Class 13 Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science Copyright 017 by D.B. Rowe 1
Agenda: Recap Chapter 6.3 6.5 Lecture Chapter 7.1 7. Review Chapter 5 for Eam 3. Problem Solving Session.
Recap Chapter 6.3-6.5 3
6: Normal Probability Distributions 6.3 Applications of Normal Distributions Eample: Assume that IQ scores are normally distributed with a mean μ of 100 and a standard deviation σ of 16. If a person is picked at random, what is the probability that his or her IQ is between 100 and 115? i.e. P(100 115)? μ μ Figures from Johnson & Kuby, 01. 4
6: Normal Probability Distributions 6.3 Applications of Normal Distributions IQ scores normally distributed μ=100 and σ=16. P(100 115) z z 1 100 115 100 100 16 1 1 0 z 115 100 16 0.94 Figures from Johnson & Kuby, 01. 5
6: Normal Probability Distributions 6.3 Applications of Normal Distributions Now we can use the table. = - P(0 z 0.94) P( z 0.94) P( z 0) 0.864.5 0.364 Figures from Johnson & Kuby, 01. 6
6: Normal Probability Distributions 6.4 Notation Eample: Let α=0.05. Let s find z(0.05). P(z>z(0.05))=0.05. P(z>z(0.05)) Same as finding P(z<z(0.05))=1-0.05. z 0.95 0.05=P(z>z(0.05)) Figures from Johnson & Kuby, 01. 7
6: Normal Probability Distributions 6.4 Notation Eample: Same as finding P(z<z(0.05))=0.95. 1.645 Figures from Johnson & Kuby, 01. 8
6: Normal Probability Distributions 6.5 Normal Approimation of the Binomial Distribution Approimate binomial probabilities with normal areas. Use a normal with np, np(1 p) n=14 p=1/ (14)(.5) 7 (14)(.5)(1.5) 3.5 Figures from Johnson & Kuby, 01. 9
6: Normal Probability Distributions 6.5 Normal Approimation of the Binomial Distribution n=14, p=1/ We then approimate binomial probabilities with normal areas. P ( 4) from the binomial formula is approimately P(3.5 4.5) from the normal with 7, 3.5 the ±.5 is called a continuity correction Figures from Johnson & Kuby, 01. 10
6: Normal Probability Distributions 6.5 Normal Approimation of the Binomial Distribution From the binomial formula 14! P(4) (.5) (1.5) 4!(14 4)! P ( 4) 0.061 4 144 P( 1.87 z 1.34) 0.0594 P( 1.87 z 1.34) Pz ( 1.34) From the Normal Distribution n=14, p=1/ P(3.5 4.5) 7, 3.5 1 3.5 7 1.87 z1-1.87 1.87 4.5 7 z 1.34 1.87 Pz ( 1.87) 0.0594 0.0901 0.0307 11
6: Normal Probability Distributions Questions? Homework: Chapter 6 # 7, 9, 13, 17, 19, 9, 31, 33, 41, 45, 47, 53, 61, 75, 95, 99 Read Chapter 7. 1
Lecture Chapter 7.1-7. 13
Chapter 7: Sample Variability Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science 14
7: Sample Variability 7. The Sampling Distribution of Sample Means When we take a random sample 1,, n from a population, one of the things that we do is compute the sample mean. The value of is not μ. Each time we take a random sample of size n, we get a different set of values 1,, n and a different value for. 15
7: Sample Variability 7. The Sampling Distribution of Sample Means Recall: When we take a sample of data 1,, n from a population, then compute an estimate of a parameter it is called a sample statistic. i.e. for μ 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. 16
7: Sample Variability 7. The Sampling Distribution of Sample Means Let s discuss the relationship between the sample mean and the population mean. Assume that we have a population of items with population mean μ and population standard deviation σ. If we take a random sample of size n and compute sample mean,. The collection of all possible means is called the sampling distribution of the sample mean. 17
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n=1 with replacement. S={ } 0 4 6 8 Prob. of each value = 18
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n=1 with replacement. Population data values: 0,, 4, 6, 8. 5 possible values 0 4 6 8 0 8 4 6 S={0,, 4, 6, 8} 0, occurs one time, occurs one time 4, occurs one time 6, occurs one time 8, occurs one time Prob. of each value = 1/5 = 0. 19
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n=1 with replacement. Population data values: 0,, 4, 6, 8. 0 4 6 8 P( ) 0 1/ 5 1/ 5 4 1/ 5 6 1/ 5 8 1/ 5 5 possible values P() 0. 0.16 0.1 0.08 0.04 0 0 4 6 8 0
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n=1 with replacement. Population data values: 0,, 4, 6, 8. 0 4 6 8 5 possible values P( ) 0 1/ 5 1/ 5 4 1/ 5 6 1/ 5 8 1/ 5 [ P( )] 1
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n=1 with replacement. Population data values: 0,, 4, 6, 8. 0 4 6 8 5 possible values P( ) [( ) P( )] 0 1/ 5 1/ 5 4 1/ 5 6 1/ 5 8 1/ 5 3
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n= with replacement. Population data values: 0 4 6 8 0 8 4 6 0 4 6 8 0,, 4, 6, 8. (0,0) (,0) (4,0) (6,0) (8,0) 0 4 6 8 0 4 6 8 0 4 6 8 0 4 6 8 (0,) (,) (4,) (6,) (8,) (0,4) (,4) (4,4) (6,4) (8,4) (0,6) (,6) (4,6) (6,6) (8,6) (0,8) (,8) (4,8) (6,8) (8,8) 5 possible samples 5
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: There are N=5 items in the population. Population data values: 0,, 4, 6, 8. Take samples of size n= (with replacement). There are 5 possible samples. Each sample has mean. (0,0) (,0) (4,0) (6,0) (8,0)????? (0,) (,) (4,) (6,) (8,) (0,4) (,4) (4,4) (6,4) (8,4) (0,6) (,6) (4,6) (6,6) (8,6) (0,8) (,8) (4,8) (6,8) (8,8)???????????????????? 6
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5, values: 0,, 4, 6, 8, n= (with replacement). 5 possible samples. Each possible sample is equally likely. Prob. of each sample = 1/5 = 0.04 P[( i, j)] 1/ 5 i 0,,4,6,8 j 0,,4,6,8 There are 5 possible samples. (0,0) (,0) (4,0) (6,0) (8,0) (0,) (,) (4,) (6,) (8,) (0,4) (,4) (4,4) (6,4) (8,4) (0,6) (,6) (4,6) (6,6) (8,6) (0,8) (,8) (4,8) (6,8) (8,8) 8
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5, values: 0,, 4, 6, 8, n= (with replacement). 5 possible samples. Each possible sample is equally likely. Prob. of each samples mean = 1/5 = 0.04????????????????????????? There are 5 possible samples. (0,0) (,0) (4,0) (6,0) (8,0) (0,) (,) (4,) (6,) (8,) (0,4) (,4) (4,4) (6,4) (8,4) (0,6) (,6) (4,6) (6,6) (8,6) (0,8) (,8) (4,8) (6,8) (8,8) 9
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5, values: 0,, 4, 6, 8, n= (with replacement). 5 possible samples. Prob. of each samples mean = 1/5 = 0.04??????????????????????????, occurs times?, occurs times?, occurs times?, occurs times?, occurs times?, occurs times?, occurs times?, occurs times?, occurs times 31
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5, values: 0,, 4, 6, 8, n= (with replacement). 5 possible samples. Prob. of each samples mean = 1/5 = 0.04????????????????????????? P (?) P (?) P (?) P (?) P (?) P (?) P (?) P (?) P (?) 33
7: Sample Variability 7. The Sampling Distribution of Sample Means Don t forget that the two values that we draw are random. That is, we may know the sample space of possible outcomes but we do not know eactly which ones we will get! Random Sample: A sample obtained in such a way that each possible sample of fied size n has an equal probability of being selected. 36
7: Sample Variability 7. The Sampling Distribution of Sample Means if samples increases the empirical dist. turns into theoretical dist. empirical distribution true distribution with population parameters, Figure from Johnson & Kuby, 01. As the number of samples increases the empirical dist. turns into theoretical dist. 37
7: Sample Variability 7. The Sampling Distribution of Sample Means Sample distribution of sample means (SDSM): If all possible random samples, each of size n, are taken from any population with mean μ and standard deviation σ, then the sampling distribution of sample means will have the following: 1. A mean equal to μ. A standard deviation equal to n Furthermore, if the sampled population has a normal distribution, then the sampling distribution of will also be normal for all samples of all sizes. Discuss Later: What if the sampled population does not have a normal distribution? 38
7: Sample Variability 7. The Sampling Distribution of Sample Means umber if samples increases the empirical dist. turns into theoretical dist. empirical distribution 0 4 6 8 6 Sample 1 0 4 Sample 1 parameter of interest, μ true distribution with population parameters, 8 Sample 3 4 6 All Other Samples many more values portion of Figure from Johnson & Kuby, 01. As the number of samples increases the empirical dist. turns into theoretical dist. 3 39
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5, values: 0,, 4, 6, 8, n= (with replacement). Instead of drawing two values with replacement and computing the sample mean, we can think of this as drawing one of the sample means with replacement. The probability for each sample mean is 40
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5, values: 0,, 4, 6, 8, n= (with replacement). P( ) P (?) P (?) P (?) P (?) P (?) P (?) P (?) P (?) P (?) 4
7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5, values: 0,, 4, 6, 8, n= (with replacement). ( ) P( ) P (?) P (?) P (?) P (?) P (?) P (?) P (?) P (?) P (?) 44
7: Sample Variability Questions? Homework: Chapter 7 # 6, 1, 3, 9, 33, 35 46
Review Chapters 5 (Eam 3 Chapter) Just the highlights! 47
5: Probability Distributions (Discrete Variables) 5. Probability Distributions of a Discrete Random Variable Random Variables: assumes a unique value for each of the outcomes in the sample space. Probability Function: A rule P() that assigns probabilities to the values of the random variable. Eample: Let = # of heads when we flip a coin twice. ={0,1,}! 1 1 P ( )!( )! 0 1 P( ) 1 4 1 1 4 48
5: Probability Distributions (Discrete Variables) 5. Mean and Variance of a Discrete Random Variable Mean of a discrete random variable (epected value): The mean, μ, of a discrete random variable is found by multiplying each possible value of by its own probability, P(), and then adding all of the products together: mean of : mu = sum of (each multiplied by its own probability) n [ ip( i)] i1 (5.1) 49
5: Probability Distributions (Discrete Variables) 5. Mean and Variance of a Discrete Random Variable n [ i P( i )] 1P( 1 ) P( )... np( n) i1 For the # of H when we flip a coin twice discrete distribution: μ = ( 1 ) P( 1 ) + ( ) P( ) + ( 3 ) P( 3 ) μ = (0) P(0) + (1) P(1) + () P() μ = (0) (1/4) + (1) (1/) + () (1/4) μ = 0 + 1/ + 1/ μ = 1 1 3 0 1 P( ) 1 4 1 1 4 P ( 1) P ( ) P ( 3) 50
5: Probability Distributions (Discrete Variables) 5. Mean and Variance of a Discrete Random Variable Variance of a discrete random variable: The variance, σ, of a discrete random variable is found by multiplying each possible value of the squared deviation, ( μ), by its own probability, P(), and then adding all of the products together: variance of : sigma squared = sum of (squared deviation times probability) n i i1 equivalent formula [( ) P( )] i n [ i P( i)] i1 (5.) (5.3b) 51
5: Probability Distributions (Discrete Variables) 5. Mean and Variance of a Discrete Random Variable n i P i 1 P 1 P n P n i1 [( ) ( )] ( ) ( ) ( ) ( )... ( ) ( ) For the # of H when we flip a coin twice discrete distribution: σ = ( 1 -μ) P( 1 ) + ( -μ) P( ) + ( 3 -μ) μ = 1 P( 3 ) σ = (0-1) P(0) + (1-1) P(1) + (-1) P() σ = (-1) (1/4) + (0) (1/) + (1) (1/4) σ = 1/4 + 0 + 1/4 σ = 1/ 1/ 1 3 0 1 P( ) 1 4 1 1 4 P ( 1) P ( ) P ( 3) 5
5: Probability Distributions (Discrete Variables) 5.3 The Binomial Probability Distribution An eperiment with only two outcomes is called a Binomial ep. Call one outcome Success and the other Failure. Each performance of ept. is called a trial and are independent. Prob of eactly successes n! P( ) p (1 p)!( n )! n = number of trials or times we repeat the eperiment. = the number of successes out of n trials. p = the probability of success on an individual trial. n num( successes) P( successes and n- failures) Bi means two like bicycle 0,..., n (5.5) n n!!( n )! 53
5: Probability Distributions (Discrete Variables) 5.3 The Binomial Probability Distribution Flip coin ten times. n! P( ) p (1 p)!( n )! n=10, =7, p=.5 7 107 10 9 8 7! 1 1 P(7) 7!3! n 7 107 10! 1 1 P(7) 1 7!(10 7)! 10 = # of Heads n()= ways to get Heads 10 33 4 1 P(7) 10 3 P(7) 104 54
5: Probability Distributions (Discrete Variables) 5.3 The Binomial Probability Distribution Flip coin ten times. = # of Heads n()= ways to get Heads 0 1 3 4 5 6 7 8 9 10 n ( ) 1 10 45 10 10 5 10 10 45 10 1 P ( ) 1 10 45 10 10 5 10 10 45 10 1 104 104 104 104 104 104 104 104 104 104 104 n=10 =0,,10 p=1/ n ( ) n!!( n )! n p (1 p) 1/104 n! P( ) p (1 p)!( n )! n Note: 1. 0 P() 1. ΣP()=1 55
5: Probability Distributions (Discrete Variables) 5.3 The Binomial Probability Distribution page page 713660 n=10, p=1/ n! P( ) p (1 p)!( n )! n Figure from Johnson & Kuby, 01. P ( ) 0 1 3 4 5 6 7 8 9 10 1 10 45 10 10 5 10 10 45 10 1 104 104 104 104 104 104 104 104 104 104 104 56
5: Probability Distributions (Discrete Variables) 5.3 The Binomial Probability Distribution Eample: n=10, p=1/ What is the probability of getting 4, 5, or 6 heads? P(4 6)=P(4)+P(5)+P(6) P(4 6)=10/104+5/104+10/104 P(4 6)=67/104 0.613 P ( ) 0 1 3 4 5 6 7 8 9 10 1 10 45 10 10 5 10 10 45 10 1 104 104 104 104 104 104 104 104 104 104 104 57
5: Probability Distributions (Discrete Variables) 5.3 Mean and Standard Deviation of the Binomial Distribution The formula for the mean μ and variance σ of Binomial is n 0 np n! p (1 p)!( n )! n 0 n n! ( ) p (1 p)!( n )! np(1 p) n np(1 p) (5.7) (5.8) 58
5: Probability Distributions (Discrete Variables) 5.3 Mean and Standard Deviation of the Binomial Distribution Eample: n Before, using, we found 1. Now using, we get. Before, using 1/ found. [ P( )] 0 np () (1/ ) 1 n 0 Now using, np(1 p) [( ) P( )] () (1/ ) (1/ ) 1/, we we get. 0 1 P( ) 1 4 1 1 4 n= =1 p=1/ 59