1. Pooled Proportions
Chapter 22 Part 2

2. 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 = #𝑆𝑢𝑐𝑐𝑒𝑠𝑠 1 + #𝑆𝑢𝑐𝑐𝑒𝑠𝑠 2 𝑛 1 + 𝑛 2
Pooled Proportions
𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 = #𝑆𝑢𝑐𝑐𝑒𝑠𝑠 1 + #𝑆𝑢𝑐𝑐𝑒𝑠𝑠 2 𝑛 1 + 𝑛 2
We can combine the counts from each sample to get an overall proportion. To find out why, read pages 511 & 512.
Everyone into the pool!

3. Important to Remember:
We always use a pooled estimate of the standard deviation when carrying out a hypothesis test whose null hypothesis is p1 = p2 but not when constructing a confidence interval for the difference in proportions.

4. Loophole for Success/Failure Condition
If np<10 for one of the samples, there is another way we can try to make it fit the conditions.
Use the pooled proportion to check that:
𝑛 1 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 >10
𝑛 2 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 >10

5. 𝑛 1 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 =7897 .0058 ≈46>10 𝑛 2 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 =7899 .0058 ≈46>10
Example:
In a clinical trial for Gardasil, a vaccine to prevent strains of HPV that are responsible for most cases of cervical cancer, 1 out of 7897 women who received the vaccine was diagnosed with HPV while 91 out of 7899 women who received a placebo were diagnosed with HPV. We want to test the effectiveness of the vaccine.
𝑛 1 𝑝 1 =1 𝑎𝑛𝑑 𝑛 2 𝑝 2 =91
We don’t have at least 10 successes for the first sample, but we still want to test the effectiveness of the vaccine. Use the loophole.
𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 = =0.0058
𝑛 1 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 = ≈46>10
𝑛 2 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 = ≈46>10

6. Two-Proportion z-Test
𝑆𝐸 𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 𝑞 𝑝𝑜𝑜𝑙𝑒𝑑 𝑛 𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 𝑞 𝑝𝑜𝑜𝑙𝑒𝑑 𝑛 2
z= 𝑝 1 − 𝑝 2 𝑆𝐸 𝑝𝑜𝑜𝑙𝑒𝑑

7. 𝐻 0 : 𝑝 30+ − 𝑝 30− =0 𝐻 𝐴 : 𝑝 30+ − 𝑝 30− ≠0 Example:
I want to know whether snoring rates differ for those under and over 30 years old. The data are from a random sample of 1010 U.S. adults surveyed in the 2002 Sleep America Poll. Of these, 995 responded to the question about snoring, indicating whether or not they snored at least a few nights a week in the past year. Among the under 30 category, 26.1% of the 184 reported snoring. Among the over 30 category, 39.2% of the 811 reported snoring.
Step 1: State the hypotheses.
𝐻 0 : 𝑝 30+ − 𝑝 30− =0
𝐻 𝐴 : 𝑝 30+ − 𝑝 30− ≠0

8. Example:
I want to know whether snoring rates differ for those under and over 30 years old. The data are from a random sample of 1010 U.S. adults surveyed in the 2002 Sleep America Poll. Of these, 995 responded to the question about snoring, indicating whether or not they snored at least a few nights a week in the past year. Among the under 30 category, 26.1% of the 184 reported snoring. Among the over 30 category, 39.2% of the 811 reported snoring.
Step 2: Check conditions and model.
Independent & random
1010 is less than 10% of all US adults
𝑛 𝑝 30+ = =318; 𝑛 30− 𝑝 30− = =48 𝑛 𝑞 30+ = =493; 𝑛 30− 𝑞 30− = =136 𝐴𝑙𝑙 𝑜𝑓 𝑡ℎ𝑒𝑠𝑒 𝑎𝑟𝑒 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 10
I will use a Normal model for the sampling distribution and perform a two-proportion z-test with: 𝜇=0 and I will used the pooled standard error (next slide)

9. Two-tailed test: P-value =2 0.0004 =0.0008 Example:
I want to know whether snoring rates differ for those under and over 30 years old.
Step 3: Mechanics
𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 = =0.3678
𝑆𝐸 𝑝𝑜𝑜𝑙𝑒𝑑 = (.6322) (.6322) 184 =
Two-tailed test:
P-value = =0.0008
z= ( 𝑝 1 − 𝑝 2 )−0 𝑆𝐸 𝑝𝑜𝑜𝑙𝑒𝑑 = 0.392− =3.33
Reject the null hypothesis that the proportions are equal.

10. Example:
I want to know whether snoring rates differ for those under and over 30 years old. The data are from a random sample of 1010 U.S. adults surveyed in the 2002 Sleep America Poll. Of these, 995 responded to the question about snoring, indicating whether or not they snored at least a few nights a week in the past year. Among the under 30 category, 26.1% of the 184 reported snoring. Among the over 30 category, 39.2% of the 811 reported snoring.
Step 4: State your conclusion.
The P-value of indicates that the likelihood of a difference observed between the two samples being due to chance is very small. Therefore, I reject the null hypothesis of no difference and conclude that there is a difference in the rate of snoring between adults over 30 and adults under 30. It appears that the adults over 30 are more likely to snore.

11. Step 1: State the hypotheses
One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him. Do these results indicate a decrease in voter support for his candidacy?
Step 1: State the hypotheses
𝐻 0 : 𝑝 𝐵 − 𝑝 𝐴 ≤0
𝐻 𝐴 : 𝑝 𝐵 − 𝑝 𝐴 >0
Let B = before news of affair, A = after news of affair

12. One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him. Do these results indicate a decrease in voter support for his candidacy?
Step 2: Check the conditions/model
Voters were randomly selected so they are independent.
630 and 1010 are both less than 10% of all voters.
𝑛 𝑝 𝐵 = =340, 𝑛 𝑞 𝐵 = =290, 𝑛 𝑝 𝐴 = =515, 𝑛 𝑞 𝐴 = =505 all of these are greater than 10

13. Use N(0, 0.02536) Step 2: Conditions/Model continued
To model the sampling distribution, use the Normal model with mean 0 and standard deviation estimated by the standard error:
𝑝 𝑝𝑜𝑜𝑙𝑒𝑑 = =0.5213
𝑆𝐸 𝑝𝑜𝑜𝑙𝑒𝑑 𝑝 𝐵 − 𝑝 𝐴 = (0.5213)(0.4788) (0.5213)(0.4788) =
Use N(0, )

14. Step 3: Mechanics
Since we are not instructed to find a confidence interval, we will find a P-value for making our decision.
𝑧= 0.54−0.51 − =1.18
𝑃 𝑧>1.18 =1−𝑃 𝑧<1.18 =
1−.881=0.119

15. Step 4: Conclusion
Since the P-value of is high, we fail to reject the null hypothesis. There is little evidence to conclude there is a decrease in the proportion of voters in support of the candidate after the news of his extramarital affair got out.

16. Today’s Assignment:
Add to HW: page 519 #4-7, 14-18
Test next Thursday, Feb. 18 over Chapters