Question

Folic Acid and Birth Defects. Refer to Exercise 11.107 and determine and interpret a $98\%$ confidence interval for the difference between the rates of major birth defects for babies born to women who have taken folic acid and those born to women who have not.

Short answer

There is $98\%$ interval for the difference between the proportions is$(-0.0188,0.008$)

Step-by-step solution

Given Information

The given values are,

$$\begin{gathered} x_1=35, \\ {n}_1=2701, \\ {x}_2=47, \\ {n}_2=2052, \\ \alpha =0.02. \end{gathered}$$

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,

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

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

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1}{n_1} \\ =\frac{35}{2701} \\ =0.013 \end{gathered}$$

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

$$\begin{gathered} {\tilde{p}}_2=\frac{x_2}{n_2} \\ =\frac{47}{2052} \\ =0.022 \end{gathered}$$

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

$z_{\frac{a}{2}}=z_{0.2/2}$

$$\begin{gathered} =z_{0.01} \\ =2.326 \end{gathered}$$

The $98\%$confidence interval is.

$=(0.013-0.022)\pm 2.326\sqrt{\frac{0.013(1-0.013)}{2701}+\frac{0.022(1-0.022)}{2052}}$

$=-0.009\pm 0.0098$

Therefore, there is$98\%$interval for the difference between the proportion is $(-0.0188,0.008)$