for the time sequence. page 735 of Elementary Statistics, 10th Edition

Example: Measuring Aircraft Altimeters
For 20 consecutive business days, four altimeters are randomly selected and tested in a pressure chamber that simulates an altitude of 1000 feet. The errors of the readings above or below 1000 feet are shown in Table 14-1. page 733 of Elementary Statistics, 10th Edition

Treating the 80 altimeter errors in Table 14-1 as a string of consecutive measurements, construct a run chart by using a vertical axis for the errors and a horizontal axis to identify the order of the sample data. Minitab Figure 14-1 Run Chart of Individual Altimeter Errors in Table 14-1 page 735 of Elementary Statistics, 10th Edition Minitab has excellent programs for generating run charts and control charts.

As time progresses from left to right, the heights of the points appear to show a pattern of increasing variation. The Federal Aviation Administration regulations require errors less than 20 ft, so the altimeters represented by the points at the left are okay, whereas several of the points farther to the right correspond to altimeters not meeting the required specifications. Minitab page 735 of Elementary Statistics, 10th Edition

Only when a process is statistically stable can its data be treated as if they came from a population with a constant mean, standard deviation, distribution, and other characteristics. Definition A process is statistically stable (or within statistical control) if it has natural variation, with no patterns, cycles, or any unusual points. page 736 of Elementary Statistics, 10th Edition

Figure 14-2 Processes That Are Not Statistically Stable
Minitab page of Elementary Statistics, 10th Edition Figure 14-2(a): There is an obvious upward trend that corresponds to values that are increasing over time.

Minitab page of Elementary Statistics, 10th Edition Figure 14-2(b): There is an obvious downward trend that corresponds to steadily decreasing values.

Minitab page of Elementary Statistics, 10th Edition Figure 14-2(c): There is an upward shift. A run chart such as this one might result from an adjustment to the filling process, making all subsequent values higher.

Minitab page of Elementary Statistics, 10th Edition Figure 14-2(d): There is a downward shift—the first few values are relatively stable, and then something happened so that the last several values are relatively stable, but at a much lower level.

Minitab page of Elementary Statistics, 10th Edition Figure 14-2(e): The process is stable except for one exceptionally high value.

Minitab page of Elementary Statistics, 10th Edition Figure 14-2(f): There is an exceptionally low value.

Minitab page of Elementary Statistics, 10th Edition Figure 14-2(g): There is a cyclical pattern (or repeating cycle). This pattern is clearly nonrandom and therefore reveals a statistically unstable process.

Minitab page of Elementary Statistics, 10th Edition Figure 14-2(h): The variation is increasing over time. This is a common problem in quality control.

A common goal of many different methods of quality control is this:
Reduce variation in a product or a service. page 738 of Elementary Statistics, 10th Edition

Definitions Random variation is due to chance; it is the type of variation inherent in any process that is not capable of producing every good or service exactly the same way every time. Assignable variation results from causes that can be identified (such factors as defective machinery, untrained employees, and so on). page 738 of Elementary Statistics, 10th Edition

Control Chart for Monitoring Variation: The R Chart
Definition A control chart of a process characteristic (such as mean or variation) consists of values plotted sequentially over time, and it includes a center line as well as a lower control limit (LCL) and an upper control limit (UCL). The centerline represents a central value of the characteristic measurements, whereas the control limits are boundaries used to separate and identify any points considered to be unusual. page 739 of Elementary Statistics, 10th Edition

Control Chart for Monitoring Variation: The R Chart
An R chart (or range chart) is a plot of the sample ranges instead of individual sample values, and it is used to monitor the variation in a process. In addition to plotting the range values, it includes a centerline located at R, which denotes the mean of all sample ranges, as well as another line for the lower control limit and a third line for the upper control limit. page 739 of Elementary Statistics, 10th Edition

Notation Given: Process data consisting of a sequence of samples all of the same size n, and the distribution of the process data is essentially normal. n = size of each sample, or subgroup R = mean of the sample ranges (that is, the sum of the sample ranges divided by the number of samples) page 739 of Elementary Statistics, 10th Edition

Monitoring Process Variation: Control Chart for R
Points plotted: Sample ranges Centerline: R Upper Control Limit (UCL): D4R (where D4 is found in Table 14-2) Lower Control Limit (LCL): D3R (where D3 is found in Table 14-2) page 739 of Elementary Statistics, 10th Edition

Table 14-2 Control Chart Constants

Example: Manufacturing
Aircraft Altimeters Refer to the altimeter errors in Table Using the samples of size n = 4 collected each day of manufacturing, construct a control chart for R. page of Elementary Statistics, 10th Edition

MINITAB Display R Chart for Errors Minitab

Interpreting Control Charts
Upper and lower control limits of a control chart are based on the actual behavior of the process, not the desired behavior. Upper and lower control limits are totally unrelated to any process specifications that may have been decreed by the manufacturer. page 741 of Elementary Statistics, 10th Edition

Interpreting Control Charts
When investigating the quality of some process, there are typically two key questions that need to be addressed: Based on the current behavior of the process, can we conclude that the process is within statistical control? Do the process goods or services meet design specifications? page 741 of Elementary Statistics, 10th Edition The methods of this chapter are intended to address the first question, but not the second.

Criteria for Determining When a Process Is Not Statistically Stable (Out of Statistical Control)
1. There is a pattern, trend, or cycle that is obviously not random. 2. There is a point lying beyond the upper or lower control limits. 3. Run of 8 Rule: There are eight consecutive points all above or all below the center line. page 742 of Elementary Statistics, 10th Edition

Additional Criteria Used
by Some Businesses There are 6 consecutive points all increasing or all decreasing. There are 14 consecutive points all alternating between up and down (such as up, down, up, down, and so on). Two out of three consecutive points are beyond control limits that are 2 standard deviations away from centerline. Four out of five consecutive points are beyond control limits that are 1 standard deviation away from the centerline. page 742 of Elementary Statistics, 10th Edition

Example: Statistical Process Control
Examine the R chart shown in the Minitab display for the preceding example and determine whether the process variation is within statistical control. 1. There is a pattern, trend, or cycle that is obviously not random: Going from left to right, there is a pattern of upward trend. 2. There is a point (the rightmost point) that lies above the upper control limit. page 742 of Elementary Statistics, 10th Edition

Example: Statistical Process Control
Examine the R chart shown in the Minitab display for the preceding example and determine whether the process variation is within statistical control. Interpretation: We conclude that the variation (not necessarily the mean) of the process is out of statistical control. Because the variation appears to be increasing with time, immediate corrective action must be taken to fix the variation among the altimeter errors. page of Elementary Statistics, 10th Edition

Control Chart for Monitoring
Means: The x Chart The x chart is a plot of the sample means and is used to monitor the center in a process. In addition to plotting the sample means, we include a centerline located at x, which denotes the mean of all sample means, as well as another line for the lower control limit and a third line for the upper control limit. page 743 of Elementary Statistics, 10th Edition

Control Chart for Monitoring where A2 is found in Table 14-2
Means: The x Chart Points plotted: Sample means Center line: x = mean of all sample means Upper Control Limit (UCL): x + A2R where A2 is found in Table 14-2 Lower Control Limit (LCL): x – A2R page 743 of Elementary Statistics, 10th Edition

Aircraft Altimeters Refer to the altimeter errors in Table Using samples of size n = 4 collected each working day, construct a control chart for x. Based on the control chart for x only, determine whether the process mean is within statistical control. Before plotting the 20 points corresponding to the 20 values of x, we must first find the value for the centerline and the values for the control limits. page 743 of Elementary Statistics, 10th Edition

Aircraft Altimeters Upper control limit: page 743 of Elementary Statistics, 10th Edition. Lower control limit:

Aircraft Altimeters Excel page 744 of Elementary Statistics, 10th Edition

Aircraft Altimeters Interpretation Examination of the control chart shows that the process mean is out of statistical control because at least one of the three out-of-control criteria is not satisfied. Specifically, the third criterion is not satisfied because there are eight (or more) consecutive points all below the center line. Also, there does appear to be a pattern of an upward trend. Again, immediate corrective action is required to fix the production process. page of Elementary Statistics, 10th Edition.

Control charts for variation and mean. Run charts determine if characteristics of a process have changed. R charts (or range charts) monitor the variation in a process. x charts monitor the center in a process.

Control Charts for Attributes
Section 14-3 Control Charts for Attributes

Key Concept This section presents a method for constructing a control chart to monitor the proportion p for some attribute, such as whether a service or manufactured item is defective or nonconforming. The control chart is interpreted by using the same three criteria from Section 14-2 to determine whether the process is statistically stable. page 748 of Elementary Statistics, 10th Edition

Control Charts for Attributes
These charts monitor the qualitative attributes of whether an item has some particular characteristic. In the previous section, the charts monitored the quantitative characteristics. The control chart for p (or p chart) is used to monitor the proportion p for some attribute. page 748 of Elementary Statistics, 10th Edition Samples must have the same size. There are ways to deal with samples of different sizes, but they are not dealt with in this text.

Notation p = pooled estimate of proportion of defective items in the process = total number of defects found among all items sampled total number of items sampled q = pooled estimate of the proportion of process items that are not defective = 1 – p n = size of each sample (not the number of samples) page 748 of Elementary Statistics, 10th Edition

Control Chart for p Center line: Upper control limit:
Lower control limit: page 749 of Elementary Statistics, 10th Edition (If the calculation for the lower control limit results in a negative value, use 0 instead. If the calculation for the upper control limit exceeds 1, use 1 instead.)

Example: Defective Aircraft Altimeters
The Altigauge Manufacturing Company produces altimeters in batches of 100, and each altimeter is tested and determined to be acceptable or defective. Listed below are the numbers of defective altimeters in successive batches of 100. Defects: total number of defects from all samples combined total number of altimeters sampled page 749 of Elementary Statistics, 10th Edition

Upper control limit: Lower control limit: page 749 of Elementary Statistics, 10th Edition Because the lower control limit is less than 0, we use 0 instead.

Interpretation We can interpret the control chart for p by considering the three out-of-control criteria listed in section 14-2. Using those criteria, we conclude that this process is out of statistical control for this reason: There appears to be an upward trend. The company should take immediate action to correct the increasing proportion of defects. page 750 of Elementary Statistics, 10th Edition

A control chart for attributes is a graph of proportions plotted sequentially over time. It includes a centerline, a lower control limit, and an upper control limit. The same three out-of-control criteria listed in Section 14-2 can be used.

by Mario F. Triola

Chapter 15 Projects, Procedures, Perspectives

Section 15-1 Projects

Key Concept This section includes suggestions for a final project in the introductory statistics course. One fantastic advantage of this course is that it deals with skills and concepts that can be applied immediately to the real world. After only one semester, students are able to conduct their own studies. page 758 of Elementary Statistics, 10th Edition Suggestions for successful projects are given in the margin of page 758

Projects Group/Individual Oral Report
Topics can be assigned to individuals, but group projects are particularly effective because they help develop the interpersonal skills that are so necessary in today’s working environment. Oral Report A 10- to 15-minute-long class presentation should involve all group members in a coordinated effort to clearly describe the important components of the study. page 758 of Elementary Statistics, 10th Edition

Projects Written Report
A brief written report should include the following: 1. List of data collected along with description of how the data were obtained. 2. Description of the method of analysis. 3. Relevant graphs and/or statistics, including STATDISK, Minitab, Excel, or TI-83/84 displays. 4. Statement of conclusion. 5. Reasons why the results might not be correct, along with a description of ways in which the study could be improved, given sufficient time and money. page 759 of Elementary Statistics, 10th Edition

Projects Large Classes or Online Classes: Posters or PowerPoint
Some classes are too large for individual projects or group projects with three or four or five students per group. Some online classes are not able to meet as a group. For such cases, reports of individual or small group projects can be presented through posters or PowerPoint presentations. Surveys can be an excellent source of data. Project Topic suggestions are listed in the text. page 758 of Elementary Statistics, 10th Edition

Created by Erin Hodgess, Houston, Texas
Section 15-2 Procedures page 763 of Elementary Statistics, 10th Edition Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

Key Concept This section describes a general procedure for conducting a statistical analysis of data. page 763 of Elementary Statistics, 10th Edition

Procedures Exploring, Comparing, Describing
After collecting data, address the following: 1. Center: Find the mean and median. 2. Variation: Find the range and standard deviation. 3. Distribution: Construct a histogram and a normal quantile plot. 4. Outliers: Identify any sample values that lie very far away from the vast majority of the others. 5. Time: Determine if the population is stable or if its characteristics are changing over time. page 763 of Elementary Statistics, 10th Edition

Procedures Inferences: Estimating Parameters and Hypothesis Testing
Here are some key questions that should be answered: What is the level of measurement (nominal, ordinal, interval, ratio) of the data? Does the study involve one, two, or more populations? What is the relevant parameter (mean, standard deviation, proportion)? Is the population standard deviation known? Is there reason to believe that the population is normally distributed? What is the basic question or issue to address? page of Elementary Statistics, 10th Edition

FIGURE 15-1 Selecting the Appropriate Procedure
Level of Measurement Interval or Ratio (such as heights, weights) What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations One Population Interval or Ratio (such as heights, weights) Two Populations More than Two Populations Chap. 12, 13-5 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations Claim or Parameter Mean One Population Variance Interval or Ratio (such as heights, weights) Two Populations More than Two Populations Chap. 12, 13-5 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations Claim or Parameter Mean One Population Variance Interval or Ratio (such as heights, weights) Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 Estimating with Confidence Interval: 7-5 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) Hypothesis Testing: 8-6 page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 Estimating with Confidence Interval: 7-5 One Population 13-2 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) Two Populations Hypothesis Testing: 8-6 More than Two Populations 13-5 page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 Estimating with Confidence Interval: 7-5 One Population 13-2 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) Independent: 13-4 Two Populations Hypothesis Testing: 8-6 Matched Pairs: 13-3 More than Two Populations 13-5 page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories)

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 Estimating with Confidence Interval: 7-5 One Population 13-2 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) Independent: 13-4 Two Populations Hypothesis Testing: 8-6 Matched Pairs: 13-3 More than Two Populations 13-5 page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories) Frequency Counts for Categories Proportions

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 Estimating with Confidence Interval: 7-5 One Population 13-2 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) Independent: 13-4 Two Populations Hypothesis Testing: 8-6 Matched Pairs: 13-3 More than Two Populations 13-5 Multinomial (one row) 11-2 page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories) Frequency Counts for Categories Contingency Table (multiple rows, columns) 11-3 Proportions

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 Estimating with Confidence Interval: 7-5 One Population 13-2 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) Independent: 13-4 Two Populations Hypothesis Testing: 8-6 Matched Pairs: 13-3 More than Two Populations 13-5 Multinomial (one row) 11-2 page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories) Frequency Counts for Categories Contingency Table (multiple rows, columns) 11-3 Two Populations: 9-2 Proportions One Population

Level of Measurement Number of Populations Claim or Parameter Inference Estimating with Confidence Interval: 7-3, 7-4 Mean One Population Variance Interval or Ratio (such as heights, weights) Hypothesis Testing with Large Sample: 8-4, 8-5 Two Populations Means: 9-3, 9-4 More than Two Populations Chap. 12, 13-5 Variances: 9-5 Correlation, Regression Chap. 10, 13-6 Estimating with Confidence Interval: 7-5 One Population 13-2 What is the level of measurement of the data? 1-2 Ordinal (such as data consisting of ranks) Independent: 13-4 Two Populations Hypothesis Testing: 8-6 Matched Pairs: 13-3 More than Two Populations 13-5 Multinomial (one row) 11-2 page 764 of Elementary Statistics, 10th Edition Nominal (data consisting of proportions or frequency counts for different categories) Frequency Counts for Categories Contingency Table (multiple rows, columns) 11-3 Estimating Proportion with Confidence Interval: 7-2 Two Populations: 9-2 Hypothesis Testing: 8-3, 13-2 Proportions One Population

Created by Erin Hodgess, Houston, Texas
Section 15-3 Perspectives Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

Key Concept No single introductory statistics course can transform anyone into an expert statistician. Know that professional help is available from expert statisticians, and this introductory statistics course will help you in discussions with one of these experts. page 765 of Elementary Statistics, 10th Edition

Perspective Successful completion of an introductory statistics course results in benefits far beyond the attainment of credit toward a college degree. Improved job marketability Ability to critically analyze reports in media and journals Understanding of the basic concepts of probability and chance Knowledge of importance of statistical procedures page 765 of Elementary Statistics, 10th Edition

Perspective Remember that expert ability in analyzing statistics is of little value if good sampling techniques are not employed to develop the sample. Although computers and calculators are good at yielding results, careful interpretation of the results are required. Successful Completion of an introductory statistics course can enable students to grow as individuals and professionals and become people who are truly educated. page of Elementary Statistics, 10th Edition

Elementary Statistics Tenth Edition

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