Question

Construct and interpret a $95\%$confidence interval for $p_1-p_2$ in Exercise $24$. Explain what additional information the confidence interval provides.

Short answer

We are $95\%$confident that the proportion difference is between $0.0175$ and $0.3075$.

Step-by-step solution

Given Information

Given

$x_1=34$

$n_1=80$

$x_2=21$

$n_2=80$

Explanation

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

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{34}{80}=0.425$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{21}{80}=0.2625$

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

$z_{\alpha /2}=1.96$

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

localid="1650451236001" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.425-0.2625)-1.96\sqrt{\frac{0.425(1-0.425)}{80}+\frac{0.2625(1-0.2625)}{80}} \\ \approx 0.0175 \end{gathered}$$

localid="1650451247912" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.425-0.2625)+1.96\sqrt{\frac{0.425(1-0.425)}{80}+\frac{0.2625(1-0.2625)}{80}} \\ \approx 0.3075 \end{gathered}$$