1 Inference about Two Populations Chapter 13

Introduction Variety of techniques are presented to compare two populations. We are interested in: –The difference between two means. –The ratio of two variances. –The difference between two proportions.

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

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

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    –  

6 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    –   When Population Variances are Unknown

7 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:

8 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

9 Inference about    –   : Unequal variances

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

11 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    –  

12 Solution: 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    –  

13 The hypotheses are: H 0 : (  1 -  2 ) = 0 H 1 : (  1 -  2 ) < 0 – – Are population variances equal? We have s 1 2 = 4103, and s 2 2 = 10,670. – – It appears that the variances are unequal. Example: Making an inference about    –  

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

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

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

17 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    –  

18 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    –  

19 Example: Making an inference about    –   Assembly times in Minutes Solution 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.

20 Example: Making an inference about    –   Compute: Manually – –The hypotheses test is: H 0 : (  1 -  2 )  0 H 1 : (  1 -  2 )  0 – –Are population variances equal? – – We have s 1 2 = , and s 2 2 = – – The two population variances appear to be equal.

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

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

23 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    –  

24 Example: Making an inference about    –   A 95% confidence interval for  1 -  2 : Thus, at 95% confidence level <  1 -  2 < Notice: “Zero” is included in the confidence interval

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?

26 The matched pairs experiment Note  D =  1 –  2. This formulation has the benefit of a smaller variability. Group 1 Group 2 Difference Mean1 =12.5 Mean2 =11.5 Mean1 – Mean2 = 1 Mean Differences = 1

27 Example 13.4 –It was suspected that salary offers were affected by students’ GPA. –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

28 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

29 Solution –Calculate t The matched pairs hypothesis test

30 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

31 The matched pairs mean difference estimation

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: It determines which of the equal-variances or unequal- variances t-test should be applied

33 Parameter to be tested is  1 2 /  2 2 Statistic used is Parameter and Statistic Sampling distribution of F: It follows the F distribution with 1 = n 1 – 1, and 2 = n 2 – 1.

34 –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

35 (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

36 – –The F statistic value is F=S 1 2 /S 2 2 =.3845 – –Conclusion: Because.3845<.58 we reject the null 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 

37 (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

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

39 Parameter and Statistic Parameter The parameter is therefore p 1 – p 2. Statistic –An unbiased estimator of p 1 – p 2 is.

40 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

41 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

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

43 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

44 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)

45 Summary of the experiment results –Supermarket purchasers of Johnson Brothers soap out of a total of 904 –Supermarket 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)

46 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)

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

48 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)

49 Summary of the experiment results –Supermarket purchasers of Johnson Brothers’ soap out of a total of 904 –Supermarket 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)

50 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)

51 Compute: Manually The rejection region is z > z  = z.05 = Conclusion: Since 1.15 < 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)

52 Find the 95% C.I. for p 1 – p 2