Chapter 22, Problem 9E

(a)

To determine

Find: that the fulfilled the assumptions and conditions needed for inference.

Explanation of Solution

There are told that both sample of men and women have been picked at random. Of all-American men and women aged 65 and older, 1012 males and 1062 females reflect less than 10 percent. It is fair to assume that both groups are independent, since the samples were chosen at random. There were 411 successes and 1011-411= 601 failures amongst these men, and there were 535 successes and 1062-535= 527 failures amongst these women. These numbers, for each group, are at least 10.

(b)

To determine

To Construct: a confidence interval of 95 percent for the difference in the proportions of older men and women who have the disease.

Answer to Problem 9E

(0.0550, 0.1404)

Explanation of Solution

Given:

  $n_W=1062$

  $n_M=1012$

Formula used:

  $SE\left({\widehat{p}}_W-{\widehat{p}}_M\right)=\sqrt{\frac{p_W{q}_W}{n_W}+\frac{p_M{q}_M}{n_M}}$

Calculation:

Observed proportions are

  $\begin{array}{l}{\widehat{p}}_W=\frac{535}{1062}=0.5038\\ {\widehat{p}}_M=\frac{411}{1012}=0.4061\end{array}$

Estimating the standard deviation

  $\begin{array}{c}SE\left({\widehat{p}}_W-{\widehat{p}}_M\right)=\sqrt{\frac{p_W{q}_W}{n_W}+\frac{p_M{q}_M}{n_M}}\\ =\sqrt{\frac{0.5038\left(1-0.5038\right)}{1062}+\frac{0.4061\left(1-0.4061\right)}{1012}}\\ =\sqrt{\frac{0.5038\left(0.4962\right)}{1062}+\frac{0.4061\left(0.5939\right)}{1012}}\\ =0.0218\end{array}$

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

  $\begin{array}{c}ME=z^{\ast }\times SE\left({\widehat{p}}_W-{\widehat{p}}_M\right)\\ =1.96\times 0.0218\\ =0.0427\end{array}$

Therefore, the 95% confidence interval is

  $\begin{array}{c}\left({\widehat{p}}_W-{\widehat{p}}_M\right)\pm ME=\left(0.5038-0.4061\right)\pm 0.0427\\ =0.0977\pm 0.0427\\ =\left(0.0550,0.1404\right)\end{array}$

A 95 percent confidence interval is (0.0550, 0.1404) for the difference in the proportions of older women and the mean that they suffered from some form of arthritis.

(c)

To determine

To Explain: the confidence interval.

Explanation of Solution

Depending on these samples, we are 95 percent sure that the real difference between 0.0550 and 0.1404 is the proportion of senior American women and American men who have suffered from any form of arthritis.

(d)

To determine

To Explain: that this confidence interval indicates that arthritis is more likely to affect women than men.

Explanation of Solution

A 95 % confidence interval is (0.0550. 0.1404) for the gap in the proportions of senior women and men who have suffered from certain form of arthritis. Since 0 is not part of the confidence interval, all limits are positive. The real difference in the proportions of senior American women and American men who have had any form of arthritis is not a reasonable value of 0. So, indeed, this confidence interval means that women are more likely to be affected by arthritis than men.