1. © 2012 McGraw-Hill Ryerson Limited1 © 2009 McGraw-Hill Ryerson Limited

2. © 2012 McGraw-Hill Ryerson Limited2 Lind Marchal Wathen Waite

3. © 2012 McGraw-Hill Ryerson Limited3 Explain why a sample is often the only feasible way to learn something about a population. Describe methods to select a sample. Define and construct a sampling distribution of the sample mean. Explain the central limit theorem. Use the central limit theorem to find probabilities of selecting possible sample means from a specified population. Define and construct a sampling distribution of a proportion. Learning Objectives LO 1 2 3 4 5 6

4. © 2012 McGraw-Hill Ryerson Limited4 Sampling Methods LO 1

5. © 2012 McGraw-Hill Ryerson Limited5 To contact the whole population would be time consuming. The cost of studying all the items in a population may be prohibitive. It is physically impossible to checking all items in the population. Certain tests are destructive in nature. The sample results are adequate. Reasons To Sample LO 1

6. © 2012 McGraw-Hill Ryerson Limited6 1.Simple random sampling 2.Systematic random sampling 3.Stratified random sampling 4.Cluster sampling Methods of Sampling LO 1

7. © 2012 McGraw-Hill Ryerson Limited7 Simple Random Sampling LO 2

8. © 2012 McGraw-Hill Ryerson Limited8 A sample selected so that each item or person in the population has the same chance of being included. To select the sample we can use tables of random numbers. Suppose a population consists of 906 employees of Nitra Industries. A sample of 16 employees is to be selected from that population randomly with the help of random table. Simple Random Sampling LO 2

9. © 2012 McGraw-Hill Ryerson Limited9 52454613527606587641456148675229683 1700147577134458613297443344526874 602737573222352483334424736044401 1959604706682942564788634125846837 2177258603109064660369056913896691 5155214875994009165416830391686374 Simple Random Sampling Starting point Third employee Second employee Fourth employee LO 2

10. © 2012 McGraw-Hill Ryerson Limited10 Simple Random Sampling in Excel LO 2 Jane and Joe Miley operate a bed-and-breakfast called the Foxtrot Inn. There are eight rooms available for rent at this B&B. Listed below are the dates in June and the number of rooms rented each night. Use Excel to select a sample of five of the nights during the month.

11. © 2012 McGraw-Hill Ryerson Limited11 Simple Random Sampling in Excel LO 2

12. © 2012 McGraw-Hill Ryerson Limited12 You Try It Out! The following class roster lists the students enrolling in an introductory course in business statistics. Three students are to be randomly selected and asked various questions regarding course content and method of instruction. a)The numbers 00 through 50 are handwritten on slips of paper and placed in a bowl. The three numbers selected are 38, 25, and 02. Which students would be included in the sample? b)Now use the table of random digits, Appendix E, to select your own sample. c)What would you do if you encountered the number 58 in the table of random digits? LO 2

13. © 2012 McGraw-Hill Ryerson Limited13 You Try It Out! LO 2

14. © 2012 McGraw-Hill Ryerson Limited14 The items or individuals of the population are arranged in some order. A starting point is selected randomly and then every kth member of the population is selected for the sample. k is calculated as the population size divided by the sample size. When the physical order is related to the population characteristic, then systematic random sampling should not be used. Systematic Random Sampling LO 2

15. © 2012 McGraw-Hill Ryerson Limited15 A population is divided into subgroups, called strata, and a sample is randomly selected from each stratum. Once the strata are defined, we can apply simple random sampling within each group or strata to collect the sample. Stratified Random Sampling LO 2

16. © 2012 McGraw-Hill Ryerson Limited16 Number Selected for a Proportional Stratified Random Sample *0.02 of 50 = 1, 0.10 of 50 = 5, etc. Stratified Random Sampling Stratum Profitability (return on equity) Number of Firms Relative Frequency Number Sampled 130 percent and over40.021* 220 up to 30 percent200.15* 310 up to 20 percent1080.5427 40 up to 10 percent660.3316 5Deficit20.011 Total2001.0050 LO 2

17. © 2012 McGraw-Hill Ryerson Limited17 A population is divided into clusters using naturally occurring geographic or other boundaries. Then, clusters are selected randomly and a sample is collected from each cluster selected randomly. Cluster Sampling LO 2

18. © 2012 McGraw-Hill Ryerson Limited18 You Try It Out! The following class roster lists the students enrolling in an introductory course in business statistics. Three students are to be randomly selected and asked various questions regarding course content and method of instruction. a)Suppose a systematic random sample will select every seventh student enrolled in the class. Initially, the third student on the list was selected at random. That student is numbered 03. Remembering that the random numbers start with 00, which students will be chosen to be members of the sample? LO 2

19. © 2012 McGraw-Hill Ryerson Limited19 You Try It Out! LO 2

20. © 2012 McGraw-Hill Ryerson Limited20 The difference between a sample statistic and its corresponding population parameter. Since the sample is a part or portion of the population, it is unlikely that the sample mean would be exactly equal to the population mean. Similarly, it is unlikely that the sample standard deviation would be exactly equal to the population standard deviation. We can therefore expect a difference between a sample statistic and its corresponding population parameter. Sampling Error LO 2

21. © 2012 McGraw-Hill Ryerson Limited21 Example – Sampling Error Johnny operates the Comfort Inn, which has eight rooms available for rent. Rentals for September are displayed. There were 95 rentals in September, so the mean number of units rented per night is 3.16. LO 2

22. © 2012 McGraw-Hill Ryerson Limited22 Example – Sampling Error SeptemberRentalsSeptemberRentalsSeptemberRentals 18116211 24122222 33134233 42144246 54150257 63160264 72175271 83183283 94193293 100202303 Continued LO 2

23. © 2012 McGraw-Hill Ryerson Limited23 Solution – Sampling Error We take three random samples of five nights rented. The first random sample of five nights resulted in the following number of rooms rented: 5, 3, 7, 3, and 4. The mean of this sample is 4.4 rooms, so the sampling error is 4.4 – 3.16 = +1.24. The second random sample of five nights resulted in the following number of rooms rented: 3, 3, 2, 3, and 6. –The mean of this sample is 3.4 rooms, so the sampling error is 3.4 – 3.16 = +0.24. In the third sample, the sampling error was found to be –1.35. LO 2

24. © 2012 McGraw-Hill Ryerson Limited24 Sampling Distribution of the Sample Mean LO 3

25. © 2012 McGraw-Hill Ryerson Limited25 The sample means in the previous example varied from one sample to the next. The mean of the first sample of 5 days was 4.4 rooms, and the second sample was 3.4 rooms. The population mean was 3.16 rooms. If we organize the means of all possible samples of five days into a probability distribution, the result is called the sampling distribution of the sample mean – a probability distribution of all possible sample means of a given sample size. Sampling Distribution of the Sample Mean LO 3

26. © 2012 McGraw-Hill Ryerson Limited26 Crown Industries has seven production employees (considered the population). The hourly earnings of each employee are given in the table 1. What is the population mean? 2. What is the sampling distribution of the sample mean for samples of size 2? 3. What is the mean of the sampling distribution? 4. What observations can be made about the population and the sampling distribution? Example – Sampling Distribution of the Sample Mean LO 3

27. © 2012 McGraw-Hill Ryerson Limited27 EmployeeHourly Earning ($) Sammy$8 Jas 9 Jimy11 Joy10 Albert 8 Lisa 9 Tom10 Example – Sampling Distribution of the Sample Mean Continued LO 3

28. © 2012 McGraw-Hill Ryerson Limited28 Solution – Sampling Distribution of the Sample Mean 1.The population mean is $9.71, found by: 2.To arrive at the sampling distribution of the sample mean, we need to select all possible samples of 2 without replacement from the population, then compute the mean of each sample. There are 21 possible samples, found by using formula (4–4). where N = 7 is the number of items in the population and n = 2 is the number of items in the sample. LO 3

29. © 2012 McGraw-Hill Ryerson Limited29 Solution – Sampling Distribution of the Sample Mean Sample Means for All Possible Samples of 2 Employees SampleEmployees Hourly Earnings ($) Sum ($) Mean ($) SampleEmployees Hourly Earnings ($) Sum ($) Mean ($) 1Samy, Jas$8, $9$17$8.512Jimy, Joy$11, $10$21$10.5 2Samy, Jimy8, 11199.513Jimy, Albert11, 8199.5 3Samy, Joy8, 1018914Jimy, Lisa11, 92010 4 Samy, Albert 8, 816815Jimy, Tom11, 102110.5 5Samy, Lisa8, 9178.516Joy, Albert10, 8189 6Samy, Tom8, 1018917Joy, Lisa10, 9199.5 7 Jas, Jimy9, 11201018Joy, Tom10, 102010 8 Jas, Joy9, 10199.519Albert, Lisa8, 9178.5 9 Jas, Albert9, 8178.520Albert, Tom8, 10189 10 Jas, Lisa9, 918921Lisa, Tom9, 10199.5 11Jas, Tom9, 10199.5 Continued LO 3

30. © 2012 McGraw-Hill Ryerson Limited30 Solution – Sampling Distribution of the Sample Mean Sampling Distribution of the Sample Mean for n = 2 Sample Mean ($) Number of Means Probability $8.0010.0476 $8.5040.1905 $9.0050.2381 $9.5060.2857 $10.0030.1429 $10.5020.0952 211.0000 Continued LO 3

31. © 2012 McGraw-Hill Ryerson Limited31 Solution – Sampling Distribution of the Sample Mean The mean of all the sample means is: Continued LO 3

32. © 2012 McGraw-Hill Ryerson Limited32 Solution – Sampling Distribution of the Sample Mean Sample mean of hourly earnings Population distribution Distribution of sample mean LO 3 Continued

33. © 2012 McGraw-Hill Ryerson Limited33 The mean of the sample means is exactly equal to the population mean. The variance of the sample means is equal to the population variance divided by n. The sampling distribution of the sample means tends to become bell-shaped and to approximate the normal probability distribution. This approximation improves with larger samples. Relationship Between Population Distribution and Sampling Distribution LO 3

34. © 2012 McGraw-Hill Ryerson Limited34 You Try It Out! The marks obtained by 5 students in statistics test are given below: LO 3 NameMarks Joe25 Sam29 Ted28 Bob22 Jan27

35. © 2012 McGraw-Hill Ryerson Limited35 You Try It Out! a)Using the combination formula, how many samples of size 2 are possible? b)List all possible samples of two students from the population and compute their means. c)Organize the means into a sampling distribution. d)Compare the population mean and the mean of the sample means. e)Compare the dispersion in the population with that in the distribution of the sample mean. f)A chart portraying the population values follows. Is the distribution of population values normally distributed (bell-shaped)? LO 3 Continued

36. © 2012 McGraw-Hill Ryerson Limited36 You Try It Out! g)Is the distribution of the sample mean computed in part (c) starting to show some tendency toward being bell- shaped? LO 3 Continued

37. © 2012 McGraw-Hill Ryerson Limited37 The Central Limit Theorem LO 4

38. © 2012 McGraw-Hill Ryerson Limited38 If all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution. This approximation improves with larger samples. The Central Limit Theorem LO 4

39. © 2012 McGraw-Hill Ryerson Limited39 Illustration LO 4

40. © 2012 McGraw-Hill Ryerson Limited40 Springer Inc. faces some major decisions regarding health care for these employees. Ed Springer decides to form a committee of five representative employees to study the health care issue carefully and make a recommendation. Ed feels the views of newer employees toward health care may differ from those of more experienced employees. If Ed randomly selects this committee, what can he expect in terms of the mean years with Springer for those on the committee? How does the shape of the distribution of years of experience of all employees (the population) compare with the shape of the sampling distribution of the mean? LO 4 Example – The Central Limit Theorem

41. © 2012 McGraw-Hill Ryerson Limited41 114182120224 34122331983 71027045114 168911251023 LO 4 Example – The Central Limit Theorem The lengths of service of the 40 employees currently on the Springer Inc. payroll are: Continued

42. © 2012 McGraw-Hill Ryerson Limited42 LO 4 Solution – The Central Limit Theorem

43. © 2012 McGraw-Hill Ryerson Limited43 Use the central limit theorem LO 5

44. © 2012 McGraw-Hill Ryerson Limited44 Solution – The Central Limit Theorem LO 5 Continued The result of selecting 25 samples of five employees each is shown in the table and chart. There are actually 658 008 possible samples of 5 from the population of 40 employees, found by the combination for 40 things taken 5 at a time.

45. © 2012 McGraw-Hill Ryerson Limited45 Solution – The Central Limit Theorem This sample mean is 4.98 years. Sample I.D.Sample DataSample Mean A500612.4 B695296.6 C3110413.8 D24535 E456845.4 F134253.0 G4527105.6 H1103886.0 I6108297.0 J386555.4 K989767.8 L093094.2 M956204.4 N1032804.6 O257233.8 P399486.6 Q933685.8 R1052355.0 S382975.8 T911153.4 U934604.4 V689315.4 W185896.2 X300593.4 Y487335.0 Z728524.8 LO 5 Continued

46. © 2012 McGraw-Hill Ryerson Limited46 LO 5 Solution – The Central Limit Theorem Continued

47. © 2012 McGraw-Hill Ryerson Limited47 Solution – The Central Limit Theorem LO 5 Continued The result of selecting 25 samples of 20 employees each is shown in the table and chart. From the table below, the mean of all sample means is 5.17.

48. © 2012 McGraw-Hill Ryerson Limited48 Solution – The Central Limit Theorem LO 5 Sample I.D.Sample DataMean A83119757129273194017 4.35 B23905863293733144383 4.30 C75736717827099791098 5.60 D94551535091742723958 4.70 E161018191551931980173 4.45 F53313787876429474828 5.30 G80966852689449384893 5.95 H13418479621577108359 5.50 I01810699999 467469545 6.50 J610442507 1 878797752 5.95 K8566413161058488917 9 5.95 L865296210127081518655 4.85 M849549515676910764783 6.15 N19281054301 519313596 4.75 O645106557045654968975 5.80 P55174023316453723635 3.75 Q269016177124961010538 4.40 R85164797785864635499 6.05 S337803589112107421828 4.60 T285236845423444106081 4.45 U52856174340245335092 3.90 V4881776610 0549068674 5.80 W89956258777922268987 6.30 X204546206296105778212 4.40 Y84393774653648429589 5.70 Z1105555816025 3881 44 5.05 Continued

49. © 2012 McGraw-Hill Ryerson Limited49 LO 5 Solution – The Central Limit Theorem Continued

50. © 2012 McGraw-Hill Ryerson Limited50 LO 5 Solution – The Central Limit Theorem What should we conclude from this example? The central limit theorem indicates that, regardless of the shape of the population distribution, the sampling distribution of the sample mean will move toward the normal probability distribution. The larger the number of observations in each sample, the stronger the convergence. The Springer Inc. example shows how the central limit theorem works. Continued

51. © 2012 McGraw-Hill Ryerson Limited51 LO 5 Standard Error of the Mean It can be demonstrated that the mean of the sampling distribution is the population mean and if the standard deviation in the population is σ, the standard deviation of the sample means is where n is the number of observations in each sample. We refer to as the standard error of the mean.

52. © 2012 McGraw-Hill Ryerson Limited52 1.The mean of the distribution of the sample mean will be exactly equal to the population mean if we are able to select all possible samples of a particular size from a given population. That is: Even if we do not select all samples, we can expect the mean of the distribution of the sample mean to be close to the population mean. Important Conclusions LO 5

53. © 2012 McGraw-Hill Ryerson Limited53 2. There will be less dispersion in the sampling distribution of the sample mean than in the population. If the standard deviation of the population is σ, the standard deviation of the distribution of the sample mean is. Note that when we increase the size of the sample, the standard error of the mean decreases. Important Conclusions LO 5

54. © 2012 McGraw-Hill Ryerson Limited54 LO 5 Using the Sampling Distribution of the Sample Mean (σ Known) If a population follows the normal distribution, the sampling distribution of the sample mean will also follow the normal distribution. To determine the probability a sample mean falls within a particular region, use:

55. © 2012 McGraw-Hill Ryerson Limited55 The quality control department for Soft Drink Inc. maintains records regarding the amount of soft drink in its “Jumbo” bottle. The actual amount of soft drink in each bottle is critical, but varies a small amount from one bottle to the next. Soft Drink Inc. does not wish to underfill the bottles, because it will have a problem with truth in labelling. On the other hand, it cannot overfill each bottle, because it would be giving soft drink away, hence reducing its profits. Example – Using the Sampling Distribution of the Sample Mean (σ Known) LO 5

56. © 2012 McGraw-Hill Ryerson Limited56 Its records indicate that the amount of soft drink follows a normal probability distribution. The mean amount per bottle is 1 L and the population standard deviation is 11.8 mL. At 9 a.m. today the quality control technician randomly selected 15 bottles from the filling line. The mean amount of soft drink contained in the bottles is 1.004 L. Is this an unlikely result? Is it likely the process is putting too much soft drink in the bottles? Example – Using the Sampling Distribution of the Sample Mean (σ Known) LO 5

57. © 2012 McGraw-Hill Ryerson Limited57 Step 1: Find the z-value corresponding to the sample mean of 1004 mL, given µ = 1000 mL and σ = 11.8 mL. Solution – Using the Sampling Distribution of the Sample Mean (σ Known) LO 5

58. © 2012 McGraw-Hill Ryerson Limited58 Step 2: Find the probability of observing a Z equal to or greater than 1.31. Sampling Distribution of the Mean Amount of Soft Drink in a Jumbo Bottle Solution – Using the Sampling Distribution of the Sample Mean (σ Known) LO 5 Continued

59. © 2012 McGraw-Hill Ryerson Limited59 What do we conclude? It is unlikely, about a 9 percent chance, we could select a sample of 15 observations from a normal population with a mean of 1 L and a population standard deviation of 11.8 mL and find the sample mean equal to or greater than 1.004 L. We conclude the process is putting too much soft drink in the bottles. The quality control technician should see the production supervisor about reducing the amount of soft drink in each bottle. Solution – Using the Sampling Distribution of the Sample Mean (σ Known) LO 5 Continued

60. © 2012 McGraw-Hill Ryerson Limited60 You Try It Out! Refer to the Soft Drink Inc. information. Suppose the quality technician selected a sample of 14 Jumbo Soft Drink bottles that averaged 0.996 L. What can you conclude about the filling process? LO 5 Continued

61. © 2012 McGraw-Hill Ryerson Limited61 Sampling Distribution of The Proportion LO 6

62. © 2012 McGraw-Hill Ryerson Limited62 For the nominal scale of measurement. A proportion is the fraction, ratio, or percent indicating the part of the sample or the population having a particular trait of interest. Sampling Distribution of The Proportion LO 6

63. © 2012 McGraw-Hill Ryerson Limited63 Atlas Corporation receives a shipment of flour every morning from its supplier. The flour is in 50 kg bags and Atlas will reject any shipment that is more than 6 percent underweight. The supervisor samples 60 bags with each shipment and if the bags average more than 6 percent underweight, the whole shipment is returned to the supplier. What is the probability that in a sample of 160 bags, the supervisor will find that less than 4 percent are underweight? Example – Using the Sampling Distribution of a Proportion LO 6

64. © 2012 McGraw-Hill Ryerson Limited64 We want to know the probability that the supervisor will find less than 4 percent underweight, so we are looking for the area to the left of z = −1.07. Using the value from Appendix A.1, we find the probability is.1423 (0.5 − 0.3577). Solution – Using the Sampling Distribution of a Proportion LO 6

65. © 2012 McGraw-Hill Ryerson Limited65 You Try It Out! Refer to the Atlas Corporation information. Compute the probability that in a sample of 200, the supervisor will find more than 5 percent of the bags underweight. LO 6

66. © 2012 McGraw-Hill Ryerson Limited66 I.There are many reasons for sampling a population: A.The results of a sample may adequately estimate the value of the population parameter, thus saving time and money. B. It may be too time consuming to contact all members of the population. C.It may be impossible to check or locate all the members of the population. D.The cost of studying all the items in the population may be prohibitive. E.Often, testing destroys the sampled item and it cannot be returned to the population. Chapter Summary

67. © 2012 McGraw-Hill Ryerson Limited67 II.In an unbiased or probability sample, all members of the population have an equal chance of being selected for the sample. There are several probability sampling methods: A.In a simple random sample, all members of the population have the same chance of being selected for the sample. B.In a systematic sample, a random starting point is selected, and then every kth item thereafter is selected for the sample. Chapter Summary

68. © 2012 McGraw-Hill Ryerson Limited68 C.In a stratified sample, the population is divided into several groups, called strata, and then a random sample is selected from each stratum. D.In cluster sampling, the population is divided into primary units, then samples are drawn from the primary units. III. The sampling error is the difference between a population parameter and a sample statistic. Chapter Summary

69. © 2012 McGraw-Hill Ryerson Limited69 IV.The sampling distribution of the sample mean is a probability distribution of all possible sample means of a given size from a population. A.For a given sample size, the mean of all possible sample means selected from a population is equal to the population mean. B.There is less variation in the distribution of the sample mean than in the population distribution. C.The standard error of the mean measures the variation in the sampling distribution of the sample mean. The standard error is found by: [7–1] Chapter Summary

70. © 2012 McGraw-Hill Ryerson Limited70 D. If the population follows a normal distribution, the sampling distribution of the sample mean will also follow a normal distribution for samples of any size. Assume the population standard deviation is known. To determine the probability that a sample mean falls in a particular region, use the following formula. Often, testing destroys the sampled item and it cannot be returned to the population. [7–2] Chapter Summary

71. © 2012 McGraw-Hill Ryerson Limited71 V.A proportion is a ratio, fraction, or percent that indicates the part of the sample or population that has a particular characteristic. A.A sample proportion is found by, the number of successes, divided by n, the number of observations. [7–3] B.The sampling distribution of the proportion follows a normal distribution if np and n(1 – p) > 5. Chapter Summary

72. © 2012 McGraw-Hill Ryerson Limited72 C.To determine the probability that a sample proportion falls in a particular region, use the following formula: [7–5] Chapter Summary