Question

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 $p$represent 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 $0.05$or less at a $99\%$confidence level?

(a) $550$

(b) $600$

(c) $650$

(d) $700$

(c)$750$

Short answer

(c)$650$

Step-by-step solution

Step-1 Given Information

Given in the question that,

Mirror of error$\left(E\right)=5\%=0.05$

confidence level $=99\%$we have to find the correct option.

Step-2 Explanation

The formula to compute the sample size is:

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

From the standard normal table, the z-score at $99\%$confidence level is $2.576$

Consider, sample proportion to be$0.05$.

The sample size is calculated as:

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

role="math" localid="1649224673729" $=0.50\left(1-0.50\right){\left(\frac{2.576}{0.05}\right)}^2$

$=664$

Here, the smallest sample size is required. Thus, the minimum sample size is $650$. Hence, the correct option is (c).