Question

Buckling Up. Refer to Exercise $11.109$and find and interpret a $99\%$confidence interval for the difference between the proportions of seat-belt users for drivers in the age groups $16-24$ years and 25-69 years.

Short answer

There is $99\%$ interval for the difference between the proportion is (-0.0937,-0.0063)

Step-by-step solution

Given Information

The given values are,

$$\begin{gathered} n_1=1000, \\ {n}_2=1100, \\ {\tilde{p}}_1=0.790, \\ {\tilde{p}}_2=0.840. \end{gathered}$$

Explanation

The formula for $z$, is given by,

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{m_2}}}$

The required value of$z_{\frac{a}{2}}$with $99\%$confidence level is $2.575.$

The $98\%$confidence interval is.

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\frac{a}{2}}\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1}+\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_2}}=(0.790-0.840)$

$\pm 2.575\sqrt{\frac{0.790(1-0.790)}{1000}+\frac{0.840(1-0.840)}{1100}}$

$=-0.05\pm 0.0437$

$=(-0.0937,-0.0063)$

Therefore, there is $99\%$interval for the difference between the proportion is $(-0.0937,-0.0063)$