1. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1 Understandable Statistics S eventh Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Chapter Seven Introduction to Sampling Distributions

2. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 2 Review of Statistical Terms Population Sample Parameter Statistic

3. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 3 Population the set of all measurements (either existing or conceptual) under consideration

4. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 4 Sample a subset of measurements from a population

5. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 5 Parameter a numerical descriptive measure of a population

6. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 6 Statistic a numerical descriptive measure of a sample

7. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 7 We use a statistic to make inferences about a population parameter.

8. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 8 Principal types of inferences Estimate the value of a population parameter Formulate a decision about the value of a population parameter

9. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 9 Sampling Distribution a probability distribution for the sample statistic we are using

10. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 10 Example of a Sampling Distribution Select samples with two elements each (in sequence with replacement) from the set {1, 2, 3, 4, 5, 6}.

11. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 11 Constructing a Sampling Distribution of the Mean for Samples of Size n = 2 List all samples and compute the mean of each sample. sample:mean:sample:mean {1,1}1.0{1,6}3.5 {1,2}1.5{2,1}1.5 {1,3}2.0{2,2}4 {1,4}2.5…... {1,5}3.0 There are 36 different samples.

12. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 12 Sampling Distribution of the Mean p 1.01/36 1.52/36 2.03/36 2.54/36 3.05/36 3.56/36 4.05/36 4.54/36 5.03/36 5.52/36 6.01/36

13. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 13 Sampling Distribution Histogram | | | | | | 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

14. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 14 Let x be a random variable with a normal distribution with mean  and standard deviation . Let be the sample mean corresponding to random samples of size n taken from the distribution.

15. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 15 Facts about sampling distribution of the mean:

16. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 16 Facts about sampling distribution of the mean: The distribution is a normal distribution.

17. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 17 Facts about sampling distribution of the mean: The distribution is a normal distribution. The mean of the distribution is  (the same mean as the original distribution).

18. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 18 Facts about sampling distribution of the mean: The distribution is a normal distribution. The mean of the distribution is  (the same mean as the original distribution). The standard deviation of the distribution is  (the standard deviation of the original distribution, divided by the square root of the sample size).

19. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 19 We can use this theorem to draw conclusions about means of samples taken from normal distributions. If the original distribution is normal, then the sampling distribution will be normal.

20. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 20 The Mean of the Sampling Distribution

21. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 21 The mean of the sampling distribution is equal to the mean of the original distribution.

22. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 22 The Standard Deviation of the Sampling Distribution

23. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 23 The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

24. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 24 The time it takes to drive between cities A and B is normally distributed with a mean of 14 minutes and a standard deviation of 2.2 minutes. Find the probability that a trip between the cities takes more than 15 minutes. Find the probability that mean time of nine trips between the cities is more than 15 minutes.

25. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 25 Mean = 14 minutes, standard deviation = 2.2 minutes Find the probability that a trip between the cities takes more than 15 minutes. 14 15 Find this area

26. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 26 Mean = 14 minutes, standard deviation = 2.2 minutes Find the probability that mean time of nine trips between the cities is more than 15 minutes.

27. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 27 Mean = 14 minutes, standard deviation = 2.2 minutes Find the probability that mean time of nine trips between the cities is more than 15 minutes. 14 15 Find this area

28. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 28 What if the Original Distribution Is Not Normal? Use the Central Limit Theorem.

29. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 29 Central Limit Theorem If x has any distribution with mean  and standard deviation , then the sample mean based on a random sample of size n will have a distribution that approaches the normal distribution (with mean  and standard deviation  divided by the square root of n) as n increases without bound.

30. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 30 How large should the sample size be to permit the application of the Central Limit Theorem? In most cases a sample size of n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.

31. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 31 Central Limit Theorem

32. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 32 Central Limit Theorem For most x distributions, if we use a sample size of 30 or larger, the distribution will be approximately normal.

33. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 33 Central Limit Theorem The mean of the sampling distribution is the same as the mean of the original distribution. The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

34. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 34 Central Limit Theorem Formula

35. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 35 Central Limit Theorem Formula

36. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 36 Central Limit Theorem Formula

37. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 37 Application of the Central Limit Theorem Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be: a.more than 530 pounds? b.less than 500 pounds? c.between 495 and 515 pounds?

38. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 38 Are we authorized to use the Normal Distribution? Yes, we are attempting to draw conclusions about means of large samples.

39. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 39 Applying the Central Limit Theorem What is the probability that their mean weight will be more than 530 pounds? Consider the distribution of sample means: P( x > 530): z = 530 – 510 = 20 = 2.22 9 9 P(z > 2.22) = _______.0132

40. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 40 Applying the Central Limit Theorem What is the probability that their mean weight will be less than 500 pounds? P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9 P(z < – 1.11) = _______.1335

41. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 41 Applying the Central Limit Theorem What is the probability that their mean weight will be between 495 and 515 pounds? P(495 < x < 515) : for 495: z = 495 – 510 =  15 =  1.67 9 9 for 515: z = 515 – 510 = 5 = 0.56 9 9 P(  1.67 < z < 0.56) = _______.6648

42. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 42 Sampling Distributions for Proportions Allow us to work with the proportion of successes rather than the actual number of successes in binomial experiments.

43. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 43 Sampling Distribution of the Proportion n= number of binomial trials r = number of successes p = probability of success on each trial q = 1 - p = probability of failure on each trial

44. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 44 Sampling Distribution of the Proportion If np > 5 and nq > 5 then p-hat = r/n can be approximated by a normal random variable (x) with:

45. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 45 The Standard Error for

46. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 46 Continuity Correction When using the normal distribution (which is continuous) to approximate p- hat, a discrete distribution, always use the continuity correction. Add or subtract 0.5/n to the endpoints of a (discrete) p-hat interval to convert it to a (continuous) normal interval.

47. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 47 Continuity Correction If n = 20, convert a p- hat interval from 5/8 to 6/8 to a normal interval. Note: 5/8 = 0.625 6/8 = 0.75 So p-hat interval is 0.625 to 0.75. Since n = 20,.5/n = 0.025 5/8 - 0.025 = 0.6 6/8 + 0.025 = 0.775 Required x interval is 0.6 to 0.775

48. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 48 Suppose 12% of the population is in favor of a new park. Two hundred citizen are surveyed. What is the probability that between 10 % and 15% of them will be in favor of the new park?

49. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 49 12% of the population is in favor of a new park. p = 0.12, q= 0.88 Two hundred citizen are surveyed. n = 200 Both np and nq are greater than five. Is it appropriate to the normal distribution?

50. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 50 Find the mean and the standard deviation

51. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 51 What is the probability that between 10 % and 15%of them will be in favor of the new park? Use the continuity correction Since n = 200,.5/n =.0025 The interval for p-hat (0.10 to 0.15) converts to 0.0975 to 0.1525.

52. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 52 Calculate z-score for x = 0.0975

53. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 53 Calculate z-score for x = 0.1525

54. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 54 P(-0.98 < z < 1.41) 0.9207 -- 0.1635 = 0.7572 There is about a 75.7% chance that between 10% and 15% of the citizens surveyed will be in favor of the park.

55. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 55 Control Chart for Proportions P-Chart

56. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 56 Constructing a P-Chart Select samples of fixed size n at regular intervals. Count the number of successes r from the n trials. Use the normal approximation for r/n to plot control limits. Interpret results.

57. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 57 Determining Control Limits for a P-Chart Suppose employee absences are to be plotted. In a daily sample of 50 employees, the number of employees absent is recorded. p/n for each day = number absent/50.For the random variable p-hat = p/n, we can find the mean and the standard deviation.

58. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 58 Finding the mean and the standard deviation

59. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 59 Is it appropriate to use the normal distribution? The mean of p-hat = p = 0.12 The value of n = 50. The value of q = 1 - p = 0.88. Both np and nq are greater than five. The normal distribution will be a good approximation of the p-hat distribution.

60. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 60 Control Limits Control limits are placed at two and three standard deviations above and below the mean.

61. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 61 Control Limits The center line is at 0.12. Control limits are placed at -0.018, 0.028, 0.212, and 0.258.

62. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 62 Control Chart for Proportions Employee Absences 0.3 +3s = 0.258 0.2+2s = 0.212 0.1 mean = 0.12 0.0 -2s = 0.028 -0.1 -3s = -0.018

63. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 63 Daily absences can now be plotted and evaluated. Employee Absences 0.3 +3s = 0.258 0.2+2s = 0.212 0.1 mean = 0.12 0.0 -2s = 0.028 -0.1 -3s = -0.018