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

2. Everyone into the Pool
The typical hypothesis test for the difference in two proportions is the one of no difference. In symbols, H0: p1 – p2 = 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. 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. Two-Proportion z-Test
We are testing the hypothesis H0: p1 - p2 = 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. Two-Proportion z-Test (cont.)
We use the pooled value to estimate the standard error:
Now we find the test statistic:

6. Number with wart successfully removed
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?
Treatment n
Number with wart successfully removed
Liquid nitrogen freezing
100 60
Duct tape
104 88

7. Number with wart successfully removed
Duct Tape Continued . . .
H0: p1 – p2 = 0
Ha: p1 – p2 < 0
Assumptions:
1) Subjects were randomly assigned to the two treatments.
Treatment n
Number with wart successfully removed
Liquid nitrogen freezing
100 60
Duct tape
104 88
Where p1 is the true proportion of warts that would be successfully removed by freezing and p2 is the true proportion of warts that would be successfully removed by duct tape
2) The sample sizes are large enough because:
n1p1 = 60 > n1(1 – p1) = 40 > 10
n2p2 = 88 > 10 n2(1 – p2) = 16 > 10
3) We assume the two groups are independent

8. Number with wart successfully removed
Duct Tape Continued . . .
H0: p1 – p2 = 0
Ha: p1 – p2 < 0
Treatment n
Number with wart successfully removed
Liquid nitrogen freezing
100 60
Duct tape
104 88
P-value ≈ 0
a = .01
Since the P-value < a, we reject H0. 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.678: #11.37, 39
(extra practice: #48)