Two-Sample Proportions Inference

1. Two-Sample Proportions Inference

2. Sampling Distributions for the difference in proportions When tossing pennies, the probability of the coin landing on heads is 0.5. However, when spinning the coin, the probability of the coin landing on heads is 0.4. Let’s investigate. Pairs of students will be given pennies and assigned to either flip or spin the penny

3. Looking at the sampling distribution of the difference in sample proportions: What is the mean of the difference in sample proportions (flip - spin)? What is the standard deviation of the difference in sample proportions (flip - spin)? Can the sampling distribution of difference in sample proportions (flip - spin) be approximated by a normal distribution? What is the probability that the difference in proportions (flipped – spun) is at least .25? Yes, since n1p1=12.5, n1(1-p1)=12.5, n2p2=10, n2(1-p2)=15 – so all are at least 5)

4. Assumptions: • Two, independent SRS’s from populations ( or randomly assigned treatments) • Populations at least 10n • Normal approximation for both

5. Formula for confidence interval: Margin of error! Standard error! Note: use p-hat when p is not known

6. Example 1: At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of 316 patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of 419 patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. What is the shape & standard error of the sampling distribution of the difference in the proportions of people with no visible scars between the two groups? Since n1p1=259, n1(1-p1)=57, n2p2=94, n2(1-p2)=325 and all > 5, then the distribution of difference in proportions is approximately normal.

7. Example 1: At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of 316 patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of 419 patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. What is a 95% confidence interval of the difference in proportion of people who had no visible scars between the plasma compress treatment & control group?

8. Assumptions: • Have 2 independent randomly assigned treatment groups • Both distributions are approximately normal since n1p1=259, n1(1-p1)=57, n2p2=94, n2(1-p2)=325 and all > 5 • Population of burn patients is at least 7350. Since these are all burn patients, we can add 316 + 419 = 735. If not the same – you MUST list separately. We are 95% confident that the true the difference in proportion of people who had no visible scars between the plasma compress treatment & control group is between 53.7% and 65.4%

9. Example 2: Suppose that researchers want to estimate the difference in proportions of people who are against the death penalty in Texas & in California. If the two sample sizes are the same, what size sample is needed to be within 2% of the true difference at 90% confidence? Since both n’s are the same size, you have common denominators – so add! n = 3383

10. Example 3: Researchers comparing the effectiveness of two pain medications randomly selected a group of patients who had been complaining of a certain kind of joint pain. They randomly divided these people into two groups, and then administered the painkillers. Of the 112 people in the group who received medication A, 84 said this pain reliever was effective. Of the 108 people in the other group, 66 reported that pain reliever B was effective. (BVD, p. 435) a) Construct separate 95% confidence intervals for the proportion of people who reported that the pain reliever was effective. Based on these intervals how do the proportions of people who reported pain relieve with medication A or medication B compare? b) Construct a 95% confidence interval for the difference in the proportions of people who may find these medications effective. SO – which is correct? CIA = (.67, .83) CIB =(.52, .70) Since the intervals overlap, it appears that there is no difference in the proportion of people who reported pain relieve between the two medicines. CI = (0.017, 0.261) Since zero is not in the interval, there is a difference in the proportion of people who reported pain relieve between the two medicines.

11. Hypothesis statements: H0: p1 - p2 = 0 Ha: p1 - p2 > 0 Ha: p1 - p2 < 0 Ha: p1 - p2 ≠ 0 H0: p1 = p2 Be sure to define both p1 & p2! Ha: p1 > p2 Ha: p1 < p2 Ha: p1 ≠ p2

12. Since we assume that the population proportions are equal in the null hypothesis, the variances are equal. Therefore, we poolthe variances!

13. Formula for Hypothesis test: Usually p1 – p2 =0

14. Example 4: A forest in Oregon has an infestation of spruce moths. In an effort to control the moth, one area has been regularly sprayed from airplanes. In this area, a random sample of 495 spruce trees showed that 81 had been killed by moths. A second nearby area receives no treatment. In this area, a random sample of 518 spruce trees showed that 92 had been killed by the moth. Do these data indicate that the proportion of spruce trees killed by the moth is different for these areas?

15. Assumptions: • Have 2 independent SRS of spruce trees • Both distributions are approximately normal since n1p1=81, n1(1-p1)=414, n2p2=92, n2(1-p2)=426 and all > 5 • Population of spruce trees is at least 10,130. H0: p1=p2 where p1 is the true proportion of trees killed by moths Ha: p1≠p2 in the treated area p2 is the true proportion of trees killed by moths in the untreated area P-value = 0.5547 a = 0.05 Since p-value > a, I fail to reject H0. There is not sufficient evidence to suggest that the proportion of spruce trees killed by the moth is different for these areas