Question

1. 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.

What if the company president demands 99% confidence instead? Determine how this would affect your answer to Question 1.

Short answer

The required sample size is 1180.

From the calculations, we know that the increase in confidence level is leading to an increase in the sample size.

Step-by-step solution

Given Information

Given that

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 $99\%$the confidence level is $2.579$

The sample size is calculated as:

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

$=0.80(1-0.80){\left(\frac{2.576}{0.03}\right)}^2$

=1179.537

$\approx 1180$

Thus, the required sample size is 1180.

From the calculations, we know that the increase in confidence level is leading to an increase in the sample size.