Question

Vasectomies and Prostate Cancer. Refer to Exercise 11.106 and determine and interpret a $98\%$ confidence interval for the difference between the prostate cancer rates of men who have had a vasectomy and those who have not.

Short answer

There is $98\%$interval for the difference between the proportion is$(-0.00334,-0.0046)$

Step-by-step solution

Given Information

The given values are,

$$\begin{gathered} x_1=69, \\ {n}_1=21300, \\ {x}_2=113, \\ {n}_2=22000, \\ {\tilde{p}}_1=0.00324 \\ ,{\tilde{p}}_2=0.00514 \end{gathered}$$

Explanation

The formula for $z$is given by,

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{m_1}+\frac{1}{m_2}}}$

Calculate the value of $z_{\frac{a}{2}}$with $98\%$confidence level is $2.33$.

The$98\%$confidence interval is.

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\frac{a}{2}}\sqrt{\frac{\tilde{p}\left(1-{\tilde{p}}_p\right)}{n_1}+\frac{\tilde{p}\left(1-{\tilde{p}}_p\right)}{n_2}}$

$=\left\{\begin{array}{l}(0.00324-0.00514)\pm \\ \left.2.33\sqrt{\frac{0.00324(1-0.00324)}{21300}+\frac{0.00514(1-0.00514)}{22000}}\right\}\end{array}\right\}$

$=-0.0019\pm 0.00144$

Therefore, there is $98\%$interval for the difference between the proportion is$(-0.00334,-0.0046)$