Chapter 8.2, Problem 49E

(a)

To determine

To explain how large a random sample is required to obtain a margin of error of at most $0.03$ with $99\%$ confidence.

Answer to Problem 49E

  $n=1816$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}E=0.03\\ \widehat{p}=44\%=0.44\end{array}$

Now, we know that the formula for sample size is as:

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

For confidence level $1-\alpha =0.99$ determine $z_{\alpha /2}=z_{0.005}$ using Table A, the z -score is then,

  $z_{\alpha /2}=2.575$

Thus, $\widehat{p}$ is known, then the sample size is calculated as:

  $\begin{array}{l}n=\frac{{[z_{\alpha /2}]}^2\widehat{p}(1-\widehat{p})}{E^2}\\ =\frac{{2.575}^2\times 0.44(1-0.44)}{{0.03}^2}\\ =1816\end{array}$

(b)

To determine

To answer the question in part (a) again but this time use the conservative guess $\widehat{p}=0.5$ and find out by how much do the two sample sizes differ.

Answer to Problem 49E

The sample size increased by $26$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}E=0.03\\ \widehat{p}=0.50\end{array}$

From part (a) we have,

  $n=1816$

Now, we know that the formula for sample size is as:

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

For confidence level $1-\alpha =0.99$ determine $z_{\alpha /2}=z_{0.005}$ using Table A, the z -score is then,

  $z_{\alpha /2}=2.575$

Thus, $\widehat{p}$ is known, then the sample size is calculated as:

  $\begin{array}{l}n=\frac{{[z_{\alpha /2}]}^2\widehat{p}(1-\widehat{p})}{E^2}\\ =\frac{{2.575}^2\times 0.50(1-0.50)}{{0.03}^2}\\ =1842\end{array}$

Thus, the difference between the two sizes is as:

  $\begin{array}{l}1842-1816\\ =26\end{array}$

Thus, we note that the sample size increased by $26$ .