Statistics Problem Set 6: Confidence Intervals and Randomized Response - Prof. Kristofer J, Assignments of Survey Sampling Techniques

Download Statistics Problem Set 6: Confidence Intervals and Randomized Response - Prof. Kristofer J and more Assignments Survey Sampling Techniques in PDF only on Docsity! Statistics 522: Problem Set No. 6 Due April 10, 2009 1. Lohr Section 6.8 Exercise 21 (page 218) 2. Lohr Section 8.9 Exercise 2 (page 283) 3. Lohr Section 8.9 Exercise 8 (page 284) 4. In a “yes/no” survey with nonresponse (1 = “yes”, 0 = “no”), let N1 denote the number responding to a certain questionnaire. Let W1 = N1N and W2 = N2 N , where N = N1 + N2. That is, W2 denotes the proportion of nonresponse in the population. One method for forming conservative confidence intervals suggests conditioning on certain extreme circumstances. Thus, the upper confidence limit is found by conditioning on the event that all non-respondents would have said “yes”, while the lower confidence limit is found by conditioning on the event that all non-respondents would have said “no”. (a) If p̂1 is the proportion of yeses in the response population and (p̂L,1, p̂U,1) is a confidence interval on p1 (ignoring nonresponse), show that this imputation implies that p̂L = W1p̂L,1 + 0 × W2 p̂U = W1p̂U,1 + 1 × W2. (b) Suppose we are interested in knowing the prevalence of the use of marijuana in a certain town. Let n = 600 be the total number of people contacted and n1 = 500 be the number who responded. If the number of people responding positively is 60, obtain a 95% confidence interval for the true proportion of people who have used in the past or are currently using marijuana. 5. Consider the randomized response scenario from Problem Set 2, Question 6. That is, suppose πA is the true population having characteristic A. A randomized device selects one of two statements, each resulting in a “yes” or “no” response to be presented to the subject and unknown by the reviewer. In this case, let the two statements be “I have the characteristic A.” (presented with known probability p) “I do not have the characteristic A.” (presented with known probability 1 − p). In a random sample of size n, suppose that m are positive responses. Then, φ̂ = m n . (a) Show that the variance of the estimator π̂A = φ̂−(1−p) 2p−1 can be written as Var(π̂A) = πA(1 − πA) n + p(1 − p) n(2p − 1)2 . Hint: assume an infinite population. 1