Question

The company’s customer satisfaction survey.

In the company’s prior-year survey, 80% of customers surveyed said they were “satisfied” or “very satisfied.” Using this value as a guess for pˆ, find the sample size needed for a margin of error of 3% at a 95% confidence level.

Short answer

The sample size is 683

Step-by-step solution

Given Information

Population proportion $\left(\hat{p}\right)=80\%=0.80$

Margin of error$\left(E\right)=3\%=0.03$

Confidence level$=95\%$

Explanation

From the standard normal table, the z-score at $95\%$the confidence level is 1.96

The sample size is calculated as:

$n=\left(\hat{p}\right)(1-\hat{p}){\left(\frac{z}{E}\right)}^2$

$=0.80(1-0.80){\left(\frac{1.96}{0.03}\right)}^2$

=682.926

$\approx 683$

Therefore the required sample size is 683