URGENT! Additional instructions in the comments

27. Bob selects independent random samples from two populations and obtains the values p1 = 0.700 and p2 = 0.500. He constructs the 95% confidence interval for p1 p2 and gets: 0.200 1.96(0.048) = 0.200 0.094. Note that 0.048 is called the estimated standard error of p1 p2 (the ESE of the estimate). Tom wants to estimate the mean of the success rates: p1 + p2 2 .

(a) Calculate Tom's point estimate.

(b) Given that the estimated standard error of (p1 + p2)/2 is 0.024, calculate the 95% confidence interval estimate of (p1 + p2)/2. Hint: The answer has our usual form: Pt. est. 1.96 ESE of the estimate.

28. Carl selects one random sample from a population and calculates three confidence intervals for p. His intervals are below. A B C p 0.080 p 0.040 p 0.072 Match each confidence interval to its level, with levels chosen from: 80%, 90%, 95%, 98%, and 99%. Note: Clearly, two of these levels will not be used. You do not need to explain your reasoning.

29. An observational study yields the following "collapsed table." Group S F Total 1 43 57 100 2 39 61 100 Total 82 118 200 Below are two (partial) component tables for these data. Complete these tables so that Simpson's Paradox is occurring (see Course Notes). Note that there is more than one possible correct answer.

Subgp A Subgp B Gp S F Tot Gp S F Tot 1 3 17 20 1 40 40 80 2 40 2 60 Tot 60 Tot 140

30. An observational study yields the following "collapsed table." Group S F Total 1 43 57 100 2 36 64 100 Total 79 121 200 Below are two (partial) component tables for these data. Explain why Simpson's Paradox cannot occur for these data. Subgp A Subgp B Gp S F Tot Gp S F Tot 1 3 17 20 1 40 40 80 2 40 2 60 Tot 60 Tot 140

31. An observational study yields the following "collapsed table." Group S F Total 1 45 55 100 2 38 62 100 Total 83 117 200 Below are two (partial) component tables for these data. Complete these tables so that Simpson's Paradox occurs (see Course Notes). Note that there might be more than one possible correct answer. Subgp A Subgp B Gp S F Tot Gp S F Tot 1 5 15 20 1 40 40 80 2 60 2 40 Tot 80 Tot 120

32. An observational study yields the following "collapsed table."Group S F Total 1 44 56 100 2 43 57 100 Total 87 113 200 Complete these tables so that Simpson's Paradox does not occur (see Course Notes). Note that there might be more than one possible correct answer. Subgp A Subgp B Gp S F Tot Gp S F Tot 1 4 16 20 1 40 40 80 2 60 2 40 Tot 80 Tot 120

33. Below is the table of population counts for a condition and its screening test. (Recall that A means the condition is present and B means the screening test is positive.) On parts (a)-(e) below, show enough work for me to understand how you obtained your answer; e.g. don't just write down "0.5." B Bc Total A 108 12 120 Ac 42 698 740 Total 150 710 860

(a) What proportion of the population is free of the condition?

(b) What proportion of the population has the condition and would test positive?

(c) Of those who have the condition, what proportion would test negative?

(d) What proportion of the population would receive a correct screening test result?

(e) Of those who would receive an incorrect screening test result, what proportion would receive a false negative?