Q.3.2 - The Practice of Statistics for AP

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. 

Step-by-Step Solution

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 1 Given Information

Given that

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

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

Confidence level $=95\%$

Step 2 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.