Chapter 22, Problem 10E

(a)

To determine

To explain: to satisfy the assumptions and conditions necessary for inference.

Explanation of Solution

there are told that male and female samples have also been randomly selected. Of all-American men and women aged 24 years, 12,460 males and 12,678 females reflect less than 10 percent. Since the samples were randomly chosen, it is fair to assume that both groups are independent. There were 84.9 percent achievements among males, which is the number of achievements. There were 88.1 percent successes among women and the number of failures, that is, the number of successes and the number of failures. These numbers, by each group, are at least 10. The assumptions and conditions required for inference are met.

(b)

To determine

To construct: a 95 percent confidence interval for the gap between males and females in graduation rates.

Answer to Problem 10E

(0.0236, 0.0404)

Explanation of Solution

Given:

  $n_F=12,678$

  $n_M=12,460$

  $\begin{array}{l}{\widehat{p}}_F=88.1\%=0.881\\ {\widehat{p}}_M=84.9\%=0.849\end{array}$

Formula used:

  $SE\left({\widehat{p}}_F-{\widehat{p}}_M\right)=\sqrt{\frac{p_F{q}_F}{n_F}+\frac{p_M{q}_M}{n_M}}$

Calculation:

Estimating the standard deviation

  $\begin{array}{c}SE\left({\widehat{p}}_F-{\widehat{p}}_M\right)=\sqrt{\frac{p_F{q}_F}{n_F}+\frac{p_M{q}_M}{n_M}}\\ =\sqrt{\frac{0.88\left(1-0.881\right)}{12,678}+\frac{0.849\left(1-0.849\right)}{12,460}}\\ =\sqrt{\frac{0.88\left(0.119\right)}{12,678}+\frac{0.849\left(0.151\right)}{12,460}}\\ =0.0043\end{array}$

For a 95% confidence level, $z^{\ast }=1.96$ , therefore

  $\begin{array}{c}ME=z^{\ast }\times SE\left({\widehat{p}}_F-{\widehat{p}}_M\right)\\ =1.96\times 0.0043\\ =0.0084\end{array}$

Therefore, the 95% confidence interval is

  $\begin{array}{c}\left({\widehat{p}}_F-{\widehat{p}}_M\right)\pm ME=\left(0.881-0.849\right)\pm 0.0084\\ =0.032\pm 0.0084\\ =\left(0.0236,0.0404\right)\end{array}$

A 95 % confidence interval is (0.0236, 0.0404) for the gap in graduation rates between American females and males aged 24 years.

(c)

To determine

To Explain: the confidence interval.

Explanation of Solution

There are 95 percent sure, based on these samples, that the true difference in graduation rates between American women and men aged 24 years is between 0.0236 and 00404.

(d)

To determine

To Explain: that this provides clear evidence that girls are more probable to graduate high school than boys.

Explanation of Solution

A 95 % confidence interval is (0.0236, 0.0404) for the disparity in graduation rates between American men and women aged 24 years. Although 0 is not in the confidence interval and both limits are positive, the real difference in graduation rates between American men and women aged 24 years is not a reasonable value of 0. So, indeed, this confidence interval suggests that girls are more likely to complete secondary school than boys.