Chapter 8.2, Problem 50E

(a)

To determine

To explain how large a sample must you test to estimate the proportion of PTC tasters within $0.04$ with $90\%$ confidence.

Answer to Problem 50E

  $n=318$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}E=0.04\\ \widehat{p}=0.75\end{array}$

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 $1-\alpha =0.90$ determine $z_{\alpha /2}=z_{0.05}$ using Table A, the z -score is then,

  $z_{\alpha /2}=1.645$

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

  $\begin{array}{l}n=\frac{{[z_{\alpha /2}]}^2\widehat{p}(1-\widehat{p})}{E^2}\\ =\frac{{1.645}^2\times 0.75(1-0.75)}{{0.04}^2}\\ =318\end{array}$

(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 50E

The sample size increased by $105$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}E=0.04\\ \widehat{p}=0.50\end{array}$

From part (a) we have,

  $n=318$

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 $1-\alpha =0.90$ determine $z_{\alpha /2}=z_{0.05}$ using Table A, the z -score is then,

  $z_{\alpha /2}=1.645$

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

  $\begin{array}{l}n=\frac{{[z_{\alpha /2}]}^2\widehat{p}(1-\widehat{p})}{E^2}\\ =\frac{{1.645}^2\times 0.50(1-0.50)}{{0.04}^2}\\ =423\end{array}$

Thus, the difference between the two sizes is as:

  $\begin{array}{l}423-318\\ =105\end{array}$

Thus, we note that the sample size increased by $105$ .