1. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Chapter Sampling Distributions 8

2. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Distribution of the Sample Mean 8.1

3. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-3 Objectives 1.Describe the distribution of the sample mean: normal population 2.Describe the distribution of the sample mean: nonnormal population

4. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-4 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.

5. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-5 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 μ and standard deviation σ.

6. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Let’s consider a population consisting of {4, 5, 9} -  ages of someone’s children If two ages are randomly selected with replacement from the population, identify the sampling distribution of the sample mean. 8-6

7. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Sample Distribution of Mean 8-7 SampleSample MeanProbability 4,44.01/9 4,54.51/9 4,96.51/9 5,44.51/9 5,55.01/9 5,97.01/9 9,46.51/9 9,57.01/9 9,991/9

8. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Sampling Distribution of Mean(condensed) Sample MeanProbability 4.01/9 4.52/9 5.01/9 6.52/9 7.02/9 9.01/9 8-8 Using this table, let’s calculate the mean of the sample mean: 4(1/9)+4.5(2/9)+5(1/9)+…+9(1/9)=6.0 The mean of the population {4,5,9} is also 6.0!

9. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Because the mean of the sample means is equal to the mean of the population, we conclude that the values of the sample mean do target the value of the population mean. The mean of the sample mean is equal to the population mean µ 8-9

10. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-10 Illustrating Sampling Distributions Step 1: Obtain a simple random sample of size n.

11. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-11 Illustrating Sampling Distributions Step 1: Obtain a simple random sample of size n. Step 2: Compute the sample mean.

12. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-12 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.

13. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-13 Objective 1 Describe the Distribution of the Sample Mean: Normal Population

14. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-14 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. Parallel Example 1: Sampling Distribution of the Sample Mean-Normal Population

15. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-15 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 2.4922.471 Note: = 2.479 for this sample

16. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-16 Sample Means for Samples of Size n = 5

17. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-17 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.

18. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-18

19. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-19 What role does n, the sample size, play in the standard deviation of the distribution of the sample mean?

20. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-20 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 increases, the standard deviation of the distribution of the sample mean decreases.

21. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-21 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 (μ = 2.46 grams and σ = 0.02 grams) Parallel Example 2: The Impact of Sample Size on Sampling Variability

22. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-22 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.

23. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-23 Suppose that a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ. 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. The Mean and Standard Deviation of the Sampling Distribution of

24. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-24 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.

25. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-25 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? This is not an individual value! Parallel Example 3: Describing the Distribution of the Sample Mean

26. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-26 is normally distributed with =2.46 and.. P(Z > 0.79) = 1 – 0.7852 = 0.2148. Solution

27. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-27 Objective 2 Describe the Distribution of the Sample Mean: Nonnormal Population

28. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-28 The following table and histogram give the probability distribution for rolling a fair die: μ = 3.5, σ = 1.708 Note that the population distribution is NOT normal Face on DieRelative Frequency 1 0.1667 2 3 4 5 6 Parallel Example 4: Sampling from a Population that is Not Normal

29. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-29 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.

30. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-30

31. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-31

32. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-32

33. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-33 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.

34. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-34 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?

35. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-35 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. =.

36. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-36 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? Solution: is approximately normally distributed with mean = 11.4 and std. dev. =. Solution:, P(Z < –0.74) = 0.23.

37. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-37 Designing Elevators When designing elevators, an obviously important consideration is the weight capacity. An Ohio college studetn died when he tried to escape from a dormitory elevator that was overloaded with 24 passengers. The elevator was rated for a capacity of 16 passengers with a total weight of 2500 lb. Weight of adults are changing over time(table below). WE assume a worst-case scenario in which all of the passengers are males. If an elevator is loaded to a capacity of 2500 lb with 16 males, the mean weight of a passenger is 156.25 lb. The distribution is Normal for both males and females. MalesFemales µ182.9 lb165.0 lb 40.8 lb45.6 lb

38. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-38 (a)Find the probablity that 1 randomly selected adult male has a weight greater than 156.25 (b)Find the probability that a sample of 16 randomly selected adult males has a maximum capacity of 156.25 lb

39. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-39 (a)Approach used for an individual Value (b)Sample has a mean: Now use Table V, to find the probabilities…

40. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Distribution of the Sample Proportion 8.2

41. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-41 Objectives 1.Describe the sampling distribution of a sample proportion 2.Compute probabilities of a sample proportion

42. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-42 Objective 1 Describe the Sampling Distribution of a Sample Proportion

43. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-43 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.

44. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-44 In a Quinnipiac University Poll conducted in May of 2008, 1745 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. Parallel Example 1: Computing a Sample Proportion

45. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-45 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. Parallel Example 1: Computing a Sample Proportion Solution:

46. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-46 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. ( ) Note: We are using simulations to create the histograms on the following slides. Parallel Example 2: Using Simulation to Describe the Distribution of the Sample Proportion

47. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-47

48. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-48

49. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-49

50. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-50 Note: The mean of the sample proportions equals the population proportion

51. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-51 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.

52. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-52 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 Sampling Distribution of

53. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-53 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 ≤ 0.05N). Regardless of whether np(1 – p) ≥ 10 or not, the mean of the sampling distribution of is p, and the standard deviation is

54. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-54 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 voters 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. Parallel Example 3: Describing the Sampling Distribution of the Sample Proportion

55. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-55 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.)

56. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-56 Objective 2 Compute Probabilities of a Sample Proportion

57. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-57 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? Parallel Example 4: Compute Probabilities of a Sample Proportion

58. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-58 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? Solution, P(Z > 0.05)=1 – 0.5199=0.4801

59. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 8-59 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? Solution, 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.