Presentation on theme: "Lecture 4 PY 427 Statistics 1 Fall 2006 Kin Ching Kong, Ph.D"— Presentation transcript:
1 Lecture 4 PY 427 Statistics 1 Fall 2006 Kin Ching Kong, Ph.D Chicago School of Professional PsychologyLecture 4Kin Ching Kong, Ph.D
2 Agenda Probability Probability & The Normal Distribution Definition Simple Random SampleProbability and Frequency DistributionsProbability & The Normal DistributionThe Normal DistributionThe Unit Normal TableProbabilities and z-ScoresProbabilities for Scores From a Normal DistributionPreview of Inferential Statistics
3 ProbabilityDefinition: When several different outcomes are possible, we define the probability for any particular outcome as a fraction or proportion. If the possible outcome are identified as A, B, C, D, and so on, then:probability of A = number of outcomes classified as A total number of possible outcomesNotation: probability of A = p(A)Inferential procedures are built around the concept of probability. That is, the relationships between samples and populations are usually defined in terms of probability.
4 Simple Random Sample Assumptions of Simple Random Sample: Each individual in the population have an equal chance of being selected. (Each individual outcome has equal probability)If more than one individual is to be selected for the sample, there must be constant probability for each and every selection.Solution: random sampling with replacement
5 Probability and Frequency Distributions In a frequency distribution graph of a population of scores, probabilities are represented by proportions of the graph. That is, the probability of any score or set of scores correspond to the proportion of the graph associated with the scores.E.g.N = 10 Scores:1, 1, 2, 3, 3, 4, 4, 4, 5, 6Figure 6.2 of your bookp (X > 4) = ?p (X = 4) = ?
6 Probability & the Normal Distribution Figure 6.3 of your bookProportions in a normal distributionFigure 6.4 of your bookProbability questions about a normal distributione.g: Adult heights form a normal distribution with a mean of 68 inches and a standard deviation of 6 inches. What is the probability of randomly selecting an individual who is taller than 6 feet 8 inches (80 inches)p(X > 80) = ?z = (X – m)/sz = (80 – 68)/6 = 2Figure 6.5 of your bookp(z > 2.00) = 2.28% or .0228
7 The Unit Normal TableThe unit normal table lists proportions of the normal distribution for a full range of z-scores (Appendix B, Table B.1).Figure 6.6 of your bookThe body (B) is always the larger part of the distribution.The tail (C) is always the smaller part of the distributionThe normal distribution is symmetrical, so the proportions on the right-hand side are exactly the same as the corresponding proportions on the left-hand side.We have defined probability = proportion, so the numbers in the unit normal table are also probabilities.
8 Using the Unit Normal Table to Find Proportions or Probabilities Exercise 1: What proportion of the normal distribution corresponds to z-score values greater than z = 1.00? Or, for a normal distribution, what is p(z > 1.00)?Exercise 2: For a normal distribution, what is the probability of selecting a z-score less than z = 1.50?Exercise 3: For a normal distribution, what is p (z < -0.50)?Figure 6.7 of your bookAnswers
9 Using the Unit Normal Table to Find z-Scores The Unit Normal Table can also be used to find z-score locations for specific proportions.Exercise 1: For a normal distribution, what z-score separates the top 10% from the rest of the distribution?Exercise 2: For a normal distribution, what z-score values form the boundaries for the middle 60% of the distribution?Figure 6.8 of your bookAnswers
10 Using the Unit Normal Table w/ X values Most of the time, the Unit Normal Table is used to find probabilities for specific X values.Step 1: Transform the X value into a z-scoreStep 2: Find the probability associated with the z-score in the unit normal table.E.g.: IQ scores form a normal distribution with m = 100 and s =15, what is the probability of randomly selecting an individual with an IQ score less than 130?p( X < 130)?z = (X – m)/s= (130 – 100)/15= 30/15z = 2p(z < 2) =.9772p(X < 130) = .9772
11 More Examples Finding probabilities between two scores: e.g.: The highway dept. found that the average speed is m = 58miles/hour with a standard deviation of s = 10 for a localsection of the interstate highway. The distribution wasapproximately normal. What proportion of the cars aretraveling between 55 and 65 miles/hour?p(55 < X <65)For X = 55: z = (X – m)/s= (55 – 58)/10= -3/10 = -0.30For X = 65 z = (65 -58)/10= 7/10 =0.70p(55 < X < 65) = p(-0.30 < z < 0.70)==Figure 6.10 of your book
12 More ExamplesTo find proportions/probabilities for specific X values and to find X values for specific proportions, finding z-scores is a necessary intermediate step.Figure 6.11 of your bookExample of going from proportion to X value:Scores on the SAT form a normal distribution with m = 500 and s = What is the minimum score necessary to be in the top 15%?Figure 6.12 of your bookp(X >?) = 0.15p(z > ?) = 0.15, z = +1.04z = (X – m)/s1.04 = (X -500)/1001.04(100) = X -500104 = X – 500= X604 = X, you must have a score of at least 604
13 Your Turn1. For a normal distribution with a mean of 80 and a standard deviation of 10, what is the probability of randomly selecting a score greater than 85?Answer2. For a normal distribution with a mean of 100 and standard deviation of 20, what is the minimum score needed to be in the top 5% of the distribution?
14 Preview of Inferential Statistics Experiment at the end of Ch. 5last slide of lecture 3Probability helps us decide exactly where to set the boundaries for what is considered extreme valuesFigure 6.14 of your book