Question

A pollster wishes to estimate the proportion of United States voters who favor capital punishment. How large a sample is needed in order to be 9898​% confident that the sample proportion will not differ from the true proportion by more than 55​%?

Answer 1

Answer:

Sample size, n = 543

Step-by-step explanation:

We are given the following in the question:

Margin of error = 5%

Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Margin of error =

$z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Since no particular proportion is given, we take

$\hat{p} = 0.5$

$z_{critical}\text{ at}~\alpha_{0.02} =\pm 2.33$

Putting values, we get,

$2.33 \times \sqrt{\dfrac{0.5(1-0.5)}{n}} = 0.05\\\\\sqrt{n} = \dfrac{2.33\times 0.5}{0.05}\\\\\sqrt{n} = 23.3\\\Rightarrow n = 542.89\approx 543$

Thus, the sample size must be 543 so the sample proportion is will not differ by 5%