Chapter 8.6, Problem 69E

To determine

Construct the 95% confidence interval for $\left(p_3-p_1\right)-\left(p_4-p_2\right)$.

Answer to Problem 69E

The 95% confidence interval for $\left(p_3-p_1\right)-\left(p_4-p_2\right)$ is $\left(-0.1117,0.5727\right)$.

Explanation of Solution

Here, $n_1=31,n_2=30,n_3=26,\text{and }{n}_4=28$.

The values of $p_1,p_2,p_3,and{p}_4$ are computed as follows:

$\begin{array}{c}{p}_1=\frac{20}{31}\\ =0.6452\\ p_2=\frac{13}{30}\\ =0.4333\\ p_3=\frac{18}{26}\\ =0.6923\\ p_4=\frac{7}{28}\\ =0.2500\end{array}$

The mean and variance of binomial distribution are given as follows:

$\begin{array}{l}E\left(Y_i\right)=n_i{p}_i\\ V\left(Y_i\right)=n{p}_i\left(1-p_i\right)\end{array}$

The unbiasedness is checked as given below:

$\begin{array}{c}E\left(\frac{Y_i}{n_i}\right)=p_i\\ E\left(\left(\frac{Y_3}{n_3}-\frac{Y_1}{n_1}\right)-\left(\frac{Y_4}{n_4}-\frac{Y_2}{n_2}\right)\right)=\left(E\left(\frac{Y_3}{n_3}\right)-E\left(\frac{Y_1}{n_1}\right)\right)-\left(E\left(\frac{Y_4}{n_4}\right)-E\left(\frac{Y_2}{n_2}\right)\right)\\ =\left(p_3-p_1\right)-\left(p_4-p_2\right)\\ V\left(\left(\frac{Y_3}{n_3}-\frac{Y_1}{n_1}\right)-\left(\frac{Y_4}{n_4}-\frac{Y_2}{n_2}\right)\right)=V\left(\frac{Y_3}{n_3}\right)+V\left(\frac{Y_1}{n_1}\right)+V\left(\frac{Y_4}{n_4}\right)+V\left(\frac{Y_2}{n_2}\right)\\ =\frac{p_3\left(1-p_3\right)}{n_3}+\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_4\left(1-p_4\right)}{n_4}+\frac{p_2\left(1-p_2\right)}{n_2}\end{array}$

Therefore, $Y_1,Y_2,Y_3,\text{and }{Y}_4$ are independent binomial distributions with parameters $p_1,p_2,p_3,\text{and }{p}_4$, respectively. Hence, the confidence interval can be obtained for this probability.

From Table 4: Normal Curve Areas, the z value corresponding to 0.025$\left(=\frac{1-0.95}{2}\right)$ is $z_{\frac{\alpha }{2}=0.025}=1.96$.

The formula for 95% confidence interval for $\left(p_3-p_1\right)-\left(p_4-p_2\right)$ is given below:

$\left[\begin{array}{c}\left(\left({\widehat{p}}_3-{\widehat{p}}_1\right)-\left({\widehat{p}}_4-{\widehat{p}}_2\right)\right)-z_{\frac{\alpha }{2}}\sqrt{\frac{p_3\left(1-p_3\right)}{n_3}+\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_4\left(1-p_4\right)}{n_4}+\frac{p_2\left(1-p_2\right)}{n_2}},\\ \left(\left({\widehat{p}}_3-{\widehat{p}}_1\right)-\left({\widehat{p}}_4-{\widehat{p}}_2\right)\right)+z_{\frac{\alpha }{2}}\sqrt{\frac{p_3\left(1-p_3\right)}{n_3}+\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_4\left(1-p_4\right)}{n_4}+\frac{p_2\left(1-p_2\right)}{n_2}}\end{array}\right]$

Some preliminary calculations are obtained as follows:

$\begin{array}{c}\left(\left({\widehat{p}}_3-{\widehat{p}}_1\right)-\left({\widehat{p}}_4-{\widehat{p}}_2\right)\right)=\left(0.6923-0.6452\right)-\left(0.25-0.4333\right)\\ =0.2305\\ \sqrt{\frac{p_3\left(1-p_3\right)}{n_3}+\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_4\left(1-p_4\right)}{n_4}+\frac{p_2\left(1-p_2\right)}{n_2}}=\sqrt{\left\{\begin{array}{l}\frac{0.6923\left(1-0.6923\right)}{26}+\frac{0.6452\left(1-0.6452\right)}{31}+\\ \frac{0.25\left(1-0.25\right)}{28}+\frac{0.4333\left(1-0.4333\right)}{30}\end{array}\right\}}\\ =\sqrt{0.0305}\\ =0.1746\end{array}$

The 95% confidence interval for $\left(p_3-p_1\right)-\left(p_4-p_2\right)$ is computed as follows:

$\begin{array}{c}\left[\begin{array}{c}\left(\left({\widehat{p}}_3-{\widehat{p}}_1\right)-\left({\widehat{p}}_4-{\widehat{p}}_2\right)\right)-z_{\frac{\alpha }{2}}\sqrt{\frac{p_3\left(1-p_3\right)}{n_3}+\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_4\left(1-p_4\right)}{n_4}+\frac{p_2\left(1-p_2\right)}{n_2}},\\ \left(\left({\widehat{p}}_3-{\widehat{p}}_1\right)-\left({\widehat{p}}_4-{\widehat{p}}_2\right)\right)+z_{\frac{\alpha }{2}}\sqrt{\frac{p_3\left(1-p_3\right)}{n_3}+\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_4\left(1-p_4\right)}{n_4}+\frac{p_2\left(1-p_2\right)}{n_2}}\end{array}\right]=\left[\begin{array}{l}0.2305-\left(1.96\right)\left(0.1746\right),\\ 0.2305+\left(1.96\right)\left(0.1746\right)\end{array}\right]\\ =\left(-0.1117,0.5727\right)\end{array}$

Therefore, the 95% confidence interval for $\left(p_3-p_1\right)-\left(p_4-p_2\right)$ is $\left(-0.1117,0.5727\right)$.