Question

Racial Crossover. Refer to Exercise $11.108$ and find and interpret a $95\%$ confidence interval for the difference between the stroke incidences of white and African-American elderly.

Short answer

There is $95\%$the interval for the difference between the proportion is$(-0.029,0.015)$

Step-by-step solution

Given Information

The given values are,

$$\begin{gathered} x_1=529, \\ {n}_1=4989, \\ {x}_2=103, \\ {n}_2=906. \end{gathered}$$

Explanation

The formula for ${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1}{n_1}$

The formula for ${\tilde{p}}_2$is given by,

${\tilde{p}}_2=\frac{x_2}{n_2}$

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 value of ${\tilde{p}}_1$is calculated as,

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1}{n_1} \\ =\frac{529}{4989} \\ =0.106 \end{gathered}$$

The value of ${\tilde{p}}_2$is calculated as,

$$\begin{gathered} {\tilde{p}}_2=\frac{x_2}{n_2} \\ =\frac{103}{906} \\ =0.113 \end{gathered}$$

The critical value of $Z$for two-tail test and $5\%$level of significance is $1.96$.

The $95\%$confidence interval is.

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\frac{a}{2}}\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1}+\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_2}}=(0.106-0.113)$

$\pm 1.96\sqrt{\frac{0.106(1-0.106)}{4989}+\frac{0.113(1-0.113)}{906}}$

$=-0.007\pm 0.022$

Therefore, there is $95\%$interval for the difference between the proportion is $(-0.029,0.015)$