91Ó°ÊÓ

$90\%$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 $1-\mathrm{Î\pm }=0.90$

$z_{\mathrm{Î\pm }/2}=z_{0.05}$

$z_{\mathrm{Î\pm }/2}=1.645$

$\hat{p}$is known,

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

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

$=138$

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

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

$=140$

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:

$\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\mathrm{Î\pm }/2}×\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\mathrm{Î\pm }/2}×\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

For confidence level $1-\mathrm{Î\pm }=0.95$,

$z_{\mathrm{Î\pm }/2}=z_{0.025}$

$1-\mathrm{Î\pm }=0.95$

The confidence interval endpoints for are then

$\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\mathrm{Î\pm }/2}×\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

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

$=0.6612$

$\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\mathrm{Î\pm }/2}×\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

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

$=0.8188$