Golf club repairs The Ping Company makes custom-built golf clubs and competes in the $\$4$billion golf equipment industry. To improve its business process, Ping decided to study the time it took to repair golf clubs sent to the company by mail. The company determined that $16\%$of a random sample of orders were sent back to the customers in $5$days or less. Ping examined the processing of repair orders and made changes. Following the changes, $90\%$of a random sample of orders were completed within $5$days. Assume that each of the estimated percent is based on a random sample of n orders.

(a) We used the sample data to construct two $90\%$ confidence intervals for the proportion of orders completed within 5 days-one before and one after the changes at Ping. The two intervals are $(0.858,0.942)$ and $(0.109,0.211)$. Find the value of $n$.

(b) Explain why the interval $(0.858-0.211,0.942-0.109)=(0.647,0.833)$ is not a $95\%$ confidence interval for the improvement in the proportion of orders sent back to customers within $5$ days following the change.

(c) Construct and interpret a correct $95\%$ confidence interval to replace the one in part (b).

$90\%$confidence interval: $(0.858,0.942)$

$90\%$confidence interval: $(0.109,0.211)$

localid="1650368097744" ${\hat{p}}_1=90\%$

$=0.90$

localid="1650368115748" ${\hat{p}}_2=16\%$

$=0.16$

The margin of error is

$E_1=0.042$

$E_2=0.051$

For confidence level $1-\alpha =0.90$

$z_{\alpha /2}=z_{0.05}$

$z_{\alpha /2}=1.645$

$\hat{p}$is known,

$n_1=\frac{{\left[z_{{\omega }_{2}2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$=\frac{1.{645}^2\times 0.90(1-0.90)}{0.{042}^2}$

$=138$

$n_2=\frac{{\left[z_{\alpha 2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$=\frac{1.{645}^2\times 0.16(1-0.16)}{0.{051}^2}$

$=140$

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

The Ping Company makes custom-built golf clubs and competes in the $\$4$ billion golf equipment industry.

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

${\hat{p}}_1=90\%=0.90$

$n=139$

The confidence interval endpoints for are then:

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

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

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

$z_{\alpha /2}=z_{0.025}$

$1-\alpha =0.95$

The confidence interval endpoints for are then

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

$=(0.90-0.16)-1.96\sqrt{\frac{0.90(1-0.90)}{139}+\frac{0.16(1-0.16)}{139}}$

$=0.6612$

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

$=(0.90-0.16)+1.96\sqrt{\frac{0.90(1-0.90)}{139}+\frac{0.16(1-0.16)}{139}}$

$=0.8188$