Question

Construct and interpret a $95\%$ confidence interval for $p_1-p_2$ in Exercise $23$. Explain what additional information the confidence interval provides.

Short answer

We are $95\%$ confident that the proportion difference is between $0.100$ and$0.382$.

Step-by-step solution

Given Information

Given

$x_1=44$

$n_1=88$

$x_2=21$

$n_2=81$

Explanation

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

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{44}{88}=0.5$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{21}{81}\approx 0.259$

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="1650451164119" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.5-0.259)-1.96\sqrt{\frac{0.5(1-0.5)}{88}+\frac{0.259(1-0.259)}{81}} \\ \approx 0.100 \end{gathered}$$

localid="1650451174743" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.5-0.259)+1.96\sqrt{\frac{0.5(1-0.5)}{88}+\frac{0.259(1-0.259)}{81}} \\ \approx 0.382 \end{gathered}$$