Chapter 10, Problem AP3.3CPT

To determine

The smallest number of the athletes that must be interviewed to estimate the true proportion.

  $\begin{array}{l}a.17\\ b.21\\ c.1608\\ d.1702\\ e.2401\end{array}$

Answer to Problem AP3.3CPT

The correct option is $d\mathrm{.1702.}$

Explanation of Solution

Given:

ME = Margin of error: $2\%=0.02$

C=confidence level: $90\%=0.90$

Concept Used:

Formula of sample size:

  $\begin{array}{l}p\text{known: n:}\frac{{\left[z_{\alpha /2}\right]}^{2}pq}{M{E}^2}=\frac{{\left[z_{\alpha /2}\right]}^{2}p\left(1-p\right)}{M{E}^2}\\ p\text{unknown: n:}\frac{{\left[z_{\alpha /2}\right]}^{2}0.25}{M{E}^2}\end{array}$

Calculation:

For confidence level $1-\alpha =0.90,$ then $z_{\alpha /2}=z0.05$ using the normal probability table $z_{\alpha /2}=1.65$

A preliminary estimate $p$ is unknown then the sample size is:

  $\text{n:}\frac{{\left[z_{\alpha /2}\right]}^{2}0.25}{M{E}^2}=\frac{{1.65}^2\times 0.25}{{0.02}^2}\approx 1702$

Conclusion:

Hence, the correct option is $d\mathrm{.1702.}$