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Chapter Eleven McGraw-Hill/Irwin © 2006 The McGraw-Hill Companies, Inc., All Rights Reserved. Two-Sample Tests of Hypothesis Pages 313-322 & 327-30.

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Presentation on theme: "Chapter Eleven McGraw-Hill/Irwin © 2006 The McGraw-Hill Companies, Inc., All Rights Reserved. Two-Sample Tests of Hypothesis Pages 313-322 & 327-30."— Presentation transcript:

1 Chapter Eleven McGraw-Hill/Irwin © 2006 The McGraw-Hill Companies, Inc., All Rights Reserved. Two-Sample Tests of Hypothesis Pages 313-322 & 327-30

2 Does the distribution of the differences in sample means have a mean of 0? Independent Samples The formula for computing the value of z is:

3 L.A.Times recently reported that the mean household income in Orange county is $38,000 with a SD of $6,000 for a sample of 40 households. The same article reported the mean income in LA county is $35,000 with a SD of $7,000 for a sample of 35 households. At the.01 significance level can we conclude the mean income in Orange county is more? Because the computed Z of 1.98 < critical Z of 2.33, the decision is to not reject the null hypothesis. We cannot conclude that the mean household income in Orange county is larger.

4 Two Sample Tests of Proportions Two Sample Tests of Proportions investigate whether two samples came from populations with an equal proportion of successes. The two samples are pooled using the following formula. where X 1 and X 2 refer to the number of successes in the respective samples of n 1 and n 2. The value of the test statistic is computed from the following formula. where X 1 and X 2 refer to the number of successes in the respective samples of n 1 and n 2.

5 Are unmarried workers more likely to be absent from work than married workers? A sample of 250 married workers showed 22 missed more than 5 days last year, while a sample of 300 unmarried workers showed 35 missed more than five days. Use a.05 significance level.

6 The null and the alternate hypotheses H 0 :  U  M The null hypothesis is rejected if the computed value of z is greater than 1.65 or the p-value <.05. The pooled proportion =.1036

7 Example 2 continued The p(z > 1.10) =.136 for a one-tailed test of significance. Because a calculated z of 1.10  of.05, the null hypothesis is not rejected. We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the married workers.

8 Hypothesis Testing with Two Samples In Chapter 10, we studied how to perform hypothesis testing with one sample. Eg. H 0 :μ = 100 H 1 :μ ≠ 100 In this chapter, we learn how to deal with two samples. Our focus in this course is on ‘dependent’ samples. This material is to help you with the case study only. There will be no questions in the exams from Chapter 11.

9 Dependent samples These are samples that are paired or related in some fashion. Eg. If you wish to measure the effectiveness of an exercise program on work absenteeism, you would count the number of days missed by the same employees before and after the exercise program.

10 When samples are dependent, use the Paired t-Test Statistic (Formula 11-7, Page 328): where is the mean of the differences is the standard deviation of the differences n is the number of pairs Note similarity to

11 Practice! Problem 41, Page 338 EmployeeBeforeAfter 165 262 371 473 543 636 753 867 Measuring effect of exercise program on absenteeism

12 Step 4 H 0 is rejected if t > 1.895; or if p-value >.05. We use the t distribution with n-1 or 7 degrees of freedom. Step 2 The stated significance level is.05. Step 3 The appropriate test statistic is the paired t-test. Step 1 H o :  d ≤ 0 H 1 :  d > 0 Step 5 Perform the calculations and make a decision.

13 d SdSd = d / (S d /√n) (for 1 tail) Solving forusing Excel Conclusion: Exercise program did not reduce absenteeism.

14 You can get Excel to do all the calculations for your case using i) Tools|Descriptive Statististics|95% ii) Tools|t-Paired Two Sample for Means Shortcut! (>0.05) Do not reject H 0


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