Chapter 8 Sampling Distributions 2010 Pearson Prentice Hall

Section 8. 1 Distribution of the Sample Mean © 2010 Pearson Prentice Hall. All

Objectives 1. Describe the distribution of the sample mean: samples from normal populations 2. Describe the distribution of the sample mean: samples from a population that is not normal © 2010 Pearson Prentice Hall. All rights reserved 3

Statistics such as are random variables since their value varies from sample to sample. As such, they have probability distributions associated with them. In this chapter we focus on the shape, center and spread of statistics such as. © 2010 Pearson Prentice Hall. All rights reserved 4

The sampling distribution of a statistic is a probability distribution for all possible values of the statistic computed from a sample of size n. The sampling distribution of the sample mean is the probability distribution of all possible values of the random variable computed from a sample of size n from a population with mean and standard deviation . © 2010 Pearson Prentice Hall. All rights reserved 5

Illustrating Sampling Distributions Step 1: Obtain a simple random sample of size n. ©

Illustrating Sampling Distributions Step 1: Obtain a simple random sample of size n. Step 2: Compute the sample mean. Step 3: Assuming we are sampling from a finite population, repeat Steps 1 and 2 until all simple random samples of size n have been obtained. © 2010 Pearson Prentice Hall. All rights reserved 8

Objective 1 • Describe the Distribution of the Sample Mean. Samples from Normal Populations

Parallel Example 1: Sampling Distribution of the Sample Mean-Normal Population The weights of pennies minted after 1982 are approximately normally distributed with mean 2. 46 grams and standard deviation 0. 02 grams. Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 5 from this population. © 2010 Pearson Prentice Hall. All rights reserved 10

The data on the following slide represent the sample means for the 200 simple random samples of size n = 5. For example, the first sample of n = 5 had the following data: 2. 493 2. 466 2. 473 2. 492 Note: =2. 479 for this sample 2. 471 © 2010 Pearson Prentice Hall. All rights reserved 11

Sample Means for Samples of Size n =5 © 2010 Pearson Prentice Hall. All

The mean of the 200 sample means is 2. 46, the same as the mean of the population. The standard deviation of the sample means is 0. 0086, which is smaller than the standard deviation of the population. The next slide shows the histogram of the sample means. © 2010 Pearson Prentice Hall. All rights reserved 13

© 2010 Pearson Prentice Hall. All rights reserved 14

What role does n, the sample size, play in the standard deviation of the distribution of the sample mean? As the size of the sample gets larger, we do not expect as much spread in the sample means since larger observations will offset smaller observations. © 2010 Pearson Prentice Hall. All rights reserved 16

Parallel Example 2: The Impact of Sample Size on Sampling Variability • Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 20 from the population of weights of pennies minted after 1982 ( =2. 46 grams and =0. 02 grams) © 2010 Pearson Prentice Hall. All rights reserved 17

The mean of the 200 sample means for n =20 is still 2. 46, but the standard deviation is now 0. 0045 (0. 0086 for n = 5). As expected, there is less variability in the distribution of the sample mean with n =20 than with n =5. © 2010 Pearson Prentice Hall. All rights reserved 18

The Mean and Standard Deviation of the Sampling Distribution of Suppose that a simple random sample of size n is drawn from a large population with mean and standard deviation . The sampling distribution of will have mean and standard deviation. The standard deviation of the sampling distribution of is called the standard error of the mean and is denoted. © 2010 Pearson Prentice Hall. All rights reserved 19

The Shape of the Sampling Distribution of If X is Normal If a random variable X is normally distributed, the distribution of the sample mean is normally distributed. © 2010 Pearson Prentice Hall. All rights reserved 20

Parallel Example 3: Describing the Distribution of the Sample Mean The weights of pennies minted after 1982 are approximately normally distributed with mean 2. 46 grams and standard deviation 0. 02 grams. What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2. 465 grams? © 2010 Pearson Prentice Hall. All rights reserved 21

Solution • is normally distributed with • =2. 46 and . . • P(Z>0.

Objective 2 • Describe the Distribution of the Sample Mean. Samples from a Population

Parallel Example 4: Sampling from a Population that is Not Normal The following table and histogram give the probability distribution for rolling a fair die: Face on Die Relative Frequency 1 0. 1667 2 0. 1667 3 0. 1667 4 0. 1667 5 0. 1667 6 0. 1667 =3. 5, =1. 708 Note that the population distribution is NOT normal © 2010 Pearson Prentice Hall. All rights reserved 24

Estimate the sampling distribution of by obtaining 200 simple random samples of size n=4 and calculating the sample mean for each of the 200 samples. Repeat for n = 10 and 30. Histograms of the sampling distribution of the sample mean for each sample size are given on the next slide. © 2010 Pearson Prentice Hall. All rights reserved 25

© 2010 Pearson Prentice Hall. All rights reserved 26

© 2010 Pearson Prentice Hall. All rights reserved 27

© 2010 Pearson Prentice Hall. All rights reserved 28

Key Points from Example 4 • The mean of the sampling distribution is equal to the mean of the parent population and the standard deviation of the sampling distribution of the sample mean is regardless of the sample size. • The Central Limit Theorem: the shape of the distribution of the sample mean becomes approximately normal as the sample size n increases, regardless of the shape of the population. © 2010 Pearson Prentice Hall. All rights reserved 29

Parallel Example 5: Using the Central Limit Theorem Suppose that the mean time for an oil change at a “ 10 -minute oil change joint” is 11. 4 minutes with a standard deviation of 3. 2 minutes. (a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean. (b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes? © 2010 Pearson Prentice Hall. All rights reserved 30

Parallel Example 5: Using the Central Limit Theorem Suppose that the mean time for an oil change at a “ 10 -minute oil change joint” is 11. 4 minutes with a standard deviation of 3. 2 minutes. (a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean. Solution: is approximately normally distributed with mean=11. 4 and std. dev. =. (b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes? Solution: , P(Z<-0. 74)=0. 23. © 2010 Pearson Prentice Hall. All rights reserved 31

Section 8. 2 Distribution of the Sample Proportion © 2010 Pearson Prentice Hall. All

Objectives 1. Describe the sampling distribution of a sample proportion 2. Compute probabilities of

Objective 1 • Describe the Sampling Distribution of a Sample Proportion © 2010 Pearson

Point Estimate of a Population Proportion Suppose that a random sample of size n is obtained from a population in which each individual either does or does not have a certain characteristic. The sample proportion, denoted (read “p-hat”) is given by where x is the number of individuals in the sample with the specified characteristic. The sample proportion is a statistic that estimates the population proportion, p. © 2010 Pearson Prentice Hall. All rights reserved 35

Parallel Example 1: Computing a Sample Proportion In a Quinnipiac University Poll conducted in May of 2008, 1, 745 registered voters nationwide were asked whether they approved of the way George W. Bush is handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approve of the way George W. Bush is handling the economy. © 2010 Pearson Prentice Hall. All rights reserved 36

Parallel Example 1: Computing a Sample Proportion In a Quinnipiac University Poll conducted in May of 2008, 1, 745 registered voters nationwide were asked whether they approved of the way George W. Bush is handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approve of the way George W. Bush is handling the economy. Solution: © 2010 Pearson Prentice Hall. All rights reserved 37

Parallel Example 2: Using Simulation to Describe the Distribution of the Sample Proportion According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion for samples of size n=10, 50, 100. © 2010 Pearson Prentice Hall. All rights reserved 38

© 2010 Pearson Prentice Hall. All rights reserved 39

© 2010 Pearson Prentice Hall. All rights reserved 40

© 2010 Pearson Prentice Hall. All rights reserved 41

Key Points from Example 2 • Shape: As the size of the sample, n, increases, the shape of the sampling distribution of the sample proportion becomes approximately normal. • Center: The mean of the sampling distribution of the sample proportion equals the population proportion, p. • Spread: The standard deviation of the sampling distribution of the sample proportion decreases as the sample size, n, increases. © 2010 Pearson Prentice Hall. All rights reserved 42

Sampling Distribution of For a simple random sample of size n with population proportion p: • The shape of the sampling distribution of is approximately normal provided np(1 -p)≥ 10. • The mean of the sampling distribution of is. • The standard deviation of the sampling distribution of is © 2010 Pearson Prentice Hall. All rights reserved 43

Sampling Distribution of • The model on the previous slide requires that the sampled values are independent. When sampling from finite populations, this assumption is verified by checking that the sample size n is no more than 5% of the population size N (n ≤ 0. 05 N). • Regardless of whether np(1 -p) ≥ 10 or not, the mean of the sampling distribution of is p, and the standard deviation is © 2010 Pearson Prentice Hall. All rights reserved 44

Parallel Example 3: Describing the Sampling Distribution of the Sample Proportion According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Suppose that we obtain a simple random sample of 50 voters and determine which believe that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion for registered voters who believe that gay and lesbian couples should be allowed to marry. © 2010 Pearson Prentice Hall. All rights reserved 45

Solution The sample of n=50 is smaller than 5% of the population size (all registered voters in the U. S. ). Also, np(1 -p)=50(0. 42)(0. 58)=12. 18≥ 10. The sampling distribution of the sample proportion is therefore approximately normal with mean=0. 42 and standard deviation=. (Note: this is very close to the standard deviation of 0. 072 found using simulation in Example 2. ) © 2010 Pearson Prentice Hall. All rights reserved 46

Objective 2 • Compute Probabilities of a Sample Proportion © 2010 Pearson Prentice Hall.

Parallel Example 4: Compute Probabilities of a Sample Proportion According to the Centers for Disease Control and Prevention, 18. 8% of school-aged children, aged 6 -11 years, were overweight in 2004. (a) In a random sample of 90 school-aged children, aged 6 -11 years, what is the probability that at least 19% are overweight? (b) Suppose a random sample of 90 school-aged children, aged 6 -11 years, results in 24 overweight children. What might you conclude? © 2010 Pearson Prentice Hall. All rights reserved 48

Solution • • • n=90 is less than 5% of the population size np(1 -p)=90(. 188)(1 -. 188)≈13. 7≥ 10 is approximately normal with mean=0. 188 and standard deviation = (a) In a random sample of 90 school-aged children, aged 6 -11 years, what is the probability that at least 19% are overweight? , P(Z>0. 05)=1 -0. 5199=0. 4801 © 2010 Pearson Prentice Hall. All rights reserved 49

Solution • is approximately normal with mean=0. 188 and standard deviation = 0. 0412 (b) Suppose a random sample of 90 school-aged children, aged 6 -11 years, results in 24 overweight children. What might you conclude? , P(Z>1. 91)=1 -0. 9719=0. 028. We would only expect to see about 3 samples in 100 resulting in a sample proportion of 0. 2667 or more. This is an unusual sample if the true population proportion is 0. 188. © 2010 Pearson Prentice Hall. All rights reserved 50