Question

A researcher plans to conduct a similar poll on Michigan salaried workers. She is writing up a proposal and would like to include a budget for the sample size needed so she can report a 90% (conservative) confidence interval for the population proportion of all Michigan salaried workers who are "satisfied" with their relations with coworkers with a margin of error of 4%. What is the minimum sample size she could budget for?

Answer 1

Answer:

The minimum sample size the researcher could budget for is 423.

Step-by-step explanation:

Assume that the proportion of all Michigan salaried workers who are "satisfied" with their relations with coworkers is 50%, i.e. p = 0.50.

The margin of error is computed using the formula:

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

The margin of error is, MOE = 0.04.

The confidence level is 90%.

The critical value is:

$z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645$

*Use the z table for critical values.

Compute the sample size as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p}{n}}\\0.04=1.645\times\sqrt{\frac{0.50(1-0.50}{n}}\\0.02432=\frac{0.50}{\sqrt{n}}\\n=[\frac{0.50}{0.02432} ]^{2}\\=422.68\\\approx423$

Thus, the minimum sample size the researcher could budget for is 423.