Question

In a survey of 500 likely voters, 271 responded that they would vote for the incumbent and 229 responded that they would vote for the challenger. Let p denote the fraction of all likely voters who preferred the incumbent at the time of the survey, and let p̂ p^ be the fraction of survey respondents who preferred the incumbent.

a. Construct a 95% confidence interval for pp. Keep in mind that you should calculate the standard error making no assumptions of truth, but using only the data. This means that you should use p̂ p^ in your standard error calculations.

b. Construct a 99% confidence interval for pp.

c. How and why are the intervals different in (a) and (b)?

d. Suppose that the survey is carried out 20 times, using independently selected likely voters in each survey. For each of these 20 surveys, a 95% confidence interval for pp is constructed. What is the probability that the true value of pp is contained in all 20 of these confidence intervals?

e. How many of these confidence intervals do you expect to contain the true value of pp?

f. In survey jargon, the "margin of error" is 1.96×SE(p̂ )1.96×SE(p^); that is, it is half the length of 95% confidence interval. Suppose you wanted to design a survey that had a margin of error of at most 2 percentage points. That is, you wanted Pr(∣∣p̂ −p∣∣>0.02)≤0.05Pr(|p^−p|>0.02)≤0.05. How large should nn be if the survey uses simple random sampling?

Answer 1

SOlutionA:

p^=proportion of voters for incumbent=x/n=271/500=0.542

Z alpha/2 for 95%=1.96

95% confidence interval for the fraction of survey respondents who preferred the incumbent.

=p^-Zcrit*sqrt(p^(1-p^)/n,p^+Zcrit*sqrt(p^(1-p^)/n

=0.542-1.96*sqrt(0.542*(1-0.542)/500,0.542+1.96*sqrt(0.542*(1-0.542)/500

=0.498,0.586

0.498<p<0.586

lower limit=0.498

upper limit=0.586

we are 95% confident that the true the proportion of survey respondents who preferred the incumbent lies in beween 0.498 and 0.586

SOlutionb:

Z alpha/2 for 99%=2.576

99% confidence interval for the fraction of survey respondents who preferred the incumbent.

=p^-Zcrit*sqrt(p^(1-p^)/n,p^+Zcrit*sqrt(p^(1-p^)/n

=0.542-2.576*sqrt(0.542*(1-0.542)/500),0.542+2.576*sqrt(0.542*(1-0.542)/500)

=0.485,0.599

0.485<p<0.599

we are 99% confident that the true the proportion of survey respondents who preferred the incumbent lies in beween 0.485 and 0.599

Solutionc:

99% confidence interval is wider than 95% as Z crit is more for 99% than 95%