91Ó°ÊÓ

Given

$x_1=34$

$n_1=1679$

$x_1=24$

$n_2=1366$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2≠0$

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{34}{1679}≈0.0203$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{24}{1366}≈0.0176$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{34+24}{1679+1366}=\frac{58}{3045}≈0.0190$

Determine the value of the test statistic:

localid="1650450339343" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.0203-0.0176}{\sqrt{0.0190(1-0.0190)}\sqrt{\frac{1}{1679}+\frac{1}{1366}}} \\ ≈0.5 \end{gathered}$$

The $p$-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the $p$-value using table A:

localid="1650450358958" $$\begin{gathered} P=P(Z<-0.54\text{or}Z>0.54) \\ =2×P(Z<-0.54) \\ =2×0.2946 \\ =0.5892 \end{gathered}$$

If the $p$-value is smaller than the significance level, reject the null hypothesis:

$P>0.05⇒\text{Fail to reject}{H}_0$

Given

$x_1=34$

$n_1=1679$

$x_1=24$

$n_2=1366$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2≠0$

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{34}{1679}≈0.0203$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{24}{1366}≈0.0176$

For confidence level $1-\mathrm{Î\pm }=0.95$, determine $z_{\mathrm{Î\pm }/2}=z_{0.025}$using table II (look up $0.025$in the table, the $z$-score is then the found $z$-score with opposite sign):

$z_{\mathrm{Î\pm }/2}=1.96$

The endpoints of the confidence interval for $p_1-p_2$are then:

localid="1650450397298" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\mathrm{Î\pm }/2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.0203-0.0176)-1.96\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366}} \\ ≈-0.0070 \end{gathered}$$

localid="1650450411041" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\mathrm{Î\pm }/2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.0203-0.0176)+1.96\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366}} \\ ≈0.0124 \end{gathered}$$

The confidence interval contains $0$and thus the confidence is consistent with the previous result (of no significant difference).