1. 1 Inference about Comparing Two Populations Chapter 13

2. 2 13.1 Introduction In previous discussions we presented methods designed to make an inference about characteristics of a single population. We estimated, for example the population mean, or hypothesized on the value of the standard deviation. However, in the real world we encounter many times the need to study the relationship between two populations. –For example, we want to compare the effects of a new drug on blood pressure, in which case we can test the relationship between the mean blood pressure of two groups of individuals: those who take the drug, and those who don’t. –Or, we are interested in the effects a certain ad has on voters’ preferences as part of an election campaign. In this case we can estimate the difference in the proportion of voters who prefer one candidate before and after the ad is televised.

3. 3 13.1 Introduction Variety of techniques are presented whose objective is to compare two populations. These techniques are designed to study the… –difference between two means. –ratio of two variances. –difference between two proportions.

4. 4 The reason we are looking at the difference between the two means is that is strongly related to a normal distribution, whose mean is  1 –  2. See next for details. Two random samples are therefore drawn from the two populations of interest and their means and are calculated. We’ll look at the relationship between the two population means by analyzing the value of  1 –  2. 13.2Inference about the Difference between Two Means: Independent Samples

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

6. 6 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. 7 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. 8 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. 9 Inference about    –   : Equal variances If the two variances  1 2 and  2 2 are equal to one another, then their estimate S 1 2 and S 2 2 estimate the same value. Therefore, we can pool the two sample variances and provide a better estimate of the common populations’ variance, based on a larger amount of information. This is done by forming the pooled variance estimate. See next.

10. 10 To get some intuition about this pooled estimate, note that we can re-write it as which has the form of a weighted average of the two sample variances. The weights are the relative sample sizes. A larger sample provides larger weight and thus influences the pooled estimate more (it might be easier to eliminate the values ‘-1’ and ‘-2’ from the formula in order to see the structure  more easily  Inference about    –   : Equal variances Calculate the pooled variance estimate by:

11. 11 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:

12. 12 Note how S p 2 replaces both S 1 2 and S 2 2. Inference about    –   : Equal variances Construct the t-statistic as follows:

13. 13 Inference about    –   : Unequal variances

14. 14 Which case to use: Equal variance or unequal variance? Whenever there is insufficient evidence that the variances are unequal, it is preferable to run 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 

15. 15

16. 16 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. 17 Solution: The data are quantitative. 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. 18 The hypotheses are: H 0 :  1 -  2 = 0 H 1 :  1 -  2 < 0 – – To check the relationships between the variances, we use a computer output to find the sample variances (Xm13-1.xls). From the data we have S 1 2 = 4103, and S 2 2 = 10,670.Xm13-1.xls – – It appears that the variances are unequal. Example: Making an inference about    –   1= mean caloric intake for fiber consumers 2= mean caloric intake for fiber non-consumers

19. 19 Solving by hand –From the data we have: Example: Making an inference about    –  

20. 20 Solving by hand –H 1 :  1 -  2 < 0 The rejection region is t < -t  df = -t.05,123  1.658 Example: Making an inference about    –  

21. 21 Example: Making an inference about    –   At 5% significance level there is sufficient evidence to reject the null hypothesis. -2.09107 < -1.65734 Xm13-1.xls.01929 <.05

22. 22 Solving by hand The confidence interval estimator for the difference between two means when the variances are unequal is Example: Making an inference about    –  

23. 23 Note that the confidence interval for the difference between the two means falls entirely in the negative region: [-56.86, -1.56]; even at best the difference between the two means is m 1 – m 2 = -1.56, so we can be 95% confident m 1 is smaller than m 2 ! This conclusion agrees with the results of the test performed before. Example: Making an inference about    –  

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

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

27. 27 Example: Making an inference about    –   Solving by hand – –The hypotheses test is: H 0 :  1 -  2  0 H 1 :  1 -  2  0 – –To check the relationship between the two variances we calculate the value of S 1 2 and S 2 2 (Xm13-02.xls).Xm13-02.xls – – From the data we have S 1 2 = 0.8478, and S 2 2 =1.3031. so  1 2 and  2 2 appear to be equal.

28. 28 Example: Making an inference about    –   Solving by hand – – To calculate the t-statistic we have:

29. 29 The 2-tail 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. 30 Example: Making an inference about    –  .35839 >.05 -2.0106 <.9273 < +2.0106 Xm13-02.xls

31. 31 Conclusion: From this experiment, it is unclear at 5% significance level if the two assembly methods are different in terms of assembly time Example: Making an inference about    –  

32. 32 Example: Making an inference about    –    Constructing a Confidence Interval A 95% confidence interval for  1 -  2 when the two variances are equal is calculated as follows: Thus, at 95% confidence level -0.3176 <  1 -  2 < 0.8616 Notice: “Zero” is included in the confidence interval and therefore the two mean values could be equal.

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

34. 34 13.4 Matched Pairs Experiment - Dependent samples What is a matched pair experiment? A matched pairs experiment is a sampling design in which every two observations share some characteristic. For example, suppose we are interested in increasing workers productivity. We establish a compensation program and want to study its efficiency. We could select two groups of workers, measure productivity before and after the program is established and run a test as we did before. But, if we believe workers’ age is a factor that may affect changes in productivity, we can divide the workers into different age groups, select a worker from each age group, and measure his or her productivity twice. One time before and one time after the program is established. Each two observations constitute a matched pair, and because they belong to the same age group they are not independent.

35. 35 13.4 Matched Pairs Experiment - Dependent samples Why matched pairs experiments are needed? The following example demonstrates a situation where a matched pair experiment is the correct approach to testing the difference between two population means.

36. 36 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. –From the data, can we infer that finance majors obtain higher salary offers than marketing majors among MBAs?. 13.4 Matched Pairs Experiment Additional example

37. 37 Solution –Compare two populations of quantitative 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. 38 Solution – continued From Xm13-3.xls we have:Xm13-3.xls 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. 39 Question –The difference between the sample means is 65624 – 60423 = 5,201. –So, why could not we reject H 0 and favor H 1 ? The effect of a large sample variability

40. 40 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 Recall that rejection of the null hypothesis occurs when ‘t’ is sufficiently large (t>t  ). A large S p 2 reduces ‘t’ and therefore it does not fall in the rejection region.

41. 41 The matched pairs experiment We are looking for hypotheses formulation where the variability of the two samples has been reduced. By taking matched pair observations and testing the differences per pair we achieve two goals: –We still test  1 –  2 (see explanation next) –The variability used to calculate the t-statistic is usually smaller (see explanation next).

42. 42 The matched pairs experiment – Are we still testing  1 –  2 ? Note that the difference between the two means is equal to the mean difference of pairs of observations A short example Group 1 Group 2 Difference 1012- 2 1511+4 Mean1 =12.5 Mean2 =11.5 Mean1 – Mean2 = 1 Mean Differences = 1

43. 43 The matched pairs experiment – Reducing the variability Observations might markedly differ... The range of observations sample B The range of observations sample A

44. 44...but the differences between pairs of observations might have much smaller variability. 0 Differences The range of the differences The matched pairs experiment – Reducing the variability

45. 45 Example 12.4 (12.3 part II) –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

46. 46 Solution (by hand) –The parameter tested is  D (=  1 –  2 ) –The hypotheses: H 0 :  D = 0 H 1 :  D > 0 –The t statistic:  The matched pairs hypothesis test    –  The rejection region is t > t.05,25-1 = 1.711

47. 47 Solution (by hand) – continue –From the data (Xm13-4.xls) calculate:Xm13-4.xls The matched pairs hypothesis test Using Descriptive Statistics in Excel we get:

48. 48 Solution (by hand) – continue –Calculate t The matched pairs hypothesis test See conclusion later

49. 49 Recall: The rejection region is t > t . Indeed, 3.809 > 1.7108.000426 <.05 The matched pairs hypothesis test Xm13-4.xls Using Data Analysis in Excel

50. 50 Conclusion: There is sufficient evidence to infer at 5% significance level that the Finance MBAs’ highest salary offer is, on the average, higher than this of the Marketing MBAs. The matched pairs hypothesis test

51. 51 The matched pairs mean difference estimation

52. 52 The matched pairs mean difference estimation Using Data Analysis Plus Xm13-4.xls First calculate the differences for each pair, then run the confidence interval procedure in Data Analysis Plus.

53. 53 Checking the required conditions for the paired observations case The validity of the results depends on the normality of the differences.

54. 54 13.5 Inferences about the ratio of two variances In this section we draw inference about the relationship between two population variances. This question is interesting because: –Variances can be used to evaluate the consistency of processes. –The relationships between variances determine the technique used to test relationships between mean values

55. 55 The parameter tested is  1 2 /  2 2 The statistic used is Parameter tested and statistic The Sampling distribution of  1 2 /  2 2 – –The statistic [ s 1 2 /  1 2 ] / [ s 2 2 /  2 2 ] follows the F distribution with… Numerator d.f. = n 1 – 1, and Denominator d.f. = n 2 – 1.

56. 56 –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 tested and statistic

57. 57

58. 58 (see example 13.1) In order to test whether having a rich-in-fiber breakfast reduces the amount of caloric intake at lunch, we need to decide whether the variances are equal or not. Example 13.6 (revisiting 13.1) Calories intake at lunch The hypotheses are: H 0 : H 1 : Testing the ratio of two population variances

59. 59 – –The F statistic value is F=S 1 2 /S 2 2 =.3845 – –Conclusion: Because.3845<.63 we can reject the null hypothesis in favor of the alternative hypothesis, and conclude that there is sufficient evidence in the data to argue at 5% significance level that the variance of the two groups differ. Testing the ratio of two population variances Solving by hand –The rejection region is     

60. 60 (see Xm13.1)Xm13.1 The hypotheses are: H 0 : H 1 : Example 13.6 (revisiting 13.1) Testing the ratio of two population variances From Data Analysis

61. 61 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:

62. 62 Example 13.7 –Determine the 95% confidence interval estimate of the ratio of the two population variances in example 12.1 –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 a/2,v1,v2 ] = (4102.98/10,669.770)[1/1.61]=.2388 UCL = (s 1 2 /s 2 2 )[ F a/2,v2,v1 ] = (4102.98/10,669.770)[1.72]=.6614 Estimating the Ratio of Two Population Variances

63. 63 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 vs.old one –Comparing market share before and after advertising campaign –Comparing defective rates between two machines

64. 64 Parameter tested and statistic Parameter –When the data is nominal, we can only count the occurrences of a certain event in the two populations, and calculate proportions. –The parameter tested is therefore p 1 – p 2. Statistic –An unbiased estimator of p 1 – p 2 is (the difference between the sample proportions).

65. 65 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 Sampling distribution of 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

66. 66 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 equal to or greater than 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

67. 67 Because p 1 and p 2 are unknown, we use their estimates instead. Thus, should all be equal to or greater than 5. The z-statistic

68. 68 Testing 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

69. 69 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 communities: Design A is distributed in Community 1. Design B is distributed in Community 2. The old design packages is still offered in both communities. –Design A is more expensive, therefore,to be financially viable it has to outsell design B. Testing p 1 – p 2 (Case I)

70. 70 Summary of the experiment results –Community 1 - 580 packages with new design A sold 324 packages with old design sold –Community 2 - 604 packages with new design B sold 442 packages with old design sold –Use 5% significance level and perform a test to find which type of packaging to use. Testing p 1 – p 2 (Case I)

71. 71 Solution –The problem objective is to compare the population of sales of the two packaging designs. –The data is qualitative (yes/no for the purchase of the new design per customer) –The hypotheses test are H 0 : p 1 - p 2 = 0 H 1 : p 1 - p 2 > 0 –We identify here case 1. Population 1 – purchases of Design A Population 2 – purchases of Design B Testing p 1 – p 2 (Case I)

72. 72 Solving by hand –For a 5% significance level the rejection region is z > z  = z.05 = 1.645 From Xm13-08.xls we have:Xm13-08.xls Testing p 1 – p 2 (Case I)

73. 73 Conclusion: At 5% significance level there sufficient evidence to infer that the proportion of sales with design A is greater that the proportion of sales with design B (since 2.89 > 1.645). Testing p 1 – p 2 (Case I)

74. 74 Excel (Data Analysis Plus) Testing p 1 – p 2 (Case I) Xm13-08.xls Conclusion Since 2.89 > 1.645, there is sufficient evidence in the data to conclude at 5% significance level, that design A will outsell design B. Additional example

75. 75 Example 13.9 (Revisit example 13.08) –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 communities: Design A is distributed in Community 1. Design B is distributed in Community 2. The old design packages is still offered in both communities. –For design A to be financially viable it has to outsell design B by at least 3%. Testing p 1 – p 2 (Case II)

76. 76 Summary of the experiment results –Community 1 - 580 packages with new design A sold 324 packages with old design sold –Community 2 - 604 packages with new design B sold 442 packages with old design sold Use 5% significance level and perform a test to find which type of packaging to use. Testing p 1 – p 2 (Case II)

77. 77 Solution –The hypotheses to test are H 0 : p 1 - p 2 =.03 H 1 : p 1 - p 2 >.03 –We identify case 2 of the test for difference in proportions (the difference is not equal to zero). Testing p 1 – p 2 (Case II)

78. 78 Solving by hand The rejection region is z > z  = z.05 = 1.645. Conclusion: Since 1.58 < 1.645 do not reject the null hypothesis. There is insufficient evidence to infer that packaging with Design A will outsell this of Design B by 3% or more. Testing p 1 – p 2 (Case II)

79. 79 Using Excel (Data Analysis Plus) Testing p 1 – p 2 (Case II) Xm13-08.xls

80. 80 Estimating p 1 – p 2 Example (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.

81. 81 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

82. 82 Solution –The problem objective: Compare the outcomes of two treatments. –The data is nominal (a patient lived/died) –The parameter estimated is p 1 – p 2. p 1 = death rate with t-PA p 2 = death rate with Streptokinase Estimating p 1 – p 2

83. 83 Solving by hand –Sample proportions: –The 95% confidence interval is Estimating p 1 – p 2

84. 84 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