91Ó°ÊÓ

The given expression is

$x_1=10,n_1=20,x_2=18,n_2=30;$

$\text{Sample proportions}{\hat{p}}_1=\frac{x_1}{n_1}=\frac{10}{20}=0.5$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{18}{30}=0.6$

$\text{Here}{x}_1=10,n_1-x_1=10,x_2=18,n_2-x_2=12$

are all $5$ or greater, so the two-proportion z-test procedure is appropriate.

The test statistic

$z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_P\left(1-{\hat{p}}_P\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\text{Where}{\hat{p}}_1\&{\hat{p}}_2\text{are sample proportions,}{\hat{p}}_p\text{is the pooled sample proportions}$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{10+18}{20+30}=\frac{28}{50}=0.56$

$z=\frac{0.5-0.6}{\sqrt{0.56(1-0.56)}\sqrt{\frac{1}{20}+\frac{1}{30}}}$

$=\frac{-0.1}{0.143}=-0.698$

$\text{Rejection region is}z<-1.282$

The test statistic does not falls in the rejection region. Thus we do not reject our hypothesis $H_0$. The test results are not statistically significant at the $10\%$level.

Interpretation: At $10\%$level of significance, the data does not provide sufficient evidence to reject the null hypothesis.

$\text{For a confidence level of}(1-\mathrm{Î\pm })\text{the confidence interval for}{p}_1-p_2\text{are}$

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

$\text{To find}80\%\text{confidence interval}80\%=100(1-0.2)\%$

$⇒\mathrm{Î\pm }=0.2$

Therefore, Confidence interval

$⇒(0.5-0.6)Â\pm 1.282×\sqrt{\frac{0.5(1-0.5)}{20}+\frac{0.6(1-0.6)}{30}}$

$⇒-0.1Â\pm 0.184$

$⇒-0.284\text{to}0.084$

We can be$80\%$confident that the difference between two proportions is somewhere between $-0.284$and$0.084$