Question

Body Mass Index. Refer to Exercise 11.111 and find and interpret a $90\%$ confidence interval for the difference between the percentages of adults in the Iwo degree categories who have an above healthy weight.

Short answer

There is $90\%$interval for the difference between the proportion is (-0.0064,0.0884)

Step-by-step solution

Given Information

The given values are,

$$\begin{gathered} x_1=386, \\ {n}_1=750, \\ {x}_2=237, \\ {n}_2=500, \\ {\tilde{p}}_1=0.515, \\ {\tilde{p}}_2=0.474 \end{gathered}$$

Explanation

The formula for $z$is given by,

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Calculate the value of $z_{\frac{a}{2}}$with $90\%$confidence level is$1.645$.

The $98\%$confidence interval is.

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

$=(-0.0064,0.0884)$

Therefore, there is $90\%$interval for the difference between the proportion is $(-0.0064,0.0884)$