1. 11- 1 Chapter Eleven McGraw-Hill/Irwin © 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.

2. 11- 2 Chapter Eleven Two-Sample Tests of Hypothesis GOALS When you have completed this chapter, you will be able to: TWO Conduct a test of hypothesis regarding the difference in two population proportions. THREE Conduct a test of hypothesis about the mean difference between paired or dependent observations. ONE Conduct a test of hypothesis about the difference between two independent population means.

3. 11- 3 Chapter Eleven continued Two Sample Tests of Hypothesis GOALS When you have completed this chapter, you will be able to: FOUR Understand the difference between dependent and independent samples.

4. 11- 4 Comparing two populations Does the distribution of the differences in sample means have a mean of 0? Comparing two populations If both samples contain at least 30 observations we use the z distribution as the test statistic. No assumptions about the shape of the populations are required. The samples are from independent populations. The formula for computing the value of z is:

5. 11- 5 EXAMPLE 1 with a standard deviation of $7,000 for a sample of 35 households. At the.01 significance level can we conclude the mean income in Bradford is more? Two cities, Bradford and Kane are separated only by the Conewango River. There is competition between the two cities. The local paper recently reported that the mean household income in Bradford is $38,000 with a standard deviation of $6,000 for a sample of 40 households. The same article reported the mean income in Kane is $35,000

6. 11- 6 Example 1 continued Step 2 State the level of significance. The.01 significance level is stated in the problem. Step 3 Find the appropriate test statistic. Because both samples are more than 30, we can use z as the test statistic. Step 1 State the null and alternate hypotheses. H 0 : µB < µK H 1 : µB > µK Step 4 State the decision rule. The null hypothesis is rejected if z is greater than 2.33 or p <.01.

7. 11- 7 Example 1 continued Step 5: Compute the value of z and make a decision. The p(z > 1.98) is.0239 for a one-tailed test of significance. Because the computed Z of 1.98  of.01, the decision is to not reject the null hypothesis. We cannot conclude that the mean household income in Bradford is larger.

8. 11- 8 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.

9. 11- 9 Example 2 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.

10. 11- 10 Example 2 continued 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

11. 11- 11 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.

12. 11- 12 Small Sample Tests of Means The required assumptions 1. Both populations must follow the normal distribution. 2. The populations must have equal standard deviations. 3. The samples are from independent populations. Small Sample Tests of Means The t distribution is used as the test statistic if one or more of the samples have less than 30 observations.

13. 11- 13 Small sample test of means continued Step Two: Determine the value of t from the following formula. Finding the value of the test statistic requires two steps. Step One: Pool the sample standard deviations.

14. 11- 14 Example 3 A recent EPA study compared the highway fuel economy of domestic and imported passenger cars. A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg. A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9. At the.05 significance level can the EPA conclude that the mpg is higher on the imported cars?

15. 11- 15 Example 3 continued Step 1 State the null and alternate hypotheses. H 0 : µ D > µ I H 1 : µ D < µ I Step 2 State the level of significance. The.05 significance level is stated in the problem. Step 3 Find the appropriate test statistic. Both samples are less than 30, so we use the t distribution.

16. 11- 16 Example 3 continued Step 4 The decision rule is to reject H 0 if t<-1.708 or if p-value <.05. There are n-1 or 25 degrees of freedom. Step 5 We compute the pooled variance.

17. 11- 17 Example 3 continued We compute the value of t as follows.

18. 11- 18 Since a computed z of –1.64 > critical z of –1.71, the p- value of.0567 >  of.05, H 0 is not rejected. There is insufficient sample evidence to claim a higher mpg on the imported cars. P(t < -1.64) =.0567 for a one- tailed t-test. Example 3 continued

19. 11- 19 Hypothesis Testing Involving Paired Observations Dependent samples are samples that are paired or related in some fashion. Independent samples are samples that are not related in any way. If you wished to buy a car you would look at the same car at two (or more) different dealerships and compare the prices. If you wished to measure the effectiveness of a new diet you would weigh the dieters at the start and at the finish of the program.

20. 11- 20 Hypothesis Testing Involving Paired Observations Use the following test when the samples are dependent: where is the mean of the differences is the standard deviation of the differences n is the number of pairs (differences)

21. 11- 21 EXAMPLE 4 An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis. A random sample of eight cities revealed the following information. At the.05 significance level can the testing agency conclude that there is a difference in the rental charged? CityHertz ($) Avis ($) Atlanta4240 Chicago5652 Cleveland4543 Denver48 Honolulu3732 Kansas City4548 Miami4139 Seattle4650

22. 11- 22 Example 4 continued Step 4 H 0 is rejected if t 2.365; 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.

23. 11- 23 Example 4 continued CityHertzAvisd d 2 Atlanta4240 2 4 Chicago5652 416 Cleveland4543 2 4 Denver4848 0 0 Honolulu3732 525 Kansas City4548-3 9 Miami4139 2 4 Seattle4650-416

24. 11- 24 Example 4 continued

25. 11- 25 Example 4 continued P(t>.894) =.20 for a one-tailed t-test at 7 degrees of freedom. Because 0.894 is less than the critical value, the p-value of.20 > a of.05, do not reject the null hypothesis. There is no difference in the mean amount charged by Hertz and Avis.

26. 11- 26 Advantage of dependent samples: Reduction in variation in the sampling distribution Disadvantage of dependent samples: Degrees of freedom are halved Comparing dependent and independent samples The same subjects measured at two different points in time. Two types of dependent samples Matched or paired observations