1. Chapter 11 Lesson 11.3b Comparing Two Populations or Treatments 11.3: Inferences Concerning the Difference Between 2 Population or Treatment Proportions

2. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Everyone into the Pool The typical hypothesis test for the difference in two proportions is the one of no difference. In symbols, H 0 : p 1 – p 2 = 0. Since we are hypothesizing that there is no difference between the two proportions, that means that the standard deviations for each proportion are the same. Since this is the case, we combine (pool) the counts to get one overall proportion.

3. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Everyone into the Pool (cont.) The pooled proportion is where and If the numbers of successes are not whole numbers, round them first. (This is the only time you should round values in the middle of a calculation.)

4. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Two-Proportion z-Test We are testing the hypothesis H 0 : p 1 - p 2 = 0. The conditions for the two-proportion z-test are the same as for the two-proportion z-interval. Since the conditions are met, we will use the Normal model to do a 2-proportion z-test. Because we hypothesize that the proportions are equal, we pool them to find

5. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Two-Proportion z-Test (cont.) We use the pooled value to estimate the standard error: Now we find the test statistic:

6. Investigators at Madigan Army Medical Center tested using duct tape to remove warts. Patients with warts were randomly assigned to either the duct tape treatment or to the more traditional freezing treatment. Those in the duct tape group wore duct tape over the wart for 6 days, then removed the tape, soaked the area in water, and used an emery board to scrape the area. This process was repeated for a maximum of 2 months or until the wart was gone. The data follows: Do these data suggest that freezing is less successful than duct tape in removing? warts? Treatmentn Number with wart successfully removed Liquid nitrogen freezing10060 Duct tape10488

7. Duct Tape Continued... H 0 : p 1 – p 2 = 0 H a : p 1 – p 2 < 0 Assumptions: 1) Subjects were randomly assigned to the two treatments. Treatmentn Number with wart successfully removed Liquid nitrogen freezing10060 Duct tape10488 Where p 1 is the true proportion of warts that would be successfully removed by freezing and p 2 is the true proportion of warts that would be successfully removed by duct tape 2) The sample sizes are large enough because: n 1 p 1 = 60 > 10 n 1 (1 – p 1 ) = 40 > 10 n 2 p 2 = 88 > 10 n 2 (1 – p 2 ) = 16 > 10 3) We assume the two groups are independent

8. Duct Tape Continued... H 0 : p 1 – p 2 = 0 H a : p 1 – p 2 < 0 Treatmentn Number with wart successfully removed Liquid nitrogen freezing10060 Duct tape10488 P-value ≈ 0  =.01 Since the P-value < , we reject H 0. There is enough evidence to suggest that the proportion of warts successfully removed with the freezing method is lower than the proportion of success using the duct tape method.

9. Practice Hd Class Example: “Would being part of a support group…”

10. Homework Pg.573: #45, (47), (49), 51