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 pp denote the fraction of all likely voters who preferred the incumbent at the time of the survey, and let p^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 p^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(p^)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. How large should nn be if the survey uses simple random sampling? [Mathematically, this would be calculating Pr(|p^−p|≤0.02)≥0.95Pr(|p^−p|≤0.02)≥0.95. But do not get caught up in that math. Just note that you want 1.96×SE(p^)=0.021.96×SE(p^)=0.02, and solve for nn. Since we do not know p^p^ before running the survey, you need to plan for the largest possible standard error.]
• 2.2 The member of Congress who used to represent the BYU campus (Utah's 3rd Congressional District), Jason Chaffetz, introduced a bill to abolish the Federal Department of Education. Suppose that you are going to conduct a survey of voters in the 3rd district to find their opinions on whether or not they support abolishing the Department of Education. You suspect that more than a majority of people in the district support the idea. Your best guess before collecting any data is that 55% of the district support the policy. You need to conduct the survey using as little money as possible to validate this conjecture.
  
  a. Given your suspicion of 55% support in the population (i.e. p=0.55p=0.55), what size sample should you collect in your survey so that you can be confident (at a 95% level) that there is more than 50% support for the proposal?
  
  b. What is the margin of error for this survey if the sample size is 150 respondents? 300 respondents? 1000 respondents? 10,000 respondents? 100,000 respondents? 1,000,000 respondents?

Answer 1

a) The sample proportion is:

$\hat{p}=\frac{x}{n}=\frac{271}{500}=0.542$

For 95% confidence level, critical value of z is $z^*=1.96$.

The 95% confidence interval for p is:

$\\\hat{p}\pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}= 0.542\pm 1.96\sqrt{\frac{0.542(1-0.542)}{500}}\\\\=0.542\pm 0.044 \ or \ {\color{Red} (0.498,0.586)}$

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b) For 99% confidence level, critical value of z is $z^*=2.575$.

The 99% confidence interval for p is:

$\\\hat{p}\pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}= 0.542\pm 2.575\sqrt{\frac{0.542(1-0.542)}{500}}\\\\=0.542\pm 0.057 \ or \ {\color{Red} (0.485,0.599)}$

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c) 99% confidence interval is wider than 95% confidence interval.

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d) There is 95% probability that p is contained in any interval, and since each survey is independent then the probability that the true value of p is contained in all 20 intervals is:

$0.95^{20}={\color{Red} 0.36}$

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e) We expect 0.95 of 20 samples containing p value, that is:

$0.95(20)={\color{Red} 19}$

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f) For 95% confidence level, critical value of z is $z^*=1.96$.

The given margin of error is $E=0.02$. The required sample size is:

$n=p(1-p)\left ( \frac{z^*}{E} \right )^2 =0.5(1-0.5)\left ( \frac{1.96}{0.02} \right )^2=2400.911763 \approx {\color{Red} 2401}$