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 true proportion p of all 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. $700$

e. $750$

Short answer

The required smallest number of consumers is $600$

The correct option is$\left(b\right)$

Step-by-step solution

Given information

Confidence level $=99\%$

Margin of error$\left(E\right)=0.05$

The objective is to find out the smallest number of consumers to have the margin of error at most 0.05at 99%the confidence level.

We know,

The formula to compute the sample size is:

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

There is no information about $\hat{p}$provided here.

Thus, assume is to be $0.50$

According to the standard norma table, the value of $z_{\frac{\alpha }{2}}$corresponding to$99\%$the confidence level is $2.576$

The sample size is calculated as follows:

$$\begin{gathered} n=\hat{p}(1-\hat{p}){\left(\frac{z_{{\displaystyle \frac{\alpha }{2}}}}{E}\right)}^2 \\ =0.50(1-0.50){\left(\frac{2.576}{0.05}\right)}^2 \\ =664 \end{gathered}$$

The smallest number of consumers is $600$

Therefore, the correct option is$\left(b\right)$