2. 1 Inference about Two Populations Chapter 13

3. 2 12.1 Introduction Variety of techniques are presented whose objective is to compare two populations. We are interested in: –The difference between two means. –The ratio of two variances. –The difference between two proportions.

4. 3 Two random samples are drawn from the two populations of interest. Because we compare two population means, we use the statistic. 13.2Inference about the Difference between Two Means: Independent Samples

5. 4 1. is normally distributed if the (original) population distributions are normal. 2. is approximately normally distributed if the (original) population is not normal, but the samples’ size is sufficiently large (greater than 30). 3. The expected value of is  1 -  2 4. The variance of is  1 2 / n 1 +  2 2 / n 2 The Sampling Distribution of

6. 5 If the sampling distribution of is normal or approximately normal we can write: Z can be used to build a test statistic or a confidence interval for  1 -  2 Making an inference about    –  

7. 6 Practically, the “Z” statistic is hardly used, because the population variances are not known. ? ? Instead, we construct a t statistic using the sample “variances” (S 1 2 and S 2 2 ). S22S22 S12S12 t Making an inference about    –  

8. 7 Two cases are considered when producing the t-statistic. –The two unknown population variances are equal. –The two unknown population variances are not equal. Making an inference about    –  

9. 8 Inference about    –   : Equal variances Example: s 1 2 = 25; s 2 2 = 30; n 1 = 10; n 2 = 15. Then, Calculate the pooled variance estimate by: n 2 = 15 n 1 = 10 The pooled variance estimator

10. 9 Inference about    –   : Equal variances Example: s 1 2 = 25; s 2 2 = 30; n 1 = 10; n 2 = 15. Then, Calculate the pooled variance estimate by: n 2 = 15 n 1 = 10 The pooled Variance estimator

11. 10 Inference about    –   : Equal variances Construct the t-statistic as follows: Perform a hypothesis test H 0 :     = 0 H 1 :     > 0 or < 0or 0 Build a confidence interval

12. 11 Inference about    –   : Unequal variances

13. 12 Inference about    –   : Unequal variances Conduct a hypothesis test as needed, or, build a confidence interval

14. 13 Which case to use: Equal variance or unequal variance? Whenever there is insufficient evidence that the variances are unequal, it is preferable to perform the equal variances t-test. This is so, because for any two given samples The number of degrees of freedom for the equal variances case The number of degrees of freedom for the unequal variances case 

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16. 15 Example 13.1 –Do people who eat high-fiber cereal for breakfast consume, on average, fewer calories for lunch than people who do not eat high-fiber cereal for breakfast? –A sample of 150 people was randomly drawn. Each person was identified as a consumer or a non-consumer of high-fiber cereal. –For each person the number of calories consumed at lunch was recorded. Example: Making an inference about    –  

17. 16 Solution: The data are interval. The parameter to be tested is the difference between two means. The claim to be tested is: The mean caloric intake of consumers (  1 ) is less than that of non-consumers (  2 ). Example: Making an inference about    –  

18. 17 The hypotheses are: H 0 : (  1 -  2 ) = 0 H 1 : (  1 -  2 ) < 0 – – To check the whether the population variances are equal, we use (Xm13-01) computer output to find the sample variances We have s 1 2 = 4103, and s 2 2 = 10,670.Xm13-01 – – It appears that the variances are unequal. Example: Making an inference about    –  

19. 18 Compute: Manually –From the data we have: Example: Making an inference about    –  

20. 19 Compute: Manually –The rejection region is t < -t  = -t.05,123  1.658 Example: Making an inference about    –  

21. 20 Example: Making an inference about    –   At the 5% significance level there is sufficient evidence to reject the null hypothesis. -2.09 < -1.6573 Xm13-01.0193 <.05

22. 21 Compute: Manually The confidence interval estimator for the difference between two means is Example: Making an inference about    –  

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24. 23 Example 13.2 –An ergonomic chair can be assembled using two different sets of operations (Method A and Method B) –The operations manager would like to know whether the assembly time under the two methods differ. Example: Making an inference about    –  

25. 24 Example 13.2 –Two samples are randomly and independently selected A sample of 25 workers assembled the chair using method A. A sample of 25 workers assembled the chair using method B. The assembly times were recorded –Do the assembly times of the two methods differs? Example: Making an inference about    –  

26. 25 Example: Making an inference about    –   Assembly times in Minutes Solution The data are interval. The parameter of interest is the difference between two population means. The claim to be tested is whether a difference between the two methods exists.

27. 26 Example: Making an inference about    –   Compute: Manually – –The hypotheses test is: H 0 : (  1 -  2 )  0 H 1 : (  1 -  2 )  0 – –To check whether the two unknown population variances are equal we calculate S 1 2 and S 2 2 (Xm13-02).Xm13-02 – – We have s 1 2 = 0.8478, and s 2 2 =1.3031. – – The two population variances appear to be equal.

28. 27 Example: Making an inference about    –   Compute: Manually – – To calculate the t-statistic we have:

29. 28 The rejection region is t t  = t. 025,48 = 2.009 The test: Since t= -2.009 < 0.93 < 2.009, there is insufficient evidence to reject the null hypothesis. For  = 0.05 2.009.093 -2.009 Rejection region Example: Making an inference about    –  

30. 29 Example: Making an inference about    –  .3584 >.05 -2.0106 <.93 < +2.0106 Xm13-02

31. 30 Conclusion: There is no evidence to infer at the 5% significance level that the two assembly methods are different in terms of assembly time Example: Making an inference about    –  

32. 31 Example: Making an inference about    –   A 95% confidence interval for  1 -  2 is calculated as follows: Thus, at 95% confidence level -0.3176 <  1 -  2 < 0.8616 Notice: “Zero” is included in the confidence interval

33. 32 Checking the required Conditions for the equal variances case (Example 13.2) The data appear to be approximately normal Design A Design B

34. 33 13.4 Matched Pairs Experiment What is a matched pair experiment? Why matched pairs experiments are needed? How do we deal with data produced in this way? The following example demonstrates a situation where a matched pair experiment is the correct approach to testing the difference between two population means.

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36. 35 Example 13.3 –To investigate the job offers obtained by MBA graduates, a study focusing on salaries was conducted. –Particularly, the salaries offered to finance majors were compared to those offered to marketing majors. –Two random samples of 25 graduates in each discipline were selected, and the highest salary offer was recorded for each one. The data are stored in file Xm13-03.Xm13-03 –Can we infer that finance majors obtain higher salary offers than do marketing majors among MBAs?. 13.4 Matched Pairs Experiment

37. 36 Solution –Compare two populations of interval data. –The parameter tested is  1 -  2 11 22 The mean of the highest salary offered to Finance MBAs The mean of the highest salary offered to Marketing MBAs – – H 0 : (  1 -  2 ) = 0 H 1 : (  1 -  2 ) > 0 13.4 Matched Pairs Experiment

38. 37 Solution – continued From the data we have: Let us assume equal variances 13.4 Matched Pairs Experiment There is insufficient evidence to conclude that Finance MBAs are offered higher salaries than marketing MBAs.

39. 38 Question –The difference between the sample means is 65624 – 60423 = 5,201. –So, why could we not reject H 0 and favor H 1 where (  1 –  2 > 0)? The effect of a large sample variability

40. 39 Answer: –S p 2 is large (because the sample variances are large) S p 2 = 311,330,926. –A large variance reduces the value of the t statistic and it becomes more difficult to reject H 0. The effect of a large sample variability

41. 40 Reducing the variability The values each sample consists of might markedly vary... The range of observations sample B The range of observations sample A

42. 41...but the differences between pairs of observations might be quite close to one another, resulting in a small variability of the differences. 0 Differences The range of the differences Reducing the variability

43. 42 The matched pairs experiment Since the difference of the means is equal to the mean of the differences we can rewrite the hypotheses in terms of  D (the mean of the differences) rather than in terms of  1 –  2. This formulation has the benefit of a smaller variability. Group 1 Group 2 Difference 1012- 2 1511+4 Mean1 =12.5 Mean2 =11.5 Mean1 – Mean2 = 1 Mean Differences = 1

44. 43 Example 13.4 –It was suspected that salary offers were affected by students’ GPA, (which caused S 1 2 and S 2 2 to increase). –To reduce this variability, the following procedure was used: 25 ranges of GPAs were predetermined. Students from each major were randomly selected, one from each GPA range. The highest salary offer for each student was recorded. –From the data presented can we conclude that Finance majors are offered higher salaries? The matched pairs experiment

45. 44 Solution (by hand) –The parameter tested is  D (=  1 –  2 ) –The hypotheses: H 0 :  D = 0 H 1 :  D > 0 –The t statistic: Finance Marketing The matched pairs hypothesis test Degrees of freedom = n D – 1 The rejection region is t > t.05,25-1 = 1.711

46. 45 Solution –From the data (Xm13-04) calculate:Xm13-04 The matched pairs hypothesis test

47. 46 Solution –Calculate t The matched pairs hypothesis test

48. 47 3.81 > 1.7109.0004 <.05 The matched pairs hypothesis test Xm13-04

49. 48 Conclusion: There is sufficient evidence to infer at 5% significance level that the Finance MBAs’ highest salary offer is, on the average, higher than that of the Marketing MBAs. The matched pairs hypothesis test

50. 49 The matched pairs mean difference estimation

51. 50 The matched pairs mean difference estimation Using Data Analysis Plus Xm13-04 First calculate the differences, then run the confidence interval procedure in Data Analysis Plus.

52. 51 Checking the required conditions for the paired observations case The validity of the results depends on the normality of the differences.

53. 52 13.5 Inference about the ratio of two variances In this section we draw inference about the ratio of two population variances. This question is interesting because: –Variances can be used to evaluate the consistency of processes. –The relationship between population variances determines which of the equal-variances or unequal- variances t-test and estimator of the difference between means should be applied

54. 53 Parameter to be tested is  1 2 /  2 2 Statistic used is Parameter and Statistic Sampling distribution of  1 2 /  2 2 – –The statistic [ s 1 2 /  1 2 ] / [ s 2 2 /  2 2 ] follows the F distribution with 1 = n 1 – 1, and 2 = n 2 – 1.

55. 54 –Our null hypothesis is always H 0 :  1 2 /  2 2 = 1 – –Under this null hypothesis the F statistic becomes F = S12/12S12/12 S22/22S22/22 Parameter and Statistic

56. 55

57. 56 (see Xm13-01)Xm13-01 In order to perform a test regarding average consumption of calories at people’s lunch in relation to the inclusion of high-fiber cereal in their breakfast, the variance ratio of two samples has to be tested first. Example 13.6 (revisiting Example 13.1) Calories intake at lunch The hypotheses are: H 0 : H 1 : Testing the ratio of two population variances

58. 57 – –The F statistic value is F=S 1 2 /S 2 2 =.3845 – –Conclusion: Because.3845<.58 we reject the null hypothesis in favor of the alternative hypothesis, and conclude that there is sufficient evidence at the 5% significance level that the population variances differ. Testing the ratio of two population variances Solving by hand –The rejection region is F>F  2, 1, 2 or F<1/F 

59. 58 (see Xm13-01)Xm13-01 In order to perform a test regarding average consumption of calories at people’s lunch in relation to the inclusion of high-fiber cereal in their breakfast, the variance ratio of two samples has to be tested first. The hypotheses are: H 0 : H 1 : Example 13.6 (revisiting Example 13.1) Testing the ratio of two population variances

60. 59 Estimating the Ratio of Two Population Variances From the statistic F = [ s 1 2 /  1 2 ] / [ s 2 2 /  2 2 ] we can isolate  1 2 /  2 2 and build the following confidence interval:

61. 60 Example 13.7 –Determine the 95% confidence interval estimate of the ratio of the two population variances in Example 13.1 –Solution We find F  /2,v1,v2 = F.025,40,120 = 1.61 (approximately) F  /2,v2,v1 = F.025,120,40 = 1.72 (approximately) LCL = (s 1 2 /s 2 2 )[1/ F  /2,v1,v2 ] = (4102.98/10,669.77)[1/1.61]=.2388 UCL = (s 1 2 /s 2 2 )[ F  /2,v2,v1 ] = (4102.98/10,669.77)[1.72]=.6614 Estimating the Ratio of Two Population Variances

62. 61 13.6 Inference about the difference between two population proportions In this section we deal with two populations whose data are nominal. For nominal data we compare the population proportions of the occurrence of a certain event. Examples –Comparing the effectiveness of new drug versus older one –Comparing market share before and after advertising campaign –Comparing defective rates between two machines

63. 62 Parameter and Statistic Parameter –When the data are nominal, we can only count the occurrences of a certain event in the two populations, and calculate proportions. –The parameter is therefore p 1 – p 2. Statistic –An unbiased estimator of p 1 – p 2 is (the difference between the sample proportions).

64. 63 Sample 1 Sample size n 1 Number of successes x 1 Sample proportion Sample 1 Sample size n 1 Number of successes x 1 Sample proportion Two random samples are drawn from two populations. The number of successes in each sample is recorded. The sample proportions are computed. Sample 2 Sample size n 2 Number of successes x 2 Sample proportion Sample 2 Sample size n 2 Number of successes x 2 Sample proportion x n 1 1 ˆ  p 1 Sampling Distribution of

65. 64 The statistic is approximately normally distributed if n 1 p 1, n 1 (1 - p 1 ), n 2 p 2, n 2 (1 - p 2 ) are all greater than or equal to 5. The mean of is p 1 - p 2. The variance of is (p 1 (1-p 1 ) /n 1 )+ (p 2 (1-p 2 )/n 2 ) Sampling distribution of

66. 65 The z-statistic Because and are unknown the standard error must be estimated using the sample proportions. The method depends on the null hypothesis

67. 66 Testing the p 1 – p 2 There are two cases to consider: Case 1: H 0 : p 1 -p 2 =0 Calculate the pooled proportion Then Case 2: H 0 : p 1 -p 2 =D (D is not equal to 0) Do not pool the data

68. 67 Example 13.8 –The marketing manager needs to decide which of two new packaging designs to adopt, to help improve sales of his company’s soap. –A study is performed in two supermarkets: Brightly-colored packaging is distributed in supermarket 1. Simple packaging is distributed in supermarket 2. –First design is more expensive, therefore,to be financially viable it has to outsell the second design. Testing p 1 – p 2 (Case 1)

69. 68 Summary of the experiment results –Supermarket 1 - 180 purchasers of Johnson Brothers soap out of a total of 904 –Supermarket 2 - 155 purchasers of Johnson Brothers soap out of a total of 1,038 –Use 5% significance level and perform a test to find which type of packaging to use. Testing p 1 – p 2 (Case 1)

70. 69 Solution –The problem objective is to compare the population of sales of the two packaging designs. –The data are nominal (Johnson Brothers or other soap) –The hypotheses are H 0 : p 1 - p 2 = 0 H 1 : p 1 - p 2 > 0 –We identify this application as case 1 Population 1: purchases at supermarket 1 Population 2: purchases at supermarket 2 Testing p 1 – p 2 (Case 1)

71. 70 Testing p 1 – p 2 (Case 1) Compute: Manually –For a 5% significance level the rejection region is z > z  = z.05 = 1.645

72. 71 Testing p 1 – p 2 (Case 1) Excel (Data Analysis Plus) Conclusion: T here is sufficient evidence to conclude at the 5% significance level, that brightly-colored design will outsell the simple design. Xm13-08

73. 72 Example 13.9 (Revisit Example 13.8) –Management needs to decide which of two new packaging designs to adopt, to help improve sales of a certain soap. –A study is performed in two supermarkets: –For the brightly-colored design to be financially viable it has to outsell the simple design by at least 3%. Testing p 1 – p 2 (Case 2)

74. 73 Summary of the experiment results –Supermarket 1 - 180 purchasers of Johnson Brothers’ soap out of a total of 904 –Supermarket 2 - 155 purchasers of Johnson Brothers’ soap out of a total of 1,038 –Use 5% significance level and perform a test to find which type of packaging to use. Testing p 1 – p 2 (Case 2)

75. 74 Solution –The hypotheses to test are H 0 : p 1 - p 2 =.03 H 1 : p 1 - p 2 >.03 –We identify this application as case 2 (the hypothesized difference is not equal to zero). Testing p 1 – p 2 (Case 2)

76. 75 Compute: Manually The rejection region is z > z  = z.05 = 1.645. Conclusion: Since 1.15 < 1.645 do not reject the null hypothesis. There is insufficient evidence to infer that the brightly-colored design will outsell the simple design by 3% or more. Testing p 1 – p 2 (Case 2)

77. 76 Testing p 1 – p 2 (Case 2) Using Excel (Data Analysis Plus) Xm13-08

78. 77 Estimating p 1 – p 2 Estimating the cost of life saved –Two drugs are used to treat heart attack victims: Streptokinase (available since 1959, costs $460) t-PA (genetically engineered, costs $2900). –The maker of t-PA claims that its drug outperforms Streptokinase. –An experiment was conducted in 15 countries. 20,500 patients were given t-PA 20,500 patients were given Streptokinase The number of deaths by heart attacks was recorded.

79. 78 Experiment results –A total of 1497 patients treated with Streptokinase died. –A total of 1292 patients treated with t-PA died. Estimate the cost per life saved by using t-PA instead of Streptokinase. Estimating p 1 – p 2

80. 79 Solution –The problem objective: Compare the outcomes of two treatments. –The data are nominal (a patient lived or died) –The parameter to be estimated is p 1 – p 2. p 1 = death rate with t-PA p 2 = death rate with Streptokinase Estimating p 1 – p 2

81. 80 Compute: Manually –Sample proportions: –The 95% confidence interval estimate is Estimating p 1 – p 2

82. 81 Interpretation –We estimate that between.51% and 1.49% more heart attack victims will survive because of the use of t-PA. –The difference in cost per life saved is 2900-460= $2440. –The total cost saved by switching to t-PA is estimated to be between 2440/.0149 = $163,758 and 2440/.0051 = $478,431 Estimating p 1 – p 2