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

Expert verified

The sample size is 683

Step by step solution

01

Given Information

Population proportion (p^)=80%=0.80

Margin of error(E)=3%=0.03

Confidence level=95%

02

Explanation

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

The sample size is calculated as:

n=(p^)(1-p^)zE2

=0.80(1-0.80)1.960.032

=682.926

683

Therefore the required sample size is 683

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Most popular questions from this chapter

Many television viewers express doubts about the validity of certain commercials. In an attempt to answer their critics, Timex Group USA wishes to estimate the proportion of consumers who believe what is shown in Timex television commercials. Let prepresent the true proportion of consumers who believe what is shown in Timex television commercials. What is the smallest number of consumers that Timex can survey to guarantee a margin of error of 0.05or less at a 99%confidence level?

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