Chapter 22, Problem 28E

(a)

To determine

To explain: the 95 percent confidence interval for the percentage of male voters who can vote and describe the interval for that candidate.

Answer to Problem 28E

(0.475, 0.565)

Explanation of Solution

Given:

52 percent of the 473 men polled said they would vote for this candidate.

  $\begin{array}{l}n=473\\ \widehat{p}=0.52\end{array}$

Formula used:

  $\begin{array}{c}SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}\\ ME=z^{\ast }\times SE\left(\widehat{p}\right)\end{array}$

For the confidence interval

  $\left(\widehat{p}\right)\pm ME$

Calculation:

the standard error is

  $\begin{array}{c}SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}\\ =\sqrt{\frac{0.52\left(1-0.52\right)}{473}}\\ =0.023\end{array}$

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

  $\begin{array}{c}ME=z^{\ast }\times SE\left(\widehat{p}\right)\\ =1.96\times 0.023\\ =0.045\end{array}$

Therefore, the 95 percent confidence interval is

  $\begin{array}{c}\left(\widehat{p}\right)\pm ME=0.52\pm 0.045\\ =\left(0.475,0.565\right)\end{array}$

If the observed percentage is 52 percent, then there are 95 percent sure that this candidate would vote for between 47.5 percent and 56.5 percent of the population percentage of men.

(b)

To determine

To explain: and to interpret a confidence interval of 95 percent for the percent of women who will vote for him.

Answer to Problem 28E

(0.407, 0.493)

Explanation of Solution

Given:

45 percent of the 522 women polled said they were going to vote for this candidate.

  $\begin{array}{l}n=522\\ \widehat{p}=0.45\end{array}$

Formula used:

  $\begin{array}{c}SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}\\ ME=z^{\ast }\times SE\left(\widehat{p}\right)\end{array}$

For the confidence interval

  $\left(\widehat{p}\right)\pm ME$

Calculation:

standard error is

  $\begin{array}{c}SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}\\ =\sqrt{\frac{0.45\left(1-0.45\right)}{522}}\\ =0.022\end{array}$

For a 95 percent confidence level, $Z^{\ast }=1.96$ , therefore

  $\begin{array}{c}ME=z^{\ast }\times SE\left(\widehat{p}\right)\\ =1.96\times 0.022\\ =0.043\end{array}$

So, the 95% confidence interval is

  $\begin{array}{c}\widehat{p}\pm ME=0.45\pm 0.043\\ =\left(0.407,0.493\right)\end{array}$

If the proportion observed is 45 percent, then there are 95 percent sure that this candidate will vote for between 40.7 percent and 49.3 percent of the population proportion of women.

(c)

To determine

To explain: the intervals for women and men differ and the sense of the gender difference.

Explanation of Solution

Yes, for male and female voters, the intervals overlap. This may mean that the proportion of male and female voters in the population is the same.

(d)

To determine

Find: the 95 percent confidence interval for the proportions difference of men and women who will vote and describe the interval for this person.

Answer to Problem 28E

(0.008, 0.132)

Explanation of Solution

Given:

  $\begin{array}{l}{n}_M=473\\ n_W=522\end{array}$

  $\begin{array}{c}{\widehat{p}}_M=0.52\\ {\widehat{p}}_W=0.45\end{array}$

Formula used:

  $\begin{array}{c}SE\left({\widehat{p}}_M-{\widehat{p}}_W\right)=\sqrt{\frac{p_M{q}_M}{n_M}+\frac{p_W{q}_W}{n_W}}\\ ME=z^{\ast }\times SE\left({\widehat{p}}_M-{\widehat{p}}_W\right)\end{array}$

For the confidence interval

  $\left({\widehat{p}}_M-{\widehat{p}}_W\right)\pm ME$

Calculation:

Estimating the standard deviation

  $\begin{array}{c}SE\left({\widehat{p}}_M-{\widehat{p}}_W\right)=\sqrt{\frac{p_M{q}_M}{n_M}+\frac{p_W{q}_W}{n_W}}\\ =\sqrt{\frac{0.52\left(1-0.52\right)}{473}+\frac{0.45\left(1-0.45\right)}{522}}\\ =\sqrt{\frac{0.52\left(0.48\right)}{473}+\frac{0.45\left(0.55\right)}{522}}\\ =0.0317\end{array}$

For the 95 percent confidence level, $Z^{\ast }=1.96$ , therefore

  $\begin{array}{c}ME=z^{\ast }\times SE\left({\widehat{p}}_M-{\widehat{p}}_W\right)\\ =1.96\times 0.0317\\ =0.062\end{array}$

So, the 95% confidence interval is

  $\begin{array}{c}\left({\widehat{p}}_M-{\widehat{p}}_W\right)\pm ME=\left(0.52-0.45\right)\pm 0.062\\ =0.07\pm 0.062\\ =\left(0.008,0.132\right)\end{array}$

A 95 percent confidence interval is (0.008, 0.132) for the proportion difference of men and women who are going to vote for this candidate. This indicates that for this candidate, there are 0.8 to 13.2 percent greater men than women who would vote.

(e)

To determine

To Explain: about the including the zero in the interval and the meaning of this.

Explanation of Solution

No, 0.0 is not included in this interval. This means 0 is not the difference 's probable value. It might assume that the proportion of men who will vote for this candidate is different from that of women.

(f)

To determine

To explain: if possible, to see if there is a gender difference between voters with respect to this candidate and the correct approach, the explanation for the results in part (c) and part (e) seems contradictory.

Explanation of Solution

There is need to calculate the standard deviation of the difference if there is required to calculate the difference in the proportions of men and women who will vote for this candidate. The difference cannot be calculated by two separate confidence intervals. If want to examine the difference between the proportions of men and women who would vote for this candidate, the two-sample approach is the best method.