Question

A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a 10% margin of error at a 95% confidence level, what size of sample is needed

Answer 1

Answer:

The desired sample size is 97.

Step-by-step explanation:

Assume that 50% people in the community that supports the political candidate.

It is provided that the candidate wants a 10% margin of error (MOE) at a 95% confidence level.

The confidence interval for the population proportion is:

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

Then the margin of error is:

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

Compute the critical value of z as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the sample size as follows:

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

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

          $=[\frac{1.96\times \sqrt{0.50(1-0.50)} }{0.10}^{2}\\\\=[9.8]^{2}\\\\=96.04\\\\\approx 97$

Thus, the desired sample size is 97.