Question

A political candidate has asked you to conduct a poll to determine what percentage of people support him. If the candidate only wants a 5% margin of error at a 97.5% confidence level, what size of sample is needed? When finding the z-value, round it to four decimal places.

Answer 1

Answer:

The sample size required is, n = 502.

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$

The margin of error is:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p\ \cdot (1-\hat p)}{n}}$

Assume that 50% of the people would support this political candidate.

The margin of error is, MOE = 0.05.

The critical value of z for 97.5% confidence level is:

z = 2.24

Compute the sample size as follows:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p\ \cdot (1-\hat p)}{n}}$

       $n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}$

           $=[\frac{2.24\times \sqrt{0.50(1-0.50)}}{0.05}]^{2}\\\\=501.76\\\\\approx 502$

Thus, the sample size required is, n = 502.