Chapter 10, Problem 8CRE

(a)

To determine

To find: the value of n.

Answer to Problem 8CRE

  $n=139$

Explanation of Solution

Given:

90% confidence interval: (0.858, 0.942)

90% confidence interval: (0.109, 0.211)

  $\begin{array}{l}{\widehat{p}}_1=90\%=0.90\\ {\widehat{p}}_2=16\%=0.16\end{array}$

Formula used:

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

Calculation:

The margin of error is

  $\begin{array}{l}{E}_1=\frac{0.942-0.858}{2}=0.042\\ E_2=\frac{0.211-0.109}{2}=0.051\end{array}$

For confidence level $1-\alpha =0.90$

  $z_{\alpha /2}=z_{0.05}$

  $z_{\alpha /2}=1.645$

  $\widehat{p}$ is known,

  $\begin{array}{l}{n}_1=\frac{{\left[z_{\alpha /2}\right]}^2\widehat{p}\left(1-\widehat{p}\right)}{E^2}\\ =\frac{{1.645}^2\times 0.90\left(1-0.90\right)}{{0.042}^2}\\ =138\end{array}$

  $\begin{array}{l}{n}_2=\frac{{\left[z_{\alpha /2}\right]}^2\widehat{p}\left(1-\widehat{p}\right)}{E^2}\\ =\frac{{1.645}^2\times 0.16\left(1-0.16\right)}{{0.051}^2}\\ =140\end{array}$

The sample size must be equal (due to rounding errors they are not equal) and therefore the sample size is most probable $n=139$ .

(b)

To determine

To Explain: the reason for the interval is not a 95 percent confidence interval for the ratio of orders returned to consumers within 5 days of the shift.

Explanation of Solution

The interval was built in order to obtain an approximation of the population proportions difference. However, the two-sample z-test used the confidence interval of the population difference, and this is not the variation of the confidence intervals derived from the one-sample z-test

(c)

To determine

To construct: and explain a valid 95 % confidence interval to replace that set out in Part (b)

Answer to Problem 8CRE

(0.6612, 0.8188)

Explanation of Solution

Given:

  $\begin{array}{l}{\widehat{p}}_1=90\%=0.90\\ {\widehat{p}}_2=16\%=0.16\\ n=139\end{array}$

Formula used:

The confidence interval endpoints for are then:

  $\left({\widehat{p}}_1-{\widehat{p}}_2\right)-z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

  $\left({\widehat{p}}_1-{\widehat{p}}_2\right)+z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Calculation:

For confidence level $1-\alpha =0.95$ ,

  $z_{\alpha /2}=z_{0.025}$

  $1-\alpha =0.95$

The confidence interval endpoints for are then

  $\begin{array}{l}\left({\widehat{p}}_1-{\widehat{p}}_2\right)-z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}\\ =\left(0.90-0.16\right)-1.96\sqrt{\frac{0.90\left(1-0.90\right)}{139}+\frac{0.16\left(1-0.16\right)}{139}}\\ =0.6612\end{array}$

  $\begin{array}{l}\left({\widehat{p}}_1-{\widehat{p}}_2\right)+z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}\\ =\left(0.90-0.16\right)+1.96\sqrt{\frac{0.90\left(1-0.90\right)}{139}+\frac{0.16\left(1-0.16\right)}{139}}\\ =0.8188\end{array}$

There is confidence in 95 percent that the real gap in population proportion is between 0.6612 and 0.8188.