1. Two-Sample Proportions Inference

2. Conditions TwoindependentTwo independent SRS’s (or randomly assigned treatments) Populations > 10n Both sampling dist.’s approx. normal:

3. Confidence Interval for a Difference in Proportions (Remember: Use p-hat when p is not known) Standard Error Margin of Error

4. Sampling Distributions for a Difference in Proportions Flipping and spinning pennies – is there a difference?

5. At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of 316 patients with minor burns received this treatment. Of these patients, 259 had no visible scars after treatment. Another random sample of 419 patients with minor burns received no plasma compress treatment. In this group, 94 had no visible scars after treatment. a) Determine the shape and standard error of the sampling distribution of the difference in the proportions of people with visible scars between the two treatments. n 1 p 1 = 259 > 5, n 1 (1 – p 1 ) = 57 > 5 n 2 p 2 = 94 > 5, n 2 (1 – p 2 ) = 325 > 5  Sampling dist. is approx. normal

6. b) Construct a 95% confidence interval for the true difference in proportion of people who had no visible scars (plasma – control group). Conditions: 2 independent randomly assigned treatments n 1 p 1 = 259 > 5, n 1 (1 – p 1 ) = 57 > 5 n 2 p 2 = 94 > 5, n 2 (1 – p 2 ) = 325 > 5  Sampling dist.’s are approx. normal Pop. of burn patients is at least 7350 Since these are all burn patients, we can add: 316 + 419 = 735. If they’re different populations, we have to list them separately.

7. We are 95% confident that the true difference in proportions of people with no visible scars (plasma – control) is between 53.7% and 65.4%. If we made lots of intervals this way, 95% of them would contain the true difference in proportions.

8. Researchers want to estimate the difference in proportions of people who are against the death penalty in Texas & 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?  n = 3383

9. Researchers comparing the effectiveness of two pain medications randomly selected a group of patients who had been complaining of joint pain. They randomly divided these people into two groups, and then administered the painkillers. Of the 112 people 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. a) Construct two separate 95% confidence intervals for the proportion of people who reported that each pain reliever was effective. Is there a significant difference between these two proportions? b) Construct a 95% confidence interval for the difference in the proportions of people who find these medications effective. Is there a significant difference between these two proportions? CI A = (.67,.83) CI B = (.52,.70) Since the intervals overlap, we do not have evidence of a significant difference in the proportion of people who report each pain reliever is effective. CI = (0.017, 0.261) Since zero is not in the interval, there is a significant difference in these proportions. So which decision is correct?

10. Hypothesis Statements H 0 : p 1 – p 2 = 0 H a : p 1 – p 2 > 0 H a : p 1 – p 2 < 0 H a : p 1 – p 2 ≠ 0 Be sure to define both p 1 & p 2 ! H 0 : p 1 = p 2 H a : p 1 > p 2 H a : p 1 < p 2 H a : p 1 ≠ p 2

11. We assume the population proportions are equal in the H0. Thus, the standard deviation ( ) is the same for both populations. Therefore, we pool the variances!

12. Test Statistic Usually p 1 – p 2 =0

13. A forest in Oregon has an infestation of spruce moths. In an effort to control the moths, one area has been regularly sprayed by airplanes. In this area, a random sample of 495 spruce trees showed that 81 had been killed by moths. A nearby area received no spray. In this area, a random sample of 518 spruce trees showed that 92 had been killed by the moths. Do these data indicate that the proportion of spruce trees killed by the moths is different in the two areas?

14. Conditions: 2 independent SRS’s n 1 p 1 = 81 > 5, n 1 (1 – p 1 ) = 414 > 5, n 2 p 2 = 92 > 5, n 2 (1 – p 2 ) = 426 > 5  approx. normal Pop. of spruce trees is at least 10,130.= H 0 : p 1 = p 2 where p 1 and p 2 are the true proportions H a : p 1 ≠ p 2 of trees killed by moths in the treated and untreated areas

15. p-value = 0.5547α = 0.05 Since p-value > α, we fail to reject H 0. There is not sufficient evidence to suggest that the proportion of spruce trees killed by the moth is different in the two areas.