Chapter 1: Statistics Note: The textbook illustrates statistical procedures using MINITAB, EXCEL 97, and the TI-83.

Chapter Goals Create an initial image of the field of statistics.
Introduce several basic vocabulary words used in studying statistics: population, variable, statistic. Learn how to obtain sample data.

1.1: What is Statistics? Statistics: The science of collecting, describing, and interpreting data. Two areas of statistics: Descriptive Statistics: collection, presentation, and description of sample data. Inferential Statistics: making decisions and drawing conclusions about populations.

Example: A recent study examined the math and verbal SAT scores of high school seniors across the country. Which of the following statements are descriptive in nature and which are inferential. The mean math SAT score was 492. The mean verbal SAT score was 475. Students in the Northeast scored higher in math but lower in verbal. 80% of all students taking the exam were headed for college. 32% of the students scored above 610 on the verbal SAT. The math SAT scores are higher than they were 10 years ago.

1.2 Introduction to Basic Terms
Population: A collection, or set, of individuals or objects or events whose properties are to be analyzed. Two kinds of populations: finite or infinite. Sample: A subset of the population.

Variable: A characteristic about each individual element of a population or sample.
Data (singular): The value of the variable associated with one element of a population or sample. This value may be a number, a word, or a symbol. Data (plural): The set of values collected for the variable from each of the elements belonging to the sample. Experiment: A planned activity whose results yield a set of data. Parameter: A numerical value summarizing all the data of an entire population. Statistic: A numerical value summarizing the sample data.

Example: A college dean is interested in learning about the average age of faculty. Identify the basic terms in this situation. The population is the age of all faculty members at the college. A sample is any subset of that population. For example, we might select 10 faculty members and determine their age. The variable is the “age” of each faculty member. One data would be the age of a specific faculty member. The data would be the set of values in the sample. The experiment would be the method used to select the ages forming the sample and determining the actual age of each faculty member in the sample. The parameter of interest is the “average” age of all faculty at the college. The statistic is the “average” age for all faculty in the sample.

Two kinds of variables:
Qualitative, or Attribute, or Categorical, Variable: A variable that categorizes or describes an element of a population. Note: Arithmetic operations, such as addition and averaging, are not meaningful for data resulting from a qualitative variable. Quantitative, or Numerical, Variable: A variable that quantifies an element of a population. Note: Arithmetic operations such as addition and averaging, are meaningful for data resulting from a quantitative variable.

Example: Identify each of the following examples as attribute (qualitative) or numerical (quantitative) variables. 1. The residence hall for each student in a statistics class. (Attribute) 2. The amount of gasoline pumped by the next 10 customers at the local Unimart. (Numerical) 3. The amount of radon in the basement of each of 25 homes in a new development. (Numerical) 4. The color of the baseball cap worn by each of 20 students. (Attribute) 5. The length of time to complete a mathematics homework assignment. (Numerical) 6. The state in which each truck is registered when stopped and inspected at a weigh station. (Attribute)

Qualitative and quantitative variables may be further subdivided:
Nominal Qualitative Ordinal Variable Discrete Quantitative Continuous

Nominal Variable: A qualitative variable that categorizes (or describes, or names) an element of a population. Ordinal Variable: A qualitative variable that incorporates an ordered position, or ranking. Discrete Variable: A quantitative variable that can assume a countable number of values. Intuitively, a discrete variable can assume values corresponding to isolated points along a line interval. That is, there is a gap between any two values. Continuous Variable: A quantitative variable that can assume an uncountable number of values. Intuitively, a continuous variable can assume any value along a line interval, including every possible value between any two values.

Note: 1. In many cases, a discrete and continuous variable may be distinguished by determining whether the variables are related to a count or a measurement. Discrete variables are usually associated with counting. If the variable cannot be further subdivided, it is a clue that you are probably dealing with a discrete variable. Continuous variables are usually associated with measurements. The values of discrete variables are only limited by your ability to measure them.

Example: Identify each of the following as examples of qualitative or numerical variables:
1. The temperature in Barrow, Alaska at 12:00 pm on any given day. 2. The make of automobile driven by each faculty member. 3. Whether or not a 6 volt lantern battery is defective. 4. The weight of a lead pencil. 5. The length of time billed for a long distance telephone call. 6. The brand of cereal children eat for breakfast. 7. The type of book taken out of the library by an adult.

Example: Identify each of the following as examples of (1) nominal, (2) ordinal, (3) discrete, or (4) continuous variables: 1. The length of time until a pain reliever begins to work. 2. The number of chocolate chips in a cookie. 3. The number of colors used in a statistics textbook. 4. The brand of refrigerator in a home. 5. The overall satisfaction rating of a new car. 6. The number of files on a computer’s hard disk. 7. The pH level of the water in a swimming pool. 8. The number of staples in a stapler.

1.3: Measure and Variability
No matter what the response variable: there will always be variability in the data. One of the primary objectives of statistics: measuring and characterizing variability. Controlling (or reducing) variability in a manufacturing process: statistical process control.

Example: A supplier fills cans of soda marked 12 ounces
Example: A supplier fills cans of soda marked 12 ounces. How much soda does each can really contain? It is very unlikely any one can contains exactly 12 ounces of soda. There is variability in any process. Some cans contain a little more than 12 ounces, and some cans contain a little less. On the average, there are 12 ounces in each can. The supplier hopes there is little variability in the process, that most cans contain close to 12 ounces of soda.

1.4: Data Collection First problem a statistician faces: how to obtain the data. It is important to obtain good, or representative, data. Inferences are made based on statistics obtained from the data. Inferences can only be as good as the data.

Biased Sampling Method: A sampling method that produces data which systematically differs from the sampled population. An unbiased sampling method is one that is not biased. Sampling methods that often result in biased samples: 1. Convenience sample: sample selected from elements of a population that are easily accessible. 2. Volunteer sample: sample collected from those elements of the population which chose to contribute the needed information on their own initiative.

Process of data collection:
1. Define the objectives of the survey or experiment. Example: Estimate the average life of an electronic component. 2. Define the variable and population of interest. Example: Length of time for anesthesia to wear off after surgery. 3. Defining the data-collection and data-measuring schemes. This includes sampling procedures, sample size, and the data-measuring device (questionnaire, scale, ruler, etc.). 4. Determine the appropriate descriptive or inferential data-analysis techniques.

Methods used to collect data:
Experiment: The investigator controls or modifies the environment and observes the effect on the variable under study. Survey: Data are obtained by sampling some of the population of interest. The investigator does not modify the environment. Census: A 100% survey. Every element of the population is listed. Seldom used: difficult and time-consuming to compile, and expensive.

Sampling Frame: A list of the elements belonging to the population from which the sample will be drawn. Note: It is important that the sampling frame be representative of the population. Sample Design: The process of selecting sample elements from the sampling frame. Note: There are many different types of sample designs. Usually they all fit into two categories: judgment samples and probability samples.

Judgment Samples: Samples that are selected on the basis of being “typical.”
Items are selected that are representative of the population. The validity of the results from a judgment sample reflects the soundness of the collector’s judgment. Probability Samples: Samples in which the elements to be selected are drawn on the basis of probability. Each element in a population has a certain probability of being selected as part of the sample.

Random Samples: A sample selected in such a way that every element in the population has a equal probability of being chosen. Equivalently, all samples of size n have an equal chance of being selected. Random samples are obtained either by sampling with replacement from a finite population or by sampling without replacement from an infinite population. Note: 1. Inherent in the concept of randomness: the next result (or occurrence) is not predictable. 2. Proper procedure for selecting a random sample: use a random number generator or a table of random numbers.

Example: An employer is interested in the time it takes each employee to commute to work each morning. A random sample of 35 employees will be selected and their commuting time will be recorded. There are 2712 employees. Each employee is numbered: 0001, 0002, 0003, etc. up to 2712. Using four-digit random numbers, a sample is identified: 1315, 0987, 1125, etc.

Systematic Sample: A sample in which every kth item of the sampling frame is selected, starting from the first element which is randomly selected from the first k elements. Note: The systematic technique is easy to execute. However, it has some inherent dangers when the sampling frame is repetitive or cyclical in nature. In these situations the results may not approximate a simple random sample. Stratified Random Sample: A sample obtained by stratifying the sampling frame and then selecting a fixed number of items from each of the strata by means of a simple random sampling technique.

Proportional Sample (or Quota Sample): A sample obtained by stratifying the sampling frame and then selecting a number of items in proportion to the size of the strata (or by quota) from each strata by means of a simple random sampling technique. Cluster Sample: A sample obtained by stratifying the sampling frame and then selecting some or all of the items from some of, but not all, the strata.

1.5: Comparison of Probability and Statistics
Probability: Properties of the population are assumed known. Answer questions about the sample based on these properties. Statistics: Use information in the sample to draw a conclusion about the population.

Example: A jar of M&M’s contains 100 candy pieces, 15 are red
Example: A jar of M&M’s contains 100 candy pieces, 15 are red. A handful of 10 is selected. Probability question: What is the probability that 3 of the 10 selected are red? Example: A handful of 10 M&M’s is selected from a jar containing 1000 candy pieces. Three M&M’s in the handful are red. Statistics question: What is the proportion of red M&M’s in the entire jar?

1.6: Statistics and the Technology
The electronic technology has had a tremendous effect on the field of statistics. Many statistical techniques are repetitive in nature: computers and calculators are good at this. Lots of statistical software packages: MINITAB, SYSTAT, STATA, SAS, Statgraphics, SPSS, and calculators.

Remember: Responsible use of statistical methodology is very important
Remember: Responsible use of statistical methodology is very important. The burden is on the user to ensure that the appropriate methods are correctly applied and that accurate conclusions are drawn and communicated to others.

Chapter 2 ~ Descriptive Analysis & Presentation of Single-Variable Data
Mean: inches Median: inches Range: 42 inches Variance: Standard deviation: inches Minimum: 36 inches Maximum: 78 inches First quartile: inches Third quartile: inches Count: 58 bears Sum: inches 10 20 30 40 50 60 70 80 Frequency Length in Inches Black Bears

Chapter Goals Learn how to present and describe sets of data
Learn measures of central tendency, measures of dispersion (spread), measures of position, and types of distributions Learn how to interpret findings so that we know what the data is telling us about the sampled population

2.1 ~ Graphic Presentation of Data
Use initial exploratory data-analysis techniques to produce a pictorial representation of the data Resulting displays reveal patterns of behavior of the variable being studied The method used is determined by the type of data and the idea to be presented No single correct answer when constructing a graphic display

Circle Graphs & Bar Graphs
Circle Graphs and Bar Graphs: Graphs that are used to summarize attribute data Circle graphs (pie diagrams) show the amount of data that belongs to each category as a proportional part of a circle Bar graphs show the amount of data that belongs to each category as proportionally sized rectangular areas

Example Example: The table below lists the number of automobiles sold last week by day for a local dealership. Describe the data using a circle graph and a bar graph: Day Number Sold Monday 15 Tuesday 23 Wednesday 35 Thursday 11 Friday 12 Saturday 42

Circle Graph Solution Automobiles Sold Last Week

Bar Graph Solution Automobiles Sold Last Week Frequency

Pareto Diagram Pareto Diagram: A bar graph with the bars arranged from the most numerous category to the least numerous category. It includes a line graph displaying the cumulative percentages and counts for the bars. Notes: Used to identify the number and type of defects that happen within a product or service Separates the “vital few” from the “trivial many” The Pareto diagram is often used in quality control applications

Example Example: The final daily inspection defect report for a cabinet manufacturer is given in the table below: Defect Number Dent 5 Stain 12 Blemish 43 Chip 25 Scratch 40 Others 10 1) Construct a Pareto diagram for this defect report. Management has given the cabinet production line the goal of reducing their defects by 50%. 2) What two defects should they give special attention to in working toward this goal?

Daily Defect Inspection Report
Solutions 1 4 2 8 6 Count Percent Blemish Scratch Chip Stain Others Dent Defect: 43 40 25 12 10 5 31.9 29.6 18.5 8.9 7.4 3.7 Cum% 61.5 80.0 88.9 96.3 100.0 Daily Defect Inspection Report 1) 2) The production line should try to eliminate blemishes and scratches. This would cut defects by more than 50%.

Key Definitions Quantitative Data: One reason for constructing a graph of quantitative data is to examine the distribution - is the data compact, spread out, skewed, symmetric, etc. Distribution: The pattern of variability displayed by the data of a variable. The distribution displays the frequency of each value of the variable. Dotplot Display: Displays the data of a sample by representing each piece of data with a dot positioned along a scale. This scale can be either horizontal or vertical. The frequency of the values is represented along the other scale.

Example : . . .:. . ..: :.::::::.. .::. ... . : . . . :. . .
Example: A random sample of the lifetime (in years) of 50 home washing machines is given below: 2.5 8.9 12.2 4.1 18.1 1.6 16.9 3.5 0.4 2.6 2.2 4.0 4.5 6.4 2.9 3.3 4.4 9.2 0.9 14.5 7.2 5.2 1.8 1.5 0.7 3.7 4.2 6.9 15.3 21.8 17.8 7.3 6.8 7.0 18.3 8.5 1.4 7.4 4.7 10.4 3.6 The figure below is a dotplot for the 50 lifetimes: . : : ..: :.::::::.. .:: : : Note: Notice how the data is “bunched” near the lower extreme and more “spread out” near the higher extreme

Stem & Leaf Display Background:
The stem-and-leaf display has become very popular for summarizing numerical data It is a combination of graphing and sorting The actual data is part of the graph Well-suited for computers Stem-and-Leaf Display: Pictures the data of a sample using the actual digits that make up the data values. Each numerical data is divided into two parts: The leading digit(s) becomes the stem, and the trailing digit(s) becomes the leaf. The stems are located along the main axis, and a leaf for each piece of data is located so as to display the distribution of the data.

Example Example: A city police officer, using radar, checked the speed of cars as they were traveling down the main street in town. Construct a stem-and-leaf plot for this data: Solution: All the speeds are in the 10s, 20s, 30s, 40s, and 50s. Use the first digit of each speed as the stem and the second digit as the leaf. Draw a vertical line and list the stems, in order to the left of the line. Place each leaf on its stem: place the trailing digit on the right side of the vertical line opposite its corresponding leading digit.

Example 20 Speeds --------------------------------------- 1 | 6 9
1 | 6 9 2 | 3 | 4 | 5 | 5 The speeds are centered around the 30s Note: The display could be constructed so that only five possible values (instead of ten) could fall in each stem. What would the stems look like? Would there be a difference in appearance?

Remember! 1. It is fairly typical of many variables to display a distribution that is concentrated (mounded) about a central value and then in some manner be dispersed in both directions. (Why?) 2. A display that indicates two “mounds” may really be two overlapping distributions 3. A back-to-back stem-and-leaf display makes it possible to compare two distributions graphically 4. A side-by-side dotplot is also useful for comparing two distributions

2.2 ~ Frequency Distributions & Histograms
Stem-and-leaf plots often present adequate summaries, but they can get very big, very fast Need other techniques for summarizing data Frequency distributions and histograms are used to summarize large data sets

Frequency Distributions
Frequency Distribution: A listing, often expressed in chart form, that pairs each value of a variable with its frequency Ungrouped Frequency Distribution: Each value of x in the distribution stands alone Grouped Frequency Distribution: Group the values into a set of classes 1. A table that summarizes data by classes, or class intervals 2. In a typical grouped frequency distribution, there are usually 5-12 classes of equal width 3. The table may contain columns for class number, class interval, tally (if constructing by hand), frequency, relative frequency, cumulative relative frequency, and class midpoint 4. In an ungrouped frequency distribution each class consists of a single value

Frequency Distribution
Guidelines for constructing a frequency distribution: 1. All classes should be of the same width 2. Classes should be set up so that they do not overlap and so that each piece of data belongs to exactly one class 3. For problems in the text, 5-12 classes are most desirable. The square root of n is a reasonable guideline for the number of classes if n is less than 150. 4. Use a system that takes advantage of a number pattern, to guarantee accuracy 5. If possible, an even class width is often advantageous

Frequency Distributions
Procedure for constructing a frequency distribution: 1. Identify the high (H) and low (L) scores. Find the range. Range = H - L 2. Select a number of classes and a class width so that the product is a bit larger than the range 3. Pick a starting point a little smaller than L. Count from L by the width to obtain the class boundaries. Observations that fall on class boundaries are placed into the class interval to the right.

Example Example: The hemoglobin test, a blood test given to diabetics during their periodic checkups, indicates the level of control of blood sugar during the past two to three months. The data in the table below was obtained for 40 different diabetics at a university clinic that treats diabetic patients: 1) Construct a grouped frequency distribution using the classes <4.7, <5.7, <6.7, etc. 2) Which class has the highest frequency?

Solutions 1) Class Frequency Relative Cumulative Class Boundaries f Frequency Rel. Frequency Midpoint, x 3.7 - < 4.7 - < 5.7 - < 6.7 - < 7.7 - < 8.7 - < 2) The class <6.7 has the highest frequency. The frequency is 16 and the relative frequency is 0.40

Histogram Histogram: A bar graph representing a frequency distribution of a quantitative variable. A histogram is made up of the following components: 1. A title, which identifies the population of interest 2. A vertical scale, which identifies the frequencies in the various classes 3. A horizontal scale, which identifies the variable x. Values for the class boundaries or class midpoints may be labeled along the x-axis. Use whichever method of labeling the axis best presents the variable. Notes: The relative frequency is sometimes used on the vertical scale It is possible to create a histogram based on class midpoints

Example Example: Construct a histogram for the blood test results given in the previous example 9 . 2 8 7 6 5 4 1 Frequency Blood Test Solution: The Hemoglobin Test

Example Example: A recent survey of Roman Catholic nuns summarized their ages in the table below. Construct a histogram for this age data: Age Frequency Class Midpoint 20 up to 30 up to 40 up to 50 up to 60 up to 70 up to 80 up to

Solution 8 5 7 6 4 3 2 1 Frequency Age Roman Catholic Nuns

Terms Used to Describe Histograms
Symmetrical: Both sides of the distribution are identical mirror images. There is a line of symmetry. Uniform (Rectangular): Every value appears with equal frequency Skewed: One tail is stretched out longer than the other. The direction of skewness is on the side of the longer tail. (Positively skewed vs. negatively skewed) J-Shaped: There is no tail on the side of the class with the highest frequency Bimodal: The two largest classes are separated by one or more classes. Often implies two populations are sampled. Normal: A symmetrical distribution is mounded about the mean and becomes sparse at the extremes

Important Reminders The mode is the value that occurs with greatest frequency (discussed in Section 2.3) The modal class is the class with the greatest frequency A bimodal distribution has two high-frequency classes separated by classes with lower frequencies Graphical representations of data should include a descriptive, meaningful title and proper identification of the vertical and horizontal scales

Cumulative Frequency Distribution
Cumulative Frequency Distribution: A frequency distribution that pairs cumulative frequencies with values of the variable The cumulative frequency for any given class is the sum of the frequency for that class and the frequencies of all classes of smaller values The cumulative relative frequency for any given class is the sum of the relative frequency for that class and the relative frequencies of all classes of smaller values

Example Example: A computer science aptitude test was given to students. The table below summarizes the data: Class Relative Cumulative Cumulative Boundaries Frequency Frequency Frequency Rel. Frequency 0 up to 4 up to 8 up to 12 up to 16 up to 20 up to 24 up to

Ogive Ogive: A line graph of a cumulative frequency or cumulative relative frequency distribution. An ogive has the following components: 1. A title, which identifies the population or sample 2. A vertical scale, which identifies either the cumulative frequencies or the cumulative relative frequencies 3. A horizontal scale, which identifies the upper class boundaries. Until the upper boundary of a class has been reached, you cannot be sure you have accumulated all the data in the class. Therefore, the horizontal scale for an ogive is always based on the upper class boundaries. Note: Every ogive starts on the left with a relative frequency of zero at the lower class boundary of the first class and ends on the right with a relative frequency of 100% at the upper class boundary of the last class.

Example Example: The graph below is an ogive using cumulative relative frequencies for the computer science aptitude data: Cumulative Relative Frequency 4 8 12 16 20 24 28 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Test Score Computer Science Aptitude Test

2.3 ~ Measures of Central Tendency
Numerical values used to locate the middle of a set of data, or where the data is clustered The term average is often associated with all measures of central tendency

Mean Mean: The type of average with which you are probably most familiar. The mean is the sum of all the values divided by the total number of values, n: x n i = + å 1 2 ( ) The population mean, , (lowercase mu, Greek alphabet), is the mean of all x values for the entire population Notes: We usually cannot measure  but would like to estimate its value A physical representation: the mean is the value that balances the weights on the number line

Example Example: The following data represents the number of accidents in each of the last 8 years at a dangerous intersection. Find the mean number of accidents: 8, 9, 3, 5, 2, 6, 4, 5: x = + 1 8 9 3 5 2 6 4 25 ( ) . Solution: In the data above, change 6 to 26: x = + 1 8 9 3 5 2 26 4 7 75 ( ) . Solution: Note: The mean can be greatly influenced by outliers

Median Median: The value of the data that occupies the middle position when the data are ranked in order according to size Notes: Denoted by “x tilde”: The population median,  (uppercase mu, Greek alphabet), is the data value in the middle position of the entire population To find the median: 1. Rank the data 2. Determine the depth of the median: 3. Determine the value of the median

Example Example: Find the median for the set of data:
{4, 8, 3, 8, 2, 9, 2, 11, 3} Solution: 1. Rank the data: 2, 2, 3, 3, 4, 8, 8, 9, 11 2. Find the depth: 3. The median is the fifth number from either end in the ranked data: Suppose the data set is {4, 8, 3, 8, 2, 9, 2, 11, 3, 15}: 1. Rank the data: 2, 2, 3, 3, 4, 8, 8, 9, 11, 15 2. Find the depth: 3. The median is halfway between the fifth and sixth observations:

Mode & Midrange Mode: The mode is the value of x that occurs most frequently Note: If two or more values in a sample are tied for the highest frequency (number of occurrences), there is no mode Midrange: The number exactly midway between a lowest value data L and a highest value data H. It is found by averaging the low and the high values:

Example Example: Consider the data set {12.7, 27.1, 35.6, 44.2, 18.0}
Midrange = + L H 2 12 7 44 28 45 . When rounding off an answer, a common rule-of-thumb is to keep one more decimal place in the answer than was present in the original data To avoid round-off buildup, round off only the final answer, not intermediate steps Notes:

2.4 ~ Measures of Dispersion
Measures of central tendency alone cannot completely characterize a set of data. Two very different data sets may have similar measures of central tendency. Measures of dispersion are used to describe the spread, or variability, of a distribution Common measures of dispersion: range, variance, and standard deviation

Range Range: The difference in value between the highest-valued (H) and the lowest-valued (L) pieces of data: Other measures of dispersion are based on the following quantity Deviation from the Mean: A deviation from the mean, , is the difference between the value of x and the mean

Example Example: Consider the sample {12, 23, 17, 15, 18}. Find 1) the range and 2) each deviation from the mean. Solutions: 1) Data Deviation from Mean _________________________ 12 23 17 15 18 2) -5 6 -2 1

Mean Absolute Deviation
Note: (Always!) å = - x ) ( Mean Absolute Deviation: The mean of the absolute values of the deviations from the mean: å - = x | 1 deviation absolute Mean n 8 . 2 5 14 ) 1 6 ( | = + - å x n For the previous example:

Sample Variance & Standard Deviation
Sample Variance: The sample variance, s2, is the mean of the squared deviations, calculated using n - 1 as the divisor: s n x 2 1 = - å ( ) where n is the sample size ( ) SS x n = - å 2 1 Note: The numerator for the sample variance is called the sum of squares for x, denoted SS(x): where Standard Deviation: The standard deviation of a sample, s, is the positive square root of the variance:

Example Example: Find the 1) variance and 2) standard deviation for the data {5, 7, 1, 3, 8}: x - ( ) 2 0.2 2.2 -3.8 -1.8 3.2 0.04 4.84 14.44 3.24 10.24 5 7 1 3 8 24 32.08 Sum: Solutions: = + 4 . First: 2 . 8 ) 32 ( 4 1 = s 1) 86 . 2 8 = s 2)

Notes The shortcut formula for the sample variance:
The unit of measure for the standard deviation is the same as the unit of measure for the data

2.5 ~ Mean & Standard Deviation of Frequency Distribution
If the data is given in the form of a frequency distribution, we need to make a few changes to the formulas for the mean, variance, and standard deviation Complete the extension table in order to find these summary statistics

To Calculate In order to calculate the mean, variance, and standard deviation for data: 1. In an ungrouped frequency distribution, use the frequency of occurrence, f, of each observation 2. In a grouped frequency distribution, we use the frequency of occurrence associated with each class midpoint:

Example Example: A survey of students in the first grade at a local school asked for the number of brothers and/or sisters for each child. The results are summarized in the table below. Find 1) the mean, 2) the variance, and 3) the standard deviation: Solutions: First: 17 23 5 2 46 20 10 1 4 62 93 Sum: 15 92 80 50 239 2) 1) 3)

TI-83 Calculations When dealing with a grouped frequency distribution, use the following technique: Input the class midpoints or data values into L1 and the frequencies into L2; then continue with Highlight: L3 Enter: L3 = L1*L2 Highlight: L4 Enter: L4 = L1*L3 Highlight: L5(1) (first position in L5 column)

TI-83 Calculations Enter: L5(1) = sum(L2) (Σf) L5(2) = sum(L3) (Σxf)
To find sum use 2nd “List”>Math>5:sum( L5(2) = sum(L3) (Σxf) L5(3) = sum(L4) (Σx2f) L5(4) = L5(2)/L5(1) to find the mean L5(5) = (L5(3)-(L5(2))2/L5(1))/(L5(1)-1) to find the variance L5(6) = 2nd √ (L5(5) to find the standard deviation Let’s work problem as an example!

Problem 2.108 Find the mean and the variance for this grouped frequency distribution: Class Boundaries f 2 – 6 – 10 – 14 – 18 – Step 1: Enter the midpoints into L1 Step 2: Enter the frequencies into L2

Problem 2.108 Highlight L3 and enter L1 * L2
Highlight L5(1) and enter Sum(L2)

Problem 2.108 Highlight L5(2) and enter Sum(L3)
Highlight L5(4) and enter L5(2)/L5(1) Highlight L5(5) and enter (L5(3)-(L5(2))2/L5(1))/(L5(1)-1) Finally highlight L5(6) and enter 2nd √(L5(5))

Problem 2.73 Mean 9.77 9.80 -0.03 Median 9.65 9.78 -0.06 Maximum 13.65
Runs At Home Runs Away Difference Mean 9.77 9.80 -0.03 Median 9.65 9.78 -0.06 Maximum 13.65 11.06 4.89 Minimum 7.64 8.67 -1.74 Midrange 10.65 9.87 1.58

Problem 2.73 cont. Teams playing the Rockies at Coors Field generated the maximum number of runs scored (13.65). On the other hand, while playing their opponents away, the Rockies and their opponents, were able to generate only 8.76 runs, which ranked second from the bottom. Collectively, these two performances produced the greatest spread (4.89) by a considerable margin between runs scored at home and runs scored away by any stadium/team combination in the major leagues. This unusually large value inflates the midrange difference. It appears the playing conditions at Coors Field, therefore, are more responsible for producing the higher combined scores than the strength of either the Rockies’ or their opponents’ bats or any weakness of the pitching staffs.

Problem 2.75 ∑ x need to be 500, therefore need any three numbers that total 330. Need two numbers smaller than 70 and one larger Need multiple 87’s Need any two numbers that total 140 for the extreme values where one is 100 or larger Need two numbers smaller than 70 and one larger than 70 so their total is 330 Need two numbers of 87 and a third number large enough so that the total of all five is 500. Mean equal to 100 requires the five data to total 500 and the midrange of 70 requires the total of L and H to be 140; 40, __, 70, ___, 100; that is a sum of 210, meaning the other two data must total One of the last two numbers must be larger than 145, which would then become H and change the midrange. Impossible. There must be two 87’s in order to have a mode, and there can only be two data larger than 70 in order for 70 to be the median. , 70, 87, 87, 100; Impossible

Problem 2.83 Range = H – L = 9 – 2 = 7 1st: find the mean: 6
s2 = ∑(x-xbar)2/(n-1) = 34/4 = 8.5 s = √s2 = √8.5 = 2.9 x x – xbar (x – xbar)2 2 -4 16 4 -2 7 1 8 9 3 ∑ 34

2.6 ~ Measures of Position Measures of position are used to describe the relative location of an observation Quartiles and percentiles are two of the most popular measures of position An additional measure of central tendency, the midquartile, is defined using quartiles Quartiles are part of the 5-number summary

Ranked data, increasing order
Quartiles Quartiles: Values of the variable that divide the ranked data into quarters; each set of data has three quartiles 1. The first quartile, Q1, is a number such that at most 25% of the data are smaller in value than Q1 and at most 75% are larger 2. The second quartile, Q2, is the median 3. The third quartile, Q3, is a number such that at most 75% of the data are smaller in value than Q3 and at most 25% are larger Ranked data, increasing order

Percentiles Percentiles: Values of the variable that divide a set of ranked data into 100 equal subsets; each set of data has 99 percentiles. The kth percentile, Pk, is a value such that at most k% of the data is smaller in value than Pk and at most (100 - k)% of the data is larger. ~ x Q P = 2 50 Notes: The 1st quartile and the 25th percentile are the same: Q1 = P25 The median, the 2nd quartile, and the 50th percentile are all the same:

Finding Pk (and Quartiles)
Procedure for finding Pk (and quartiles): 1. Rank the n observations, lowest to highest 2. Compute A = (nk)/100 3. If A is an integer: d(Pk) = A.5 (depth) Pk is halfway between the value of the data in the Ath position and the value of the next data If A is a fraction: d(Pk) = B, the next larger integer Pk is the value of the data in the Bth position

Example Example: The following data represents the pH levels of a random sample of swimming pools in a California town. Find: 1) the first quartile, 2) the third quartile, and 3) the 37th percentile: Solutions: 1) k = 25: (20) (25) / 100 = 5, depth = 5.5, Q1 = 6 2) k = 75: (20) (75) / 100 = 15, depth = 15.5, Q3 = 6.95 3) k = 37: (20) (37) / 100 = 7.4, depth = 8, P37 = 6.2

Midquartile Midquartile: The numerical value midway between the first and third quartile: Example: Find the midquartile for the 20 pH values in the previous example: 475 . 6 2 95 12 e midquartil 3 1 = + Q Note: The mean, median, midrange, and midquartile are all measures of central tendency. They are not necessarily equal. Can you think of an example when they would be the same value?

5-Number Summary 5-Number Summary: The 5-number summary is composed of: 1. L, the smallest value in the data set 2. Q1, the first quartile (also P25) 3. , the median (also P50 and 2nd quartile) 4. Q3, the third quartile (also P75) 5. H, the largest value in the data set Notes: The 5-number summary indicates how much the data is spread out in each quarter The interquartile range is the difference between the first and third quartiles. It is the range of the middle 50% of the data

Box-and-Whisker Display
Box-and-Whisker Display: A graphic representation of the 5-number summary: The five numerical values (smallest, first quartile, median, third quartile, and largest) are located on a scale, either vertical or horizontal The box is used to depict the middle half of the data that lies between the two quartiles The whiskers are line segments used to depict the other half of the data One line segment represents the quarter of the data that is smaller in value than the first quartile The second line segment represents the quarter of the data that is larger in value that the third quartile

Example Example: A random sample of students in a sixth grade class was selected. Their weights are given in the table below. Find the 5-number summary for this data and construct a boxplot: Solution:

Boxplot for Weight Data
Weights from Sixth Grade Class 1 9 8 7 6 Weight

z-Score z-Score: The position a particular value of x has relative to the mean, measured in standard deviations. The z-score is found by the formula: Notes: Typically, the calculated value of z is rounded to the nearest hundredth The z-score measures the number of standard deviations above/below, or away from, the mean z-scores typically range from to +3.00 z-scores may be used to make comparisons of raw scores

Example Example: A certain data set has mean 35.6 and standard deviation Find the z-scores for 46 and 33: Solutions: 46 is 1.46 standard deviations above the mean z x s = - 33 35 6 7 1 37 . 33 is 0.37 standard deviations below the mean.

2.7 ~ Interpreting & Understanding Standard Deviation
Standard deviation is a measure of variability, or spread Two rules for describing data rely on the standard deviation: Empirical rule: applies to a variable that is normally distributed Chebyshev’s theorem: applies to any distribution

Empirical Rule Empirical Rule: If a variable is normally distributed, then: 1. Approximately 68% of the observations lie within 1 standard deviation of the mean 2. Approximately 95% of the observations lie within 2 standard deviations of the mean 3. Approximately 99.7% of the observations lie within 3 standard deviations of the mean Notes: The empirical rule is more informative than Chebyshev’s theorem since we know more about the distribution (normally distributed) Also applies to populations Can be used to determine if a distribution is normally distributed

Illustration of the Empirical Rule
68% 95% 99.7%

Example Example: A random sample of plum tomatoes was selected from a local grocery store and their weights recorded. The mean weight was 6.5 ounces with a standard deviation of 0.4 ounces. If the weights are normally distributed: 1) What percentage of weights fall between 5.7 and 7.3? 2) What percentage of weights fall above 7.7? Solutions: ( , ) . (0. ), )) x s - + = 2 6 5 4 7 3 Approximat ely 95% of the weigh ts fall be tween 5.7 and 7.3 1) ( , ) . (0. ), )) x s - + = 3 6 5 4 7 Approximat ely 99.7% of the wei ghts fall between 5. 3 and 7.7 ely 0.3% of the weigh ts fall out side (5.3, 7.7) ely (0.3 / 2) =0.15% of t he weights fall abov e 7.7

A Note about the Empirical Rule
Note: The empirical rule may be used to determine whether or not a set of data is approximately normally distributed 1. Find the mean and standard deviation for the data 2. Compute the actual proportion of data within 1, 2, and 3 standard deviations from the mean 3. Compare these actual proportions with those given by the empirical rule 4. If the proportions found are reasonably close to those of the empirical rule, then the data is approximately normally distributed

Chebyshev’s Theorem Chebyshev’s Theorem: The proportion of any distribution that lies within k standard deviations of the mean is at least 1 - (1/k2), where k is any positive number larger than 1. This theorem applies to all distributions of data. Illustration:

Important Reminders! Chebyshev’s theorem is very conservative and holds for any distribution of data Chebyshev’s theorem also applies to any population The two most common values used to describe a distribution of data are k = 2, 3 The table below lists some values for k and 1 - (1/k2):

Example Example: At the close of trading, a random sample of 35 technology stocks was selected. The mean selling price was and the standard deviation was Use Chebyshev’s theorem (with k = 2, 3) to describe the distribution. Solutions: Using k=2: At least 75% of the observations lie within 2 standard deviations of the mean: Using k=3: At least 89% of the observations lie within 3 standard deviations of the mean:

2.8 ~ The Art of Statistical Deception
Good Arithmetic, Bad Statistics Misleading Graphs Insufficient Information

Good Arithmetic, Bad Statistics
The mean can be greatly influenced by outliers Example: The mean salary for all NBA players is $15.5 million Misleading graphs: 1. The frequency scale should start at zero to present a complete picture. Graphs that do not start at zero are used to save space. 2. Graphs that start at zero emphasize the size of the numbers involved 3. Graphs that are chopped off emphasize variation

Flight Cancellations 2 1 9 98 6 3 5 Number of Cancellations Year

Flight Cancellations 3 5 4 2 1 9 8 7 98 6 Year Number of Cancellations

Insufficient Information
Example: An admissions officer from a state school explains that the average tuition at a nearby private university is $13,000 and only $4500 at his school. This makes the state school look more attractive. If most students pay the full tuition, then the state school appears to be a better choice However, if most students at the private university receive substantial financial aid, then the actual tuition cost could be much lower!

Chapter 3: Descriptive Analysis and Presentation of Bivariate Data

Chapter Goals To be able to present bivariate data in tabular and graphic form. To gain an understanding of the distinction between the basic purposes of correlation analysis and regression analysis. To become familiar with the ideas of descriptive presentation.

3.1: Bivariate Data Bivariate Data: Consists of the values of two different response variables that are obtained from the same population of interest. Three combinations of variable types: 1. Both variables are qualitative (attribute). 2. One variable is qualitative (attribute) and the other is quantitative (numerical). 3. Both variables are quantitative (both numerical).

Two Qualitative Variables:
When bivariate data results from two qualitative (attribute or categorical) variables, the data is often arranged on a cross-tabulation or contingency table. Example: A survey was conducted to investigate the relationship between preferences for television, radio, or newspaper for national news, and gender. The results are given in the table below.

This table may be extended to display the marginal totals (or marginals). The total of the marginal totals is the grand total. Contingency tables often show percentages (relative frequencies). These percentages are based on the entire sample or on the subsample (row or column) classifications.

Percentages based on the grand total (entire sample):
The previous contingency table may be converted to percentages of the grand total by dividing each frequency by the grand total and multiplying by 100. For example, 175 becomes 13.3%

These same statistics (numerical values describing sample results) can be shown in a (side-by-side) bar graph.

Percentages based on row (column) totals:
The entries in a contingency table may also be expressed as percentages of the row (column) totals by dividing each row (column) entry by that row’s (column’s) total and multiplying by The entries in the contingency table below are expressed as percentages of the column totals. These statistics may also be displayed in a side-by-side bar graph.

One Qualitative and One Quantitative Variable:
1. When bivariate data results from one qualitative and one quantitative variable, the quantitative values are viewed as separate samples. 2. Each set is identified by levels of the qualitative variable. 3. Each sample is described using summary statistics, and the results are displayed for side-by-side comparison. 4. Statistics for comparison: measures of central tendency, measures of variation, 5-number summary. 5. Graphs for comparison: dotplot, boxplot.

Example: A random sample of households from three different parts of the country was obtained and their electric bill for June was recorded. The data is given in the table below. The part of the country is a qualitative variable with three levels of response. The electric bill is a quantitative variable. The electric bills may be compared with numerical and graphical techniques.

Comparison using dotplots:
: Northeast . :..:. .. Midwest : West The electric bills in the Northeast tend to be more spread out than those in the Midwest. The bills in the West tend to be higher than both those in the Northeast and Midwest.

Comparison using Box-and-Whisker plots:

Two Quantitative Variables:
1. Expressed as ordered pairs: (x, y) 2. x: input variable, independent variable. y: output variable, dependent variable. Scatter Diagram: A plot of all the ordered pairs of bivariate data on a coordinate axis system. The input variable x is plotted on the horizontal axis, and the output variable y is plotted on the vertical axis. Note: Use scales so that the range of the y-values is equal to or slightly less than the range of the x-values. This creates a window that is approximately square.

Example: In a study involving children’s fear related to being hospitalized, the age and the score each child made on the Child Medical Fear Scale (CMFS) are given in the table below. Construct a scatter diagram for this data.

Scatter diagram: age = input variable, CMFS = output variable

3.2: Linear Correlation Measure the strength of a linear relationship between two variables. As x increases, no definite shift in y: no correlation. As x increase, a definite shift in y: correlation. Positive correlation: x increases, y increases. Negative correlation: x increases, y decreases. If the ordered pairs follow a straight-line path: linear correlation.

Example: no correlation.
As x increases, there is no definite shift in y.

Example: positive correlation.
As x increases, y also increases.

Example: negative correlation.
As x increases, y decreases.

Note: 1. Perfect positive correlation: all the points lie along a line with positive slope. 2. Perfect negative correlation: all the points lie along a line with negative slope. 3. If the points lie along a horizontal or vertical line: no correlation. 4. If the points exhibit some other nonlinear pattern: no linear relationship, no correlation. 5. Need some way to measure correlation.

Coefficient of linear correlation: r, measures the strength of the linear relationship between two variables. Pearson’s product moment formula: Note: 1. 2. r = +1: perfect positive correlation 3. r = -1 : perfect negative correlation

Alternate formula for r:

Example: The table below presents the weight (in thousands of pounds) x and the gasoline mileage (miles per gallon) y for ten different automobiles. Find the linear correlation coefficient.

To complete the calculation for r:

Note: 1. r is usually rounded to the nearest hundredth. 2. r close to 0: little or no linear correlation. 3. As the magnitude of r increases, towards -1 or +1, there is an increasingly stronger linear correlation between the two variables. 4. Method of estimating r based on the scatter diagram. Window should be approximately square. Useful for checking calculations.

3.3: Linear Regression Regression analysis finds the equation of the line that best describes the relationship between two variables. One use of this equation: to make predictions.

Models or prediction equations:
Some examples of various possible relationships. Linear: Quadratic: Exponential: Logarithmic: Note: What would a scatter diagram look like to suggest each relationship?

Method of least squares:
Equation of the best-fitting line: Predicted value: Least squares criterion: Find the constants b0 and b1 such that the sum is as small as possible.

Observed and predicted values of y:

The equation of the line of best fit:
Determined by b0: slope b1: y-intercept Values that satisfy the least squares criterion:

Example: A recent article measured the job satisfaction of subjects with a 14-question survey. The data below represents the job satisfaction scores, y, and the salaries, x, for a sample of similar individuals. 1. Draw a scatter diagram for this data. 2. Find the equation of the line of best fit.

Preliminary calculations needed to find b1 and b0:

Finding b1 and b0:

Scatter diagram:

Note: 1. Keep at least three extra decimal places while doing the calculations to ensure an accurate answer. 2. When rounding off the calculated values of b0 and b1, always keep at least two significant digits in the final answer. 3. The slope b1 represents the predicted change in y per unit increase in x. 4. The y-intercept is the value of y where the line of best fit intersects the y-axis. 5. The line of best fit will always pass through the point

Making predictions: 1. One of the main purposes for obtaining a regression equation is for making predictions. 2. For a given value of x, we can predict a value of y, 3. The regression equation should be used to make predictions only about the population from which the sample was drawn. 4. The regression equation should be used only to cover the sample domain on the input variable. You can estimate values outside the domain interval, but use caution and use values close to the domain interval. 5. Use current data. A sample taken in 1987 should not be used to make predictions in 1999.

Limiting Relative Frequency
Chapter 4 ~ Probability Limiting Relative Frequency Relative Frequency Trials 0.1 0.2 0.3 0.4 0.5 0.6 200 400 600 800 1000 1200

Chapter Goals Learn the basic concepts of probability
Learn the rules that apply to the probability of both simple and compound events In order to make inferences, we need to study sample results in situations in which the population is known

4.1 ~ The Nature of Probability
Example: Consider an experiment in which we roll two six- sided fair dice and record the number of 3s face up. The only possible outcomes are zero 3s, one 3, or two 3s. Here are the results after 100 rolls, and after 1000 rolls:

Using a Histogram We can express these results (from the 1000 rolls) in terms of relative frequencies and display the results using a histogram: 1 2 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Relative Frequency Three’s Face Up

Continuing the Experiment
If we continue this experiment for several thousand more rolls: 1. The frequencies will have approximately a 25:10:1 ratio in totals 2. The relative frequencies will settle down Use random number tables Use a computer to randomly generate number values representing the various experimental outcomes Key to either method is to maintain the probabilities Note: We can simulate many probability experiments:

4.2 ~ Probability of Events
Probability that an Event Will Occur: The relative frequency with which that event can be expected to occur The probability of an event may be obtained in three different ways: Empirically Theoretically Subjectively

Experimental or Empirical Probability
1. The observed relative frequency with which an event occurs 2. Prime notation is used to denote empirical probabilities: 3. n(A): number of times the event A has occurred 4. n: number of times the experiment is attempted Question: What happens to the observed relative frequency as n increases?

Example Example: Consider tossing a fair coin. Define the event H as the occurrence of a head. What is the probability of the event H, P(H)? 1. In a single toss of the coin, there are two possible outcomes 2. Since the coin is fair, each outcome (side) should have an equally likely chance of occurring 3. Intuitively, P(H) = 1/2 (the expected relative frequency) Notes: This does not mean exactly one head will occur in every two tosses of the coin In the long run, the proportion of times that a head will occur is approximately 1/2

Long-Run Behavior To illustrate the long-run behavior:
1. Consider an experiment in which we toss the coin several times and record the number of heads 2. A trial is a set of 10 tosses 3. Graph the relative frequency and cumulative relative frequency of occurrence of a head 4. A cumulative graph demonstrates the idea of long-run behavior 5. This cumulative graph suggests a stabilizing, or settling down, effect on the observed cumulative probability 6. This stabilizing effect, or long-term average value, is often referred to as the law of large numbers

Experiment Experimental results of tossing a coin 10 times each trial:

Relative Frequency Expected value = 1/2 Trial

Cumulative Relative Frequency
Expected value = 1/2 Trial

Law of Large Numbers Law of Large Numbers: If the number of times an experiment is repeated is increased, the ratio of the number of successful occurrences to the number of trials will tend to approach the theoretical probability of the outcome for an individual trial Interpretation: The law of large numbers says: the larger the number of experimental trials n, the closer the empirical probability P(A) is expected to be to the true probability P(A) In symbols: As

4.3 ~ Simple Sample Spaces We need to talk about data collection and experimentation more precisely With many activities, like tossing a coin, rolling a die, selecting a card, there is uncertainty as to what will happen We will study and characterize this uncertainty

Experiment & Outcome Experiment: Any process that yields a result or an observation Outcome: A particular result of an experiment Example: Suppose we select two students at random and ask each if they have a car on campus: 1. A list of possible outcomes: (Y, Y), (Y, N), (N, Y), (N, N) 2. This is called ordered pair notation 3. The outcomes may be displayed using a tree diagram

Tree Diagram Student 1 Student 2 Outcomes Y Y, Y Y N Y, N Y N, Y N N N, N 1. This diagram consists of four branches: 2 first generation branches and 4 second generation branches 2. Each branch shows a possible outcome

Sample Space & Event Sample Space: The set of all possible outcomes of an experiment. The sample space is typically called S and may take any number of forms: a list, a tree diagram, a lattice grid system, etc. The individual outcomes in a sample space are called sample points. n(S) is the number of sample points in the sample space. Event: any subset of the sample space. If A is an event, then n(A) is the number of sample points that belong to A Example: For the student car example above: S = { (Y, Y), (Y, N), (N, Y), (N, N) } n(S) = 4

Examples Example: An experiment consists of two trials. The first is tossing a penny and observing a head or a tail; the second is rolling a die and observing a 1, 2, 3, 4, 5, or 6. Construct the sample space: S = { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 } Example: Three voters are randomly selected and asked if they favor an increase in property taxes for road construction in the county. Construct the sample space: S = { NNN, NNY, NYN, NYY, YNN, YNY, YYN, YYY}

Example Example: An experiment consists of selecting electronic parts from an assembly line and testing each to see if it passes inspection (P) or fails (F). The experiment terminates as soon as one acceptable part is found or after three parts are tested. Construct the sample space: Outcome F FFF F F P FFP P FP P P S = { FFF, FFP, FP, P }

Example Example: The 1200 students at a local university have been cross tabulated according to resident and college status: The experiment consists of selecting one student at random from the entire student body n(S) = 1200

Example Example: On the way to work, some employees at a certain company stop for a bagel and/or a cup of coffee. The accompanying Venn diagram summarizes the behavior of the employees for a randomly selected work day: Coffee Bagel The experiment consists of selecting one employee at random n(S) = 77

Notes 1. The outcomes in a sample space can never overlap
2. All possible outcomes must be represented 3. These two characteristics are called mutually exclusive and all inclusive

4.4 ~ Rules of Probability Consider the concept of probability and relate it to the sample space Recall: the probability of an event is the relative frequency with which the event could be expected to occur, the long-term average

Equally Likely Events 1. In a sample space, suppose all sample points are equally likely to occur 2. The probability of an event A is the ratio of the number of sample points in A to the number of sample points in S 3. In symbols: 4. This formula gives a theoretical probability value of event A’s occurrence 5. The use of this formula requires the existence of a sample space in which each outcome is equally likely

Example Example: A fair coin is tossed 5 times, and a head (H) or a tail (T) is recorded each time. What is the probability of: A = {exactly one head in 5 tosses}, and B = {exactly 5 heads}? The outcomes consist of a sequence of 5 Hs and Ts A typical outcome includes a mixture of Hs and Ts, like: HHTTH There are 32 possible outcomes, all equally likely A = {HTTTT, THTTT, TTHTT, TTTHT, TTTTH} B = {HHHHH}

Subjective Probability
1. Suppose the sample space elements are not equally likely, and empirical probabilities cannot be used 2. Only method available for assigning probabilities may be personal judgment 3. These probability assignments are called subjective probabilities 4. Personal judgment of the probability is expressed by comparing the likelihood among the various outcomes

Basic Probability Ideas
1. Probability represents a relative frequency 2. P(A) is the ratio of the number of times an event can be expected to occur divided by the number of trials 3. The numerator of the probability ratio must be a positive number or zero 4. The denominator of the probability ratio must be a positive number (greater than zero) 5. The number of times an event can be expected to occur in n trials is always less than or equal to the total number of trials, n

Properties 1. The probability of any event A is between 0 and 1:
2. The sum of the probabilities of all outcomes in the sample space is 1: Notes: The probability is zero if the event cannot occur The probability is one if the event occurs every time (a sure thing)

Example Example: On the way to work Bob’s personal judgment is that he is four times more likely to get caught in a traffic jam (TJ) than have an easy commute (EC). What values should be assigned to P(TJ) and P(EC)?

Odds Odds: another way of expressing probabilities
If the odds in favor of an event A are a to b, then: 1. The odds against A are b to a 2. The probability of event A is: 3. The probability that event A will not occur is

Example Example: The odds in favor of you passing an introductory statistics class are 11 to 3. Find the probability you will pass and the probability you will fail. Using the preceding notation: a = 11 and b = 3:

Complement of An Event Complement of an Event: The set of all sample points in the sample space that do not belong to event A. The complement of event A is denoted by (read “A complement”).

Example Example: 1. The complement of the event “success” is “failure”
2. The complement of the event “rain” is “no rain” 3. The complement of the event “at least 3 patients recover” out of 5 patients is “2 or fewer recover” Notes: Every event A has a complementary event Complementary probabilities are very useful when the question asks for the probability of “at least one.” P ( ) A for any event + = 1

Example Example: A fair coin is tossed 5 times, and a head(H) or a tail (T) is recorded each time. What is the probability of 1) A = {at least one head in 5 tosses} 2) B = {at most 3 heads in 5 tosses} Solutions: 1) 2)

Example Example: A local automobile dealer classifies purchases by number of doors and transmission type. The table below gives the number of each classification. If one customer is selected at random, find the probability that: 1) The selected individual purchased a car with automatic transmission 2) The selected individual purchased a 2-door car

Solutions 1) 2)

Chapter 5 ~ Probability Distributions (Discrete Variables)
Express Checkout Number of Items Purchased

Chapter Goals Combine the ideas of frequency distributions and probability to form probability distributions Investigate discrete probability distributions and study measures of central tendency and dispersion Study the binomial random variable

5.1 ~ Random Variables Bridge between experimental outcomes and statistical analysis Each outcome in an experiment is assigned to a number This suggests the idea of a function

Random Variable Random Variable: A variable that assumes a unique numerical value for each of the outcomes in the sample space of a probability experiment Notes: Used to denote the outcomes of a probability experiment Each outcome in a probability experiment is assigned to a unique value Illustration:

Examples of Random Variable
1. Let the number of computers sold per day by a local merchant be a random variable. Integer values ranging from zero to about 50 are possible values. 2. Let the number of pages in a mystery novel at a bookstore be a random variable. The smallest number of pages is 125 while the largest number of pages is 547. 3. Let the time it takes an employee to get to work be a random variable. Possible values are 15 minutes to over 2 hours. 4. Let the volume of water used by a household during a month be a random variable. Amounts range up to several thousand gallons. 5. Let the number of defective components in a shipment of 1000 be a random variable. Values range from 0 to 1000.

Discrete & Random Variables
Discrete Random Variable: A quantitative random variable that can assume a countable number of values Intuitively, a discrete random variable can assume values corresponding to isolated points along a line interval. That is, there is a gap between any two values. Note: Usually associated with counting Continuous Random Variable: A quantitative random variable that can assume an uncountable number of values Intuitively, a continuous random variable can assume any value along a line interval, including every possible value between any two values Note: Usually associated with a measurement

Example Example: Determine whether the following random variables are discrete or continuous 1. The barometric pressure at 12:00 PM 2. The length of time it takes to complete a statistics exam 3. The number of items in the shopping cart of the person in front of you at the checkout line 4. The weight of a home grown zucchini 5. The number of tickets issued by the PA State Police during a 24 hour period 6. The number of cans of soda pop dispensed by a machine placed in the Mathematics building on campus 7. The number of cavities the dentist discovers during your next visit

5.2 ~ Probability Distributions of a Discrete Random Variable
Need a complete description of a discrete random variable This includes all the values the random variable may assume and all of the associated probabilities This information may be presented in a variety of ways

Probability Distribution & Function
Probability Distribution: A distribution of the probabilities associated with each of the values of a random variable. The probability distribution is a theoretical distribution; it is used to represent populations. Notes: The probability distribution tells you everything you need to know about the random variable. The probability distribution may be presented in the form of a table, chart, function, etc. Probability Function: A rule that assigns probabilities to the values of the random variable

Example Example: The number of people staying in a randomly selected room at a local hotel is a random variable ranging in value from 0 to 4. The probability distribution is known and is given in various forms below: P x ( ) the random variable equals 2 = 2 5 15 Notes: This chart implies the only values x takes on are 0, 1, 2, 3, and 4

A Line Representation Hotel Room Probability Distribution

Histogram A histogram may be used to present a probability distribution: 1 2 3 4 0.0 0.1 0.2 0.3 0.4 Hotel Room Probability Distribution

Notes The histogram of a probability distribution uses the physical area of each bar to represent its assigned probability In the Hotel Room probability distribution: the width of each bar is 1, so the height of each bar is equal to the assigned probability, which is the area of each bar. The idea of area representing probability is important in the study of continuous random variables

Reminder! Every probability function must satisfy the two basic properties of probability: 1. The probability assigned to each value of the random variable must be between 0 and 1, inclusive: 2. The sum of the probabilities assigned to all the values of the random variable must equal 1:

5.3 ~ Mean and Variance of a Discrete Probability Distribution
Describe the center and spread of a population m, s, s2 : population parameters Population parameters are usually unknown values (we would like to estimate)

Important Notes 1. is the mean of the sample
2. s2 and s are the variance and standard deviation of the sample 3. , s2, and s are called sample statistics 4. m (lowercase Greek letter “mu”) is the mean of the population 5. s2 (“sigma squared”) is the variance of the population 6. s (lowercase Greek letter “sigma”) is the standard deviation of the population 7. m, s2, and s are called population parameters. (A parameter is a constant. m, s2, and s are typically unknown values.)

Mean of a Discrete Random Variable
The mean, m, of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together: m = å [ ( )] xP x Notes: The mean is the average value of the random variable, what happens on average The mean is not necessarily a value of the random variable

Discrete Random Variables
Variance of a Discrete Random Variable: Variance, s2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x - m)2, by its own probability and then adding all the products together: { } s m 2 = - å [( ) ( )] [ x P xP Standard Deviation of a Discrete Random Variable: The positive square root of the variance: s = 2

Example Example: The number of standby passengers who get seats on a daily commuter flight from Boston to New York is a random variable, x, with probability distribution given below (in an extensions table). Find the mean, variance, and standard deviation:

Solution Using the formulas for mean, variance, and standard deviation: m = å [ ( )] . xP x 1 55 { } s 2 4 45 1 55 4025 0475 = - å [ ( )] . ) x P xP 43 Note: 1.55 is not a value of the random variable (in this case). It is only what happens on average.

Example Example: The probability distribution for a random variable x is given by the probability function: Find the mean, variance, and standard deviation Solutions: Find the probability associated with each value by using the probability function:

Finding the Population Parameters
Use an extensions table to find the population parameters: m = å [ ( )] . xP x 65 15 4 33 { } s 2 305 15 65 1 56 = - æ è ç ö ø ÷ å [ ( )] . x P xP s = 2 1 56 25 .

5.4 ~ The Binomial Probability Distribution
One of the most important discrete distributions Based on a series of repeated trials whose outcomes can be classified in one of two categories: success or failure Distribution based on a binomial probability experiment

Binomial Probability Experiment
Binomial Probability Experiment: An experiment that is made up of repeated trials that possess the following properties: 1. There are n repeated independent trials 2. Each trial has two possible outcomes (success, failure) 3. P(success) = p, P(failure) = q, and p + q = 1 4. The binomial random variable x is the count of the number of successful trials that occur; x may take on any integer value from zero to n

Notes Properties 1 and 2 are the two basic properties of any binomial experiment Property 3 concerns the algebraic notation for each trial Property 4 concerns the algebraic notation for the complete experiment Both x and p must be associated with “success” Independent trials mean that the result of one trial does not affect the probability of success of any other trial in the experiment. The probability of “success” remains constant throughout the entire experiment.

Example Example: It is known that 40% of all graduating seniors on the campus of a very large university have taken a statistics class. Five seniors are selected at random and asked if they have taken a statistics class. This approximates a binomial experiment: 1. A trial is asking one student, repeated 5 times. The trials are independent since the probability of taking a statistics class for any one student is not affected by the results from any other student. 2. Two outcomes on each trial: taken a statistics class (success), not taken a statistics class (failure) 3. p = P(taken a statistics class) = q = P(not taken a statistics class) = 0.60 4. x = number of students who have taken a statistics class

Binomial Probability Function
For a binomial experiment, let p represent the probability of a “success” and q represent the probability of a “failure” on a single trial; then P(x), the probability that there will be exactly x successes on n trials is: Notes: The number of ways that exactly x successes can occur in n trials: The probability of exactly x successes: px The probability that failure will occur on the remaining (n - x) trials: qn - x

Notes The number of ways that exactly x successes can occur in a set of n trials is represented by the symbol: 1. Must always be a positive integer 2. Called the binomial coefficient 3. Found by using the formula: n! is an abbreviation for n factorial:

Factorial Calculations with TI
Example: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 Enter the number 6 and then use MATH > PRB> 4 to enter the !, then press enter. The display reads 720 Try these values next: 10! ! 12! 20! Answers: 10! = 3,628, ! = 1 12! = 479,001,600 20! = 2.43E18 As you can see, the factorials grow very rapidly

Example Example: According to a recent study, 65% of all homes in a certain county have high levels of radon gas leaking into their basements. Four homes are selected at random and tested for radon. The random variable x is the number of homes with high levels of radon (out of the four). Properties: 1. There are 4 repeated trials: n = 4. The trials are independent. 2. Each test for radon is a trial, and each test has two outcomes: radon or no radon 3. p = P(radon) = 0.65, q = P(no radon) = 0.35 p + q = 1 4. x is the number of homes with high levels of radon, possible values: 0, 1, 2, 3, 4

Binomial Probability Function
x ( ) (0. , = æ è ç ö ø ÷ - 4 65 35 for 0, 1, 2, 3, P ( ) (0. )( . 4 65 35 1 0150 = æ è ç ö ø ÷ P ( ) (0. )( . 1 4 65 35 0429 1115 3 = æ è ç ö ø ÷ P ( ) (0. )( . 2 4 65 35 6 4225 1225 3105 = æ è ç ö ø ÷ P ( ) (0. )( . 3 4 65 35 2746 3845 1 = æ è ç ö ø ÷ P ( ) (0. )( . 4 65 35 1 1785 = æ è ç ö ø ÷

Example P x ( ) (0. , = æ è ç ö ø ÷ 15 7 3 for 0, 1, 2, . ,15
Example: In a certain automobile dealership, 70% of all customers purchase an extended warranty with their new car. For 15 customers selected at random: 1) Find the probability that exactly 12 will purchase an extended warranty 2) Find the probability at most 13 will purchase an extended warranty P x ( ) (0. , = æ è ç ö ø ÷ - 15 7 3 for 0, 1, 2, . ,15 Solutions: Let x be the number of customers who purchase an extended warranty. x is a binomial random variable. The probability function associated with this experiment:

Solutions Continued [ ]
1) Probability exactly 12 purchase an extended warranty: P ( ) (0. 0. 12 15 7 3 1700 = æ è ç ö ø ÷ [ ] 9648 0. 0352 1 0047 0305 [0. ) 3 (0. 7 15 14 ( 13 = - + ú û ù ê ë é ÷ ø ö ç è æ + ... P x 2) Probability at most 13 purchase an extended warranty:

Notes Many graphing calculators also have built-in functions for computing binomial probabilities and cumulative probabilities TI Calculators: Choose 2nd >DISTR> 0 or A:binompdf( Enter: n, p) where n is the number of independent trials and p is the probability of success

5.5 ~ Mean & Standard Deviation of the Binomial Distribution
Population parameters of the binomial distribution help to describe the distribution Mean and standard deviation indicate where the distribution is centered and the spread of the distribution

Mean & Standard Deviation
The mean and standard deviation of a theoretical binomial distribution can be found by using the following two formulas: m s = np npq Notes: Mean is intuitive: number of trials multiplied by the probability of a success The variance of a binomial probability distribution is: ( ) s 2 = npq

Example Example: Find the mean and standard deviation of the binomial distribution when n = 18 and p = 0.75 Solutions: 1) n = 18, p = 0.75, q = = 0.25 m = np ( )(0. ) . 18 75 13 5 s npq 25 3 375 1 8371 P x ( ) (0. = æ è ç ö ø ÷ - 18 75 25 for 0, 1, 2, . , 2) The probability function is:

Table of Values & Probabilities
x P(x)

Histogram 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22

Chapter 6 ~ Normal Probability Distributions

Chapter Goals Learn about the normal, bell-shaped, or Gaussian distribution How probabilities are found How probabilities are represented How normal distributions are used in the real world

6.1 ~ Normal Probability Distributions
The normal probability distribution is the most important distribution in all of statistics Many continuous random variables have normal or approximately normal distributions Need to learn how to describe a normal probability distribution

Normal Probability Distribution
1. A continuous random variable 2. Description involves two functions: a. A function to determine the ordinates of the graph picturing the distribution b. A function to determine probabilities 3. Normal probability distribution function: This is the function for the normal (bell-shaped) curve f x e ( ) = - 1 2 s p m 4. The probability that x lies in some interval is the area under the curve

The Normal Probability Distribution

Probabilities for a Normal Distribution
Illustration

Notes The definite integral is a calculus topic
We will use the TI83/84 to find probabilities for normal distributions We will learn how to compute probabilities for one special normal distribution: the standard normal distribution We will learn to transform all other normal probability questions to this special distribution Recall the empirical rule: the percentages that lie within certain intervals about the mean come from the normal probability distribution We need to refine the empirical rule to be able to find the percentage that lies between any two numbers

Percentage, Proportion & Probability
Basically the same concepts Percentage (30%) is usually used when talking about a proportion (3/10) of a population Probability is usually used when talking about the chance that the next individual item will possess a certain property Area is the graphic representation of all three when we draw a picture to illustrate the situation

6.2 ~ The Standard Normal Distribution
There are infinitely many normal probability distributions They are all related to the standard normal distribution The standard normal distribution is the normal distribution of the standard variable z (the z-score)

Standard Normal Distribution
Properties: The total area under the normal curve is equal to 1 The distribution is mounded and symmetric; it extends indefinitely in both directions, approaching but never touching the horizontal axis The distribution has a mean of 0 and a standard deviation of 1 The mean divides the area in half, 0.50 on each side Nearly all the area is between z = and z = 3.00 Notes: Table 3, Appendix B lists the probabilities associated with the intervals from the mean (0) to a specific value of z Probabilities of other intervals are found using the table entries, addition, subtraction, and the properties above

Table 3, Appendix B Entries
The table contains the area under the standard normal curve between 0 and a specific value of z

Example Example: Find the area under the standard normal curve between z = 0 and z = 1.45 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 1.4 0.4265 A portion of Table 3: .

Using the TI 83/84 To find the area between 0 and 1.45, do the following: 2nd DISTR 2 which is normalcdf( Enter the lower bound of 0 Enter a comma Then enter 1.45 Close the parentheses if you like or hit “Enter” The value of is shown as the answer! Interpretation of the result: The probability that Z lies between 0 and 1.45 is 0.426

Example Example: Find the area under the normal curve to the right of z = 1.45; P(z > 1.45) Area asked for

Using the TI 83/84 To find the area between 1.45 and ∞, do the following: 2nd DISTR 2 which is normalcdf( Enter the lower bound of 1.45 Enter a comma Then enter 1 2nd EE 99 Close the parentheses if you like or hit “Enter” The value of is shown as the answer! Interpretation of result: The probability that Z is greater than 1.45 is 0.074

Example Example: Find the area to the left of z = 1.45; P(z < 1.45)

Using The TI 83/84 To find the area between - ∞ and 1.45, do the following: 2nd DISTR 2 which is normalcdf( Enter the lower bound of -1 2nd EE 99 Enter a comma Then enter 1.45 Close the parentheses if you like or hit “Enter” The value of is shown as the answer! Interpretation of result: The probability that Z is less than 1.45 is 0.926

Notes The addition and subtraction used in the previous examples are correct because the “areas” represent mutually exclusive events The symmetry of the normal distribution is a key factor in determining probabilities associated with values below (to the left of) the mean. For example: the area between the mean and z = is exactly the same as the area between the mean and z = When finding normal distribution probabilities, a sketch is always helpful

Example Example: Find the area between the mean (z = 0) and z = -1.26
Area asked for

Using the TI 83/84 Find the area to the left of z = -0.98
Use -1E99 for - ∞ and enter 2nd DISTR Normalcdf (-1e99, -0.98) which gives .164 - 0. 98 Area asked for

Example Example: Find the area between z = -2.30 and z = 1.80 P z ( .
) - < = + 2 30 1 80 4893 4641 9534

Using the TI 83/84 Find the area between z = -2.30 and z = 1.80
Enter 2nd DISTR, normalcdf (-2.3, 1.80) and press enter .953 is given as the answer. Remember, the function normalcdf is of the form: Normalcdf(lower limit, upper limit, mean, standard deviation) and if you’re working with distributions other than the standard normal (recall mean = 0, stddev = 1), you must enter the values for mean and standard deviation

Normal Distribution Note
The normal distribution table may also be used to determine a z-score if we are given the area (working backwards) Example: What is the z-score associated with the 85th percentile?

Using the TI 83/84 There is another function in the DISTR list that is used to find the value of z (or x) when the probability is given. For the previous problem, we are actually asking what is the value of z such that 85% of the distribution lies below it.

Using the TI 83/84 Use 2nd DISTR invNorm( to calculate this value
2nd DISTR invNorm(.85) “ENTER” gives us a value of which is shown

Example Example: What z-scores bound the middle 90% of a standard normal distribution?

Using the TI 83/84 The TI 83/84 calculates areas from -∞ to the value of z we are interested in. Therefore, we must get a little creative to solve some problems. Using the idea that the total area equals one comes in very handy here! For the example given, where we are interested in the value of z that bounds the middle 90%, the tails therefore represent a total of 10%. Divide this in two since it is symmetric and this gives 5% in each tail.

Using the TI 83/84 Now use the 2nd DISTR invNorm with .05 in the argument like this: Which gives an answer of Since the distribution is symmetric, the upper limit is 1.645, so 90% of the distribution lies between (-1.645, 1.645)

Using the TI 83/84 Now let’s work the problems on page 279

6.3 ~ Applications of Normal Distributions
Apply the techniques learned for the z distribution to all normal distributions Start with a probability question in terms of x-values Convert, or transform, the question into an equivalent probability statement involving z-values

Standardization Suppose x is a normal random variable with mean m and standard deviation s The random variable has a standard normal distribution

Example Example: A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of Assume the amount of fill is normally distributed and a bottle is selected at random: 1) Find the probability the bottle contains between oz and oz 2) Find the probability the bottle contains more than oz When x z = - 32.00 32.0 0.00 ; 0.02 m s Solutions: 1) When x z = - 32 025 32.025 32.0 1 25 . ; 0.02 m s

Solution Continued Area asked for P x z ( . ) 0. 32.0 32 025 02 1 25
1 25 3944 < = - æ è ç ö ø ÷

Example, Part 2 P x z ( . ) > = - æ è ç ö ø ÷ + 31 97 32.0 0. 02 1
2) 32.0 - 1 50 . P x z ( . ) > = - æ è ç ö ø ÷ + 31 97 32.0 0. 02 1 50) 5000 4332 9332

Notes The normal table may be used to answer many kinds of questions involving a normal distribution Often we need to find a cutoff point: a value of x such that there is a certain probability in a specified interval defined by x Example: The waiting time x at a certain bank is approximately normally distributed with a mean of 3.7 minutes and a standard deviation of 1.4 minutes. The bank would like to claim that 95% of all customers are waited on by a teller within c minutes. Find the value of c that makes this statement true.

Solution P x c z ( ) 0. . = - æ è ç ö ø ÷ 95 3 7 1 4

Example Example: A radar unit is used to measure the speed of automobiles on an expressway during rush-hour traffic. The speeds of individual automobiles are normally distributed with a mean of 62 mph. Find the standard deviation of all speeds if 3% of the automobiles travel faster than 72 mph.

Solution P x ( ) . > = 72 03 P z ( . ) > = 1 88 03 - 72 62 s
03 P z ( . ) > = 1 88 03 - 72 62 s 1.88 = x - m ; z = s 1 . 88 s = 10 / . = 10 1 88 5 32 s

Notation If x is a normal random variable with mean m and standard deviation s, this is often denoted: x ~ N(m, s) Example: Suppose x is a normal random variable with m = 35 and s = 6. A convenient notation to identify this random variable is: x ~ N(35, 6).

6.4 ~ Notation z-score used throughout statistics in a variety of ways
Need convenient notation to indicate the area under the standard normal distribution z(a) is the algebraic name, for the z-score (point on the z axis) such that there is a of the area (probability) to the right of z(a)

Illustrations z(0.10) represents the value of z such that the area to the right under the standard normal curve is 0.10 z(0.10) z(0.80) represents the value of z such that the area to the right under the standard normal curve is 0.80 z(0.80)

Example z(0.10) = 1.28 Example: Find the numerical value of z(0.10):
0.10 (area information from notation) Table shows this area (0.4000) z(0.10) z(0.10) = 1.28

Example z(0.80) = -0.84 Example: Find the numerical value of z(0.80):
Look for ; remember that z must be negative z(0.80) Use Table 3: look for an area as close as possible to z(0.80) = -0.84

Notes The values of z that will be used regularly come from one of the following situations: 1. The z-score such that there is a specified area in one tail of the normal distribution 2. The z-scores that bound a specified middle proportion of the normal distribution

Example Example: Find the numerical value of z(0.99): 0.01 z(0.99)
Because of the symmetrical nature of the normal distribution, z(0.99) = -z(0.01)

Example Example: Find the z-scores that bound the middle 0.99 of the normal distribution: z(0.995) -z(0.005) z(0.005) z(0.005) = and z(0.995) = -z(0.005) =

6.5 ~ Normal Approximation of the Binomial
Recall: the binomial distribution is a probability distribution of the discrete random variable x, the number of successes observed in n repeated independent trials Binomial probabilities can be reasonably estimated by using the normal probability distribution

Background & Histogram
Background: Consider the distribution of the binomial variable x when n = 20 and p = 0.5 Histogram: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 The histogram may be approximated by a normal curve

Notes The normal curve has mean and standard deviation from the binomial distribution: Can approximate the area of the rectangles with the area under the normal curve The approximation becomes more accurate as n becomes larger

Two Problems 1. As p moves away from 0.5, the binomial distribution is less symmetric, less normal-looking Solution: The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 - p) both equal or exceed 5 2. The binomial distribution is discrete, and the normal distribution is continuous Solution: Use the continuity correction factor. Add or subtract 0.5 to account for the width of each rectangle.

Example Example: Research indicates 40% of all students entering a certain university withdraw from a course during their first year. What is the probability that fewer than 650 of this year’s entering class of 1800 will withdraw from a class? Let x be the number of students that withdraw from a course during their first year x has a binomial distribution: n = 1800, p = 0.4 The probability function is given by:

Solution Use the normal approximation method:

Random Number Generation
With each rand execution, the TI-84 Plus generates the same random-number sequence for a given seed value. The TI-84 Plus factory-set seed value for rand is 0. To generate a different random-number sequence, store any nonzero seed value to rand. To restore the factory-set seed value, store 0 to rand or reset the defaults (Chapter 18). Note: The seed value also affects randInt(, randNorm(, and randBin( instructions.

Chapter 7 ~ Sample Variability
6.8 7.2 7.6 8.0 8.4 8.8 9.2 9.6 10.0 10.4 10.8 11.2 Sample Mean 1 2 3 4 5 6 7 8 9 Frequency Empirical Distribution of Sample Means

Chapter Goals Investigate the variability in sample statistics from sample to sample Find measures of central tendency for sample statistics Find measures of dispersion for sample statistics. Find the pattern of variability for sample statistics

7.1 ~ Sampling Distributions
To make inferences about a population, we need to understand sampling The sample mean varies from sample to sample The sample mean has a distribution; we need to understand how the sample mean varies and the pattern (if any) in the distribution

Sampling Distribution of a Sample Statistic
Sampling Distribution of a Sample Statistic: The distribution of values for a sample statistic obtained from repeated samples, all of the same size and all drawn from the same population Example: Consider the set {1, 2, 3, 4}: 1) Make a list of all samples of size 2 that can be drawn from this set (Sample with replacement) 2) Construct the sampling distribution for the sample mean for samples of size 2 3) Construct the sampling distribution for the minimum for samples of size 2

Table of All Possible Samples
This table lists all possible samples of size 2, the mean for each sample, the minimum for each sample, and the probability of each sample occurring (all equally likely)

Sampling Distribution
Summarize the information in the previous table to obtain the sampling distribution of the sample mean and the sample minimum: Sampling Distribution of the Sample Mean Histogram: Sampling Distribution of the Sample Mean

Sampling Distribution
Sampling Distribution of the Sample Minimum: Histogram: Sampling Distribution of the Sample Minimum:

The Population: Theoretical Probability Distribution
Example 1 Example: Consider the population consisting of six equally likely integers: 1, 2, 3, 4, 5, and 6. Empirically investigate the sampling distribution of the sample mean. Select 50 samples of size 5, find the mean for each sample, and construct the empirical distribution of the sample mean. 1 2 3 4 5 6 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 m s = 7078 . The Population: Theoretical Probability Distribution

Empirical Distribution of the Sample Mean
Samples of Size 5 x 1.8 2.3 2.8 3.3 3.8 4.3 4.8 5.3 2 4 6 8 10 12 14 Frequency s = 3 352 714 . Sample Mean

Important Notes & Random Sample
: the mean of the sample means : the standard deviation of the sample means 3. The theory involved with sampling distributions described in the remainder of this chapter requires random sampling x s x Random Sample: A sample obtained in such a way that each possible sample of a fixed size n has an equal probability of being selected (Every possible handful of size n has the same probability of being selected)

7.2 ~ Where Does This Lead Us?
Describing the most important idea in all of statistics Describes the sampling distribution of the sample mean Examples suggest: the sample mean (and sample total) tend to be normally distributed

Important Definition & Theorem
Sampling Distribution of Sample Means If all possible random samples, each of size n, are taken from any population with a mean m and a standard deviation s, the sampling distribution of sample means will: 1. have a mean equal to m 2. have a standard deviation equal to Further, if the sampled population has a normal distribution, then the sampling distribution of will also be normal for samples of all sizes n s m x Central Limit Theorem The sampling distribution of sample means will become normal as the sample size increases.

Summary The mean of the sampling distribution of is equal to the mean of the original population: x m = The standard deviation of the sampling distribution of (also called the standard error of the mean) is equal to the standard deviation of the original population divided by the square root of the sample size: Notes: The distribution of becomes more compact as n increases. (Why?) The variance of : x s n = 2 The distribution of is (exactly) normal when the original population is normal x The CLT says: the distribution of is approximately normal regardless of the shape of the original distribution, when the sample size is large enough! x

Standard Error of the Mean
Standard Error of the Mean: The standard deviation of the sampling distribution of sample means: s x n = Notes: The n in the formula for the standard error of the mean is the size of the sample The proof of the Central Limit Theorem is beyond the scope of this course The following example illustrates the results of the Central Limit Theorem

Graphical Illustration of the Central Limit Theorem
Original Population 30 Distribution of x: n = 2 30 Distribution of x: n = 10 Distribution of x: n = 30

7.3 ~ Applications of the Central Limit Theorem
When the sampling distribution of the sample mean is (exactly) normally distributed, or approximately normally distributed (by the CLT), we can answer probability questions using the standard normal distribution, using the TI 83/84 functions for dealing with the normal distribution, like normcdf and invNorm.

Example 2 Example: Consider a normal population with m = 50 and s = 15. Suppose a sample of size 9 is selected at random. Find: P x ( ) 45 60 . 47 5 1) 2) Solutions: Since the original population is normal, the distribution of the sample mean is also (exactly) normal 1) m x = 50 s x n = 15 9 3 5 2)

Example 2 50 45 60 x - 1.00 2.00 P x z ( ) . 45 60 50 5 1.00 2.00) 3413 4772 8185 = - æ è ç ö ø ÷ + z = ; x -  s n

Example 2 P x z ( . ) £ = - æ è ç ö ø ÷ 47 5 50 5000 1915 3085 z = ;
47.5 x -0.50 P x z ( . ) = - æ è ç ö ø ÷ 47 5 50 5000 1915 3085 z = ; x -  s n

Example 3 Example: A recent report stated that the day-care cost per week in Boston is $109. Suppose this figure is taken as the mean cost per week and that the standard deviation is known to be $20. 1) Find the probability that a sample of 50 day-care centers would show a mean cost of $105 or less per week. 2) Suppose the actual sample mean cost for the sample of 50 day-care centers is $120. Is there any evidence to refute the claim of $109 presented in the report? x Solutions: The shape of the original distribution is unknown, but the sample size, n, is large. The CLT applies. The distribution of is approximately normal m s n = 109 20 50 2 83 .

Example 3 1) x P z ( ) . £ = - æ è ç ö ø ÷ 105 109 2 83 1 41 5000 4207
5000 4207 0793 z = ; x -  s n

Example 3 2) To investigate the claim, we need to examine how likely an observation is the sample mean of $120 Consider how far out in the tail of the distribution of the sample mean is $120 P x z ( ) . = - æ è ç ö ø ÷ 120 109 2 83 3 89 = z = ; x -  s n Since the probability is so small, this suggests the observation of $120 is very rare (if the mean cost is really $109) There is evidence (the sample) to suggest the claim of  = $109 is likely wrong

Chapter 8 ~ Introduction to Statistical Inferences
= z(a/2) s

Chapter Goals Learn the basic concepts of estimation and hypothesis testing Consider questions about a population mean using two methods that assume the population standard deviation is known Consider: what value or interval of values can we use to estimate a population mean? Consider: is there evidence to suggest the hypothesized mean is incorrect?

8.1 ~ The Nature of Estimation
Discuss estimation more precisely What makes a statistic good ? Assume the population standard deviation, s, is known throughout this chapter Concentrate on learning the procedures for making statistical inferences about a population mean m

Point Estimate for a Parameter
Point Estimate for a Parameter: The value of the corresponding statistic Example: is a point estimate (single number value) for the mean m of the sampled population x = 14 7 . How good is the point estimate? Is it high? Or low? Would another sample yield the same result? Note: The quality of an estimation procedure is enhanced if the sample statistic is both less variable and unbiased

Unbiased Statistic Unbiased Statistic: A sample statistic whose sampling distribution has a mean value equal to the value of the population parameter being estimated. A statistic that is not unbiased is a biased statistic. Example: The figures on the next slide illustrate the concept of being unbiased and the effect of variability on a point estimate Assume A is the parameter being estimated

Illustrations Negative bias Under estimate High variability Unbiased
On target estimate Positive bias Over estimate Low variability

Notes x 1. The sample mean, ,is an unbiased statistic because the mean value of the sampling distribution is equal to the population mean: m = 2. Sample means vary from sample to sample. We don’t expect the sample mean to be exactly equal the population mean m. 3. We do expect the sample mean to be close to the population mean 4. Since closeness is measured in standard deviations, we expect the sample mean to be within 2 standard deviations of the population mean

Important Definitions
Interval Estimate: An interval bounded by two values and used to estimate the value of a population parameter. The values that bound this interval are statistics calculated from the sample that is being used as the basis for the estimation. Level of Confidence 1 - a : The probability that the sample to be selected yields an interval that includes the parameter being estimated Confidence Interval: An interval estimate with a specified level of confidence

Summary To construct a confidence interval for a population mean m, use the CLT Use the point estimate as the central value of an interval x x Since the sample mean ought to be within 2 standard deviations of the population mean (95% of the time), we can find the bounds to an interval centered at : - + 2 ( ) s to The level of confidence for the resulting interval is approximately 95%, or 0.95 We can be more accurate in determining the level of confidence

Illustration m x Distribution of The interval is an approximate 95% confidence interval for the population mean m based on this x x - + 2 s to

8.2 ~ Estimation of Mean m (s Known)
Formalize the interval estimation process as it applies to estimating the population mean m based on a random sample Assume the population standard deviation s is known The assumptions are the conditions that need to exist in order to correctly apply a statistical procedure

The Assumption... x The assumption for estimating the mean m using a known s : The sampling distribution of has a normal distribution Assumption satisfied by: 1. Knowing that the sampled population is normally distributed, or 2. Using a large enough random sample (CLT) Note: The CLT may be applied to smaller samples (for example n = 15) when there is evidence to suggest a unimodal distribution that is approximately symmetric. If there is evidence of skewness, the sample size needs to be much larger.

The 1-a Confidence Interval of m
A 1-a confidence interval for m is found by x n - + s to z(a/2) Notes: is the point estimate and the center point of the confidence interval x 2. z(a/2) : confidence coefficient, the number of multiples of the standard error needed to construct an interval estimate of the correct width to have a level of confidence 1- a - z(a/2) z(a/2)

Notes Continued 3. : standard error of the mean
The standard deviation of the distribution of s / n x : maximum error of estimate E One-half the width of the confidence interval (the product of the confidence coefficient and the standard error) n ( / ) s z(a/2) : lower confidence limit (LCL) : upper confidence limit (UCL) x - + n ( / ) s z(a/2)

The Confidence Interval
A Five-Step Model: 1. Describe the population parameter of concern 2. Specify the confidence interval criteria a. Check the assumptions b. Identify the probability distribution and the formula to be used c. Determine the level of confidence, 1 - a 3. Collect and present sample information 4. Determine the confidence interval a. Determine the confidence coefficient b. Find the maximum error of estimate c. Find the lower and upper confidence limits 5. State the confidence interval

Example Example: The weights of full boxes of a certain kind of cereal are normally distributed with a standard deviation of 0.27 oz. A sample of 18 randomly selected boxes produced a mean weight of 9.87 oz. Find a 95% confidence interval for the true mean weight of a box of this cereal. Solution: 1. Describe the population parameter of concern The mean, m, weight of all boxes of this cereal 2. Specify the confidence interval criteria a. Check the assumptions The weights are normally distributed, the distribution of is normal b. Identify the probability distribution and formula to be used Use the standard normal variable z with s = 0.27 c. Determine the level of confidence, 1 - a The question asks for 95% confidence: 1 - a = 0.95 x

Solution Continued 3. Collect and present information The sample information is given in the statement of the problem Given: n x = 18 9 87 ; . 1.15 1.28 1.65 1.96 2.33 2.58 0.75 0.80 0.90 0.95 0.98 0.99 1-  4. Determine the confidence interval a. Determine the confidence coefficient The confidence coefficient is found using Table 4B: z(a/2)

Solution Continued b. Find the maximum error of estimate Use the maximum error part of the formula for a CI E n = s 1 96 27 18 1247 . = z(a/2) c. Find the lower and upper confidence limits Use the sample mean and the maximum error: x n - + s 9 87 1247 7453 9947 75 10 00 to . - z(a/2) + z(a/2) 5. State the confidence interval 9.75 to is a 95% confidence interval for the true mean weight, , of cereal boxes

Example Example: A random sample of the test scores of 100 applicants for clerk-typist positions at a large insurance company showed a mean score of Determine a 99% confidence interval for the mean score of all applicants at the insurance company. Assume the standard deviation of test scores is 10.5. Solution: 1. Parameter of concern The mean test score, m, of all applicants at the insurance company 2. Confidence interval criteria a. Assumptions: The distribution of the variable, test score, is not known. However, the sample size is large enough (n = 100) so that the CLT applies b. Probability distribution: standard normal variable z with  = 10.5 c. The level of confidence: 99%, or 1 -  = 0.99

Solution Continued x = E n ( / ) )( s 10 5 100 709
3. Sample information Given: n = 100 and = 72.6 x 4. The confidence interval a. Confidence coefficient: b. Maximum error: c. The lower and upper limits: z(0.005) . 2 58 = z(a/2) E n ( / ) )( s 10 5 100 709 5. Confidence interval With 99% confidence we say, “The mean test score is between 69.9 and 75.3”, or “69.9 to 75.3 is a 99% confidence interval for the true mean test score” Note: The confidence is in the process. 99% confidence means: if we conduct the experiment over and over, and construct lots of confidence intervals, then 99% of the confidence intervals will contain the true mean value m.

Sample Size Problem: Find the sample size necessary in order to obtain a specified maximum error and level of confidence (assume the standard deviation is known) E = n s z(a/2) Solve this expression for n: n E = × é ë ê ù û ú s 2 z(a/2)

Example Example: Find the sample size necessary to estimate a population mean to within 0.5 with 95% confidence if the standard deviation is 6.2 Solution: Therefore, n = 591 n = é ë ê ù û ú = ( . )( ) 0. 1 96 6 2 5 [24 304] 590 684 E × s z(a/2) Note: When solving for sample size n, always round up to the next largest integer (Why?)

8.3 ~ The Nature of Hypothesis Testing
Formal process for making an inference Consider many of the concepts of a hypothesis test and look at several decision-making situations The entire process starts by identifying something of concern and then formulating two hypotheses about it

Hypothesis Hypothesis: A statement that something is true
Statistical Hypothesis Test: A process by which a decision is made between two opposing hypotheses. The two opposing hypotheses are formulated so that each hypothesis is the negation of the other. (That way one of them is always true, and the other one is always false). Then one hypothesis is tested in hopes that it can be shown to be a very improbable occurrence thereby implying the other hypothesis is the likely truth.

Null & Alternative Hypothesis
There are two hypotheses involved in making a decision: Null Hypothesis, Ho: The hypothesis to be tested. Assumed to be true. Usually a statement that a population parameter has a specific value. The “starting point” for the investigation. Alternative Hypothesis, Ha: A statement about the same population parameter that is used in the null hypothesis. Generally this is a statement that specifies the population parameter has a value different, in some way, from the value given in the null hypothesis. The rejection of the null hypothesis will imply the likely truth of this alternative hypothesis.

Notes 1. Basic idea: proof by contradiction
Assume the null hypothesis is true and look for evidence to suggest that it is false 2. Null hypothesis: the status quo A statement about a population parameter that is assumed to be true 3. Alternative hypothesis: also called the research hypothesis Generally, what you are trying to prove? We hope experimental evidence will suggest the alternative hypothesis is true by showing the unlikeliness of the truth of the null hypothesis

Example Example: Suppose you are investigating the effects of a new pain reliever. You hope the new drug relieves minor muscle aches and pains longer than the leading pain reliever. State the null and alternative hypotheses. Solutions: Ho: The new pain reliever is no better than the leading pain reliever Ha: The new pain reliever lasts longer than the leading pain reliever

Example Example: You are investigating the presence of radon in homes being built in a new development. If the mean level of radon is greater than 4 then send a warning to all home owners in the development. State the null and alternative hypotheses. Solutions: Ho: The mean level of radon for homes in the development is 4 (or less) Ha: The mean level of radon for homes in the development is greater than 4

Hypothesis Test Outcomes
Null Hypothesis Decision True False Fail to reject H o Type A correct decision Type II error Reject Type I error Type B correct decision Type A correct decision: Null hypothesis true, decide in its favor Type B correct decision: Null hypothesis false, decide in favor of alternative hypothesis Type I error: Null hypothesis true, decide in favor of alternative hypothesis Type II error: Null hypothesis false, decide in favor of null hypothesis

Example Example: A calculator company has just received a large shipment of parts used to make the screens on graphing calculators. They consider the shipment acceptable if the proportion of defective parts is 0.01 (or less). If the proportion of defective parts is greater than 0.01 the shipment is unacceptable and returned to the manufacturer. State the null and alternative hypotheses, and describe the four possible outcomes and the resulting actions that would occur for this test. Solutions: Ho: The proportion of defective parts is 0.01 (or less) Ha: The proportion of defective parts is greater than 0.01

Fail To Reject Ho Null Hypothesis Is True: Type A correct decision
Truth of situation: The proportion of defective parts is 0.01 (or less) Conclusion: It was determined that the proportion of defective parts is 0.01 (or less) Action: The calculator company received parts with an acceptable proportion of defectives Null Hypothesis Is False: Type II error Truth of situation: The proportion of defective parts is greater than 0.01 Conclusion: It was determined that the proportion of defective parts is 0.01 (or less) Action: The calculator company received parts with an unacceptable proportion of defectives

Reject Ho Null hypothesis is true: Type I error
Truth of situation: The proportion of defectives is 0.01 (or less) Conclusion: It was determined that the proportion of defectives is greater than 0.01 Action: Send the shipment back to the manufacturer. The proportion of defectives is acceptable Null hypothesis is false: Type B correct decision Truth of situation: The proportion of defectives is greater than 0.01 Conclusion: It was determined that the proportion of defectives is greater than 0.01 Action: Send the shipment back to the manufacturer. The proportion of defectives is unacceptable

Errors Notes: 1. The type II error sometimes results in what represents a lost opportunity 2. Since we make a decision based on a sample, there is always the chance of making an error Probability of a type I error = a Probability of a type II error = b Error in Decision Type Probability Rejection of a true H o I a Failure to reject a false II b Correct Decision Type Probability Failure to reject a true H o A 1 - a Rejection of a false B b

Notes 1. Would like a and b to be as small as possible
2. a and b are inversely related 3. Usually set a (and don’t worry too much about b. Why?) 4. Most common values for a and b are 0.01 and 0.05 b : the power of the statistical test A measure of the ability of a hypothesis test to reject a false null hypothesis 6. Regardless of the outcome of a hypothesis test, we never really know for sure if we have made the correct decision

Interrelationship Interrelationship between the probability of a type I error (a), the probability of a type II error (b), and the sample size (n) a error) I (type P b II

Level of Significance & Test Statistic
Level of Significance, a : The probability of committing the type I error Test Statistic: A random variable whose value is calculated from the sample data and is used in making the decision fail to reject Ho or reject Ho Notes: The value of the test statistic is used in conjunction with a decision rule to determine fail to reject Ho or reject Ho The decision rule is established prior to collecting the data and specifies how you will reach the decision

The Conclusion a. If the decision is reject Ho, then the conclusion should be worded something like, “There is sufficient evidence at the a level of significance to show that (the meaning of the alternative hypothesis)” b. If the decision is fail to reject Ho, then the conclusion should be worded something like, “There is not sufficient evidence at the a level of significance to show that (the meaning of the alternative hypothesis)” Notes: The decision is about Ho The conclusion is a statement about Ha There is always the chance of making an error

8.4 ~ Hypothesis Test of Mean  ( known): A Probability-Value Approach
The concepts and much of the reasoning behind hypothesis tests are given in the previous sections Formalize the hypothesis test procedure as it applies to statements concerning the mean  of a population ( known): a probability-value approach

The Assumption... The assumption for hypothesis tests about a mean m using a known s : The sampling distribution of has a normal distribution x Recall: 1. The distribution of has mean m 2. The distribution of has standard deviation x n s Hypothesis test: 1. A well-organized, step-by-step procedure used to make a decision 2. Probability-value approach (p-value approach): a procedure that has gained popularity in recent years. Organized into five steps.

The Probability-Value Hypothesis Test
A Five-Step Procedure: 1. The Set-Up a. Describe the population parameter of concern b. State the null hypothesis (Ho) and the alternative hypothesis (Ha) 2. The Hypothesis Test Criteria a. Check the assumptions b. Identify the probability distribution and the test statistic formula to be used c. Determine the level of significance, a 3. The Sample Evidence a. Collect the sample information b. Calculate the value of the test statistic 4. The Probability Distribution a. Calculate the p-value for the test statistic b. Determine whether or not the p-value is smaller than a 5. The Results a. State the decision about Ho b. State a conclusion about Ha

Example Example: A company advertises the net weight of its cereal is 24 ounces. A consumer group suspects the boxes are underfilled. They cannot check every box of cereal, so a sample of cereal boxes will be examined. A decision will be made about the true mean weight based on the sample mean. State the consumer group’s null and alternative hypotheses. Assume s = 0.2 Solution: 1. The Set-Up a. Describe the population parameter of concern The population parameter of interest is the mean , the mean weight of the cereal boxes

Solution Continued b. State the null hypothesis (Ho) and the alternative hypothesis (Ha) Formulate two opposing statements concerning m Ho: m = 24 ( ) (the mean is at least 24) Ha: m < 24 (the mean is less than 24) Note: The trichotomy law from algebra states that two numerical values must be related in exactly one of three possible relationships: <, =, or >. All three of these possibilities must be accounted for between the two opposing hypotheses in order for the hypotheses to be negations of each other.

Possible Statements of Null & Alternative Hypotheses
Notes: The null hypothesis will be written with just the equal sign (a value is assigned) When equal is paired with less than or greater than, the combined symbol is written beside the null hypothesis as a reminder that all three signs have been accounted for in these two opposing statements.

Examples Example: An automobile manufacturer claims a new model gets at least 27 miles per gallon. A consumer groups disputes this claim and would like to show the mean miles per gallon is lower. State the null and alternative hypotheses. Solution: Ho: m = 27 (³) and Ha: m < 27 Example: A freezer is set to cool food to If the temperature is higher, the food could spoil, and if the temperature is lower, the freezer is wasting energy. Random freezers are selected and tested as they come off the assembly line. The assembly line is stopped if there is any evidence to suggest improper cooling. State the null and alternative hypotheses. Solution: Ho: m = and Ha: m ¹ 10

Common Phrases & Their Negations
) ( : o H = at least less than no less than not less than at most more than no more than not greater than greater than is is not not different from different from same as not same as

Example Continued Example Continued: Weight of cereal boxes
Recall: Ho: m = 24 (³) (at least 24) Ha: m < 24 (less than 24) 2. The Hypothesis Test Criteria a. Check the assumptions The weight of cereal boxes is probably mound shaped. A sample size of 40 should be sufficient for the CLT to apply. The sampling distribution of the sample mean can be expected to be normal. b. Identify the probability distribution and the test statistic to be used To test the null hypothesis, ask how many standard deviations away from m is the sample mean n x z s m - = * : statistic test

Solution Continued c. Determine the level of significance 40 and 95 .
Let a = 0.05 3. The Sample Evidence a. Collect the sample information A random sample of 40 cereal boxes is examined 40 and 95 . 23 = n x 5811 1 2 0. 24 * - z s m b. Calculate the value of the test statistic (s = 0.2) 4. The Probability Distribution a. Calculate the p-value for the test statistic

Probability-Value or p-Value
Probability-Value, or p-Value: The probability that the test statistic could be the value it is or a more extreme value (in the direction of the alternative hypothesis) when the null hypothesis is true (Note: the symbol P will be used to represent the p-value, especially in algebraic situations) P 0571 . 4429 0.5000 ) 58 1 ( *) P = - > < z

Solution Continued b. Determine whether or not the p-value is smaller than a The p-value (0.0571) is greater than a (0.05) 5. The Results Decision Rule: a. If the p-value is less than or equal to the level of significance a, then the decision must be to reject Ho b. If the p-value is greater than the level of significance a, then the decision must be to fail to reject Ho a. State the decision about Ho Decision about Ho : Fail to reject Ho b. Write a conclusion about Ha There is not sufficient evidence at the 0.05 level of significance to show that the mean weight of cereal boxes is less than 24 ounces

Notes If we fail to reject Ho, there is no evidence to suggest the null hypothesis is false. This does not mean Ho is true. The p-value is the area, under the curve of the probability distribution for the test statistic, that is more extreme than the calculated value of the test statistic. There are 3 separate cases for p-values. The direction (or sign) of the alternative hypothesis is the key.

Finding p-Values 1. Ha contains > (Right tail)
p-value = P(z > z*) 2. Ha contains < (Left tail) p-value = P(z < z*) 3. Ha contains (Two-tailed) p-value = P(z < -|z*|) + P(z > |z*|)

Example Example: The mean age of all shoppers at a local jewelry store is 37 years (with a standard deviation of 7 years). In an attempt to attract older adults with more disposable income, the store launched a new advertising campaign. Following the advertising, a random sample of 47 shoppers showed a mean age of Is there sufficient evidence to suggest the advertising campaign has succeeded in attracting older customers? Solution: 1. The Set-Up a. Parameter of concern: the mean age, m, of all shoppers b. The hypotheses: Ho: m = 37 (£) Ha: m > 37

Solution Continued 2. The Hypothesis Test Criteria
a. The assumptions: The distribution of the age of shoppers is unknown. However, the sample size is large enough for the CLT to apply. b. The test statistic: The test statistic will be z* c. The level of significance: none given We will find a p-value 3 . 39 , 47 = x n 25 2 7 37 * - z s m 3. The Sample Evidence a. Sample information: b. Calculated test statistic:

Solution Continued 4. The Probability Distribution a. The p-value:
b. Determine whether or not the p-value is smaller than a A comparison is not possible, no a given 5. The Results Because the p-value is so small (P < 0.05), there is evidence to suggest the mean age of shoppers at the jewelry store is greater than 37

p-Value The idea of the p-value is to express the degree of belief in the null hypothesis: 1. When the p-value is minuscule (like ), the null hypothesis would be rejected by everyone because the sample results are very unlikely for a true Ho 2. When the p-value is fairly small (like 0.01), the evidence against Ho is quite strong and Ho will be rejected by many 3. When the p-value begins to get larger (say, 0.02 to 0.08), there is too much probability that data like the sample involved could have occurred even if Ho were true, and the rejection of Ho is not an easy decision 4. When the p-value gets large (like 0.15 or more), the data is not at all unlikely if the Ho is true, and no one will reject Ho

p-Value Advantages & Disadvantage
Advantages of p-value approach: 1. The results of the test procedure are expressed in terms of a continuous probability scale from 0.0 to 1.0, rather than simply on a reject or fail to reject basis 2. A p-value can be reported and the user of the information can decide on the strength of the evidence as it applies to his/her own situation 3. Computers can do all the calculations and report the p-value, thus eliminating the need for tables Disadvantage: 1. Tendency for people to put off determining the level of significance

Example Example: The active ingredient for a drug is manufactured using fermentation. The standard process yields a mean of grams (assume s = 3.2). A new mixing technique during fermentation is implemented. A random sample of 32 batches showed a sample mean Is there any evidence to suggest the new mixing technique has changed the yield? Solution: 1. The Set-Up a. The parameter of interest is the mean yield of active ingredient, m b. The null and alternative hypotheses: H0: m = 26.5 Ha: m ¹ 26.5

Solution Continued 2 The Hypothesis Test Criteria
. a. Assumptions: A sample of size 32 is large enough to satisfy the CLT b. The test statistic: z* c. The level of significance: find a p-value 06 . 1 32 2 3 5 26 27 * = - n x z s m 3. The Sample Evidence a. From the sample: b. The calculated test statistic: ,

Solution Continued 4. The Probability Distribution a. The p-value:
b. The p-value is large There is no a given in the statement of the problem 5. The Results Because the p-value is large (P = ), there is no evidence to suggest the new mixing technique has changed the mean yield

8.5 ~ Hypothesis Test of mean  ( known): A Classical Approach
Concepts and reasoning behind hypothesis testing given in Section 8.3 Formalize the hypothesis test procedure as it applies to statements concerning m of a population with known s : a classical approach

The Assumption... x The assumption for hypothesis tests about mean m using a known s : The sampling distribution of has a normal distribution Recall: 1. The distribution of has mean m 2. The distribution of has standard deviation n s x Hypothesis Test: A well-organized, step-by-step procedure used to make a decision. The classical approach is the hypothesis test process that has enjoyed popularity for many years.

The Classical Hypothesis Test
A Five-Step Procedure: 1. The Set-Up a. Describe the population parameter of concern b. State the null hypothesis (Ho) and the alternative hypothesis (Ha) 2. The Hypothesis Test Criteria a. Check the assumptions b. Identify the probability distribution and the test statistic to be used c. Determine the level of significance, a 3. The Sample Evidence a. Collect the sample information b. Calculate the value of the test statistic 4. The Probability Distribution a. Determine the critical region(s) and critical value(s) b. Determine whether or not the calculated test statistic is in the critical region 5. The Results a. State the decision about Ho b. State the conclusion about Ha

Example 1. The Set-Up a. Describe the population parameter of concern
Example: A company advertises the net weight of its cereal is 24 ounces. A consumer group suspects the boxes are underfilled. They cannot check every box of cereal, so a sample of cereal boxes will be examined. A decision will be made about the true mean weight based on the sample mean. State the consumer group’s null and alternative hypotheses. Assume s = 0.2 Solution: 1. The Set-Up a. Describe the population parameter of concern The population parameter of interest is the mean, m, the mean weight of the cereal boxes

Solution Continued b. State the null hypothesis (Ho) and the alternative hypothesis (Ha) Formulate two opposing statements concerning the m: Ho: m = 24 ( ) (the mean is at least 24) Ha: m < 24 (the mean is less than 24) Note: The trichotomy law from algebra states that two numerical values must be related in exactly one of three possible relationships: <, =, or >. All three of these possibilities must be accounted for between the two opposing hypotheses in order for the hypotheses to be negations of each other.

Possible Statements of Null & Alternative Hypotheses
Notes: The null hypothesis will be written with just the equal sign (a value is assigned) When equal is paired with less than or greater than, the combined symbol is written beside the null hypothesis as a reminder that all three signs have been accounted for in these two opposing statements

Examples Example: An automobile manufacturer claims a new model gets at least 27 miles per gallon. A consumer groups disputes this claim and would like to show the mean miles per gallon is lower. State the null and alternative hypotheses. Solution: Ho: m = 27 (³) and Ha: m < 27 Example: A freezer is set to cool food to If the temperature is higher, the food could spoil, and if the temperature is lower, the freezer is wasting energy. Random freezers are selected and tested as they come off the assembly line. The assembly line is stopped if there is any evidence to suggest improper cooling. State the null and alternative hypotheses. Solution: Ho: m = and Ha: m ¹ 10

Common Phrases & Their Negations
) ( : o H = at least less than no less than not less than at most more than no more than not greater than greater than is is not not different from different from same as not same as

Example Continued m s n x z - = * : statistic test Solution Continued:
Example (continued): Weight of cereal boxes Recall: Ho: m = 24 (>) (at least 24) Ha: m < 24 (less than 24) n x z s m - = * : statistic test 2. The Hypothesis Test Criteria a. Check the assumptions The weight of cereal boxes is probably mound shaped. A sample size of 40 should be sufficient for the CLT to apply. The sampling distribution of the sample mean can be expected to be normal. Solution Continued: b. Identify the probability distribution and the test statistic to be used To test the null hypothesis, ask how many standard deviations away from m is the sample mean

Solution Continued 40 and 95 . 23 = n x
c. Determine the level of significance Consider the four possible outcomes and their consequences Let a = 0.05 3. The Sample Evidence a. Collect the sample information A random sample of 40 cereal boxes is examined b. Calculate the value of the test statistic (s = 0.2) 40 and 95 . 23 = n x 5811 1 2 0. 24 * - z s m 4. The Probability Distribution a. Determine the critical region(s) and critical value(s)

Critical Region & Critical Value(s)
Critical Region: The set of values for the test statistic that will cause us to reject the null hypothesis. The set of values that are not in the critical region is called the noncritical region (sometimes called the acceptance region). Critical Value(s): The first or boundary value(s) of the critical region(s)

Critical Region & Critical Value(s)
Illustration: Critical Region Critical Value

Solution Continued * 4. The Probability Distribution (Continued)
b. Determine whether or not the calculated test statistic is in the critical region The calculated value of z, z* = -1.58, is in the noncritical region * 5. The Results We need a decision rule

Decision Rule Decision Rule:
a. If the test statistic falls within the critical region, we will reject Ho (the critical value is part of the critical region) b. If the test statistic is in the noncritical region, we will fail to reject Ho a. State the decision about Ho Decision: Fail to reject Ho b. State the conclusion about Ha Conclusion: There is not enough evidence at the 0.05 level of significance to show that the mean weight of cereal boxes is less than 24

Notes 1. The null hypothesis specifies a particular value of a population parameter 2. The alternative hypothesis can take three forms. Each form dictates a specific location of the critical region(s) 3. For many hypothesis tests, the sign in the alternative hypothesis points in the direction in which the critical region is located 4. Significance level: a

Example Example: The mean water pressure in the main water pipe from a town well should be kept at 56 psi. Anything less and several homes will have an insufficient supply, and anything greater could burst the pipe. Suppose the water pressure is checked at 47 random times. The sample mean is (Assume s = 7). Is there any evidence to suggest the mean water pressure is different from 56? Use a = 0.01 Solution: 1. The Set-Up a. Describe the parameter of concern: The mean water pressure in the main pipe b. State the null and alternative hypotheses Ho: m = 56 Ha: m ¹ 56

a. Check the assumptions: A sample of n = 47 is large enough for the CLT to apply b. Identify the test statistic The test statistic is z* c. Determine the level of significance: a = 0.01 (given) 3. The Sample Evidence a. The sample information: b. Calculate the value of the test statistic: 47 , 1 . 57 = n x 077 7 56 * - z s m

Solution Continued * 4. The Probability Distribution
a. Determine the critical regions and the critical values b. Determine whether or not the calculated test statistic is in the critical region The calculated value of z, z* = 1.077, is in the noncritical region *

Solution Continued 5. The Results a. State the decision about Ho:
Fail to reject Ho b. State the conclusion about Ha: There is no evidence to suggest the water pressure is different from 56 at the 0.01 level of significance

Example Solution: 1. The Set-Up
Example: An elementary school principal claims students receive no more than 30 minutes of homework each night. A random sample of 36 students showed a sample mean of minutes spent doing homework (assume s = 7.5). Is there any evidence to suggest the mean time spent on homework is greater than 30 minutes? Use a = 0.05 Solution: 1. The Set-Up The parameter of concern: m, the mean time spent doing homework each night Ho: m = 30 (£) Ha: m > 30

a. The sample size is n = 36, the CLT applies b. The test statistic is z* c. The level of significance is given: a = 0.01 44 . 5 36 7 30 8 * = - n x z s m 3. The Sample Evidence

Solution Continued 4. The Probability Distribution *
The calculated value of z, z* = 5.44, is in the critical region

Solution Continued 5. The Results Decision: Reject Ho
Conclusion: There is sufficient evidence at the 0.01 level of significance to conclude the mean time spent on homework by the elementary students is more than 30 minutes Note: Suppose we took repeated sample of size 36. What would you expect to happen?

Chapter 1: Statistics Note: The textbook illustrates statistical procedures using MINITAB, EXCEL 97, and the TI-83.

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Chapter 1: Statistics Note: The textbook illustrates statistical procedures using MINITAB, EXCEL 97, and the TI-83.

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Presentation on theme: "Chapter 1: Statistics Note: The textbook illustrates statistical procedures using MINITAB, EXCEL 97, and the TI-83."— Presentation transcript:

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