Chapter 8.2, Problem 43E

(a)

To determine

To calculate: The total size of the sample for PTC test when the sample proportion $\stackrel{⌢}{p}$ is 75%.

Answer to Problem 43E

The sample size of 318.has tested PTC when the sample proportion $\stackrel{⌢}{p}$ is 75%.

Explanation of Solution

Given information:

Margin of error $E=0.04$

Guessed value $\stackrel{⌢}{p}=75\%$

Confidence interval = 90%

Formula used:

Sample size $n=\frac{{\left[z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}^2\stackrel{⌢}{p}\left(1-\stackrel{⌢}{p}\right)}{E^2}$

Calculation:

Confidence interval is 90%.

Convert 90% into decimal.

  $\frac{90}{100}=0.90$

For confidence interval 0.90, use table A.

  $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1.645$

Guessed value $\stackrel{⌢}{p}=75\%$

Convert 75% into decimal.

  $\frac{75}{100}=0.75$

Now, find the sample size. Use the formula $n=\frac{{\left[z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}^2\stackrel{⌢}{p}\left(1-\stackrel{⌢}{p}\right)}{E^2}$ .

  $\begin{array}{c}n=\frac{{\left[z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}^2\stackrel{⌢}{p}\left(1-\stackrel{⌢}{p}\right)}{E^2}\\ n=\frac{{1.645}^2\times 0.75\times \left(1-0.75\right)}{{0.04}^2}\\ =318\end{array}$

Hence, the sample size is 318.

(b)

To determine

To calculate: The difference of the sample size when $\stackrel{⌢}{p}$ is changes to 0.5.

Answer to Problem 43E

The sample size is increased by 105 when the guessed value $\stackrel{⌢}{p}$ is 0.5.

Explanation of Solution

Given information:

Margin of error $E=0.04$

Sample proportion $\stackrel{⌢}{p}=0.5$

Confidence interval = 90%

Formula used:

Sample size $n=\frac{{\left[z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}^2\stackrel{⌢}{p}\left(1-\stackrel{⌢}{p}\right)}{E^2}$

Calculation:

Confidence interval is 90%.

Convert 90% into decimal.

  $\frac{90}{100}=0.90$

For confidence interval 0.90, use table A.

  $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1.645$

From part(a) sample size n=318

Find the new sample size. Use the formula $n=\frac{{\left[z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}^2\stackrel{⌢}{p}\left(1-\stackrel{⌢}{p}\right)}{E^2}$ .

  $\begin{array}{c}{n}^{\prime }=\frac{{\left[z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}^2\stackrel{⌢}{p}\left(1-\stackrel{⌢}{p}\right)}{E^2}\\ n^{\prime }=\frac{{1.645}^2\times 0.5\times \left(1-0.5\right)}{{0.04}^2}\\ =423\end{array}$

Thus, the new sample size is 423.

For difference between the sample sizes,subtract $n$ from $n^{\prime }$

  $\begin{array}{c}{n}^{\prime }-n=423-318\\ =105\end{array}$

Hence, sample size is increased by 105.