18. A survey is taken of 100 undergraduates, 100 graduate students, and 100 continuing education students at a university. Assume a simple random sample within each group. Each student is asked to rate his or her satisfaction (on a 1–10 scale) with his or her experiences. Write the estimate and standard error of the average satisfaction of all the students at the university. Introduce notation as necessary for all the information needed to solve the problem.
Solution to question 17
From yesterday:
17. In a survey of n people, half are asked if they support “the health care law recently passed by Congress” and half are asked if they support “the law known as Obamacare.” The goal is to estimate the effect of the wording on the proportion of Yes responses. How large must n be for the effect to be estimated within a standard error of 5 percentage points?
Solution: se is sqrt(.5*.5/(n/2)+.5*.5/(n/2)) = 1/sqrt(n). Solve 1/sqrt(n) = .05, you get n = (1/.05)^2 = 400.
Hi Andrew –
Can you briefly explain the the se formula here, specifically how do you interpret the variance needed for the se calculation?
p*(1-p)/n is maximized at p=.5.
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