Chapter 22, Problem 7E

(a)

To determine

To find: the sampling style of what his employees are planning to use.

Explanation of Solution

Here, workers will divide the entire population into two strata, males and females, and then pick 300 males from the male stratum and 300 females from the female stratum. So, his workers are preparing to use the design of stratified sampling.

(b)

To determine

To find: the difference that the poll will expect to reveal.

Answer to Problem 7E

6 percent difference

Explanation of Solution

59% -53% =6%. Since there is 6 percent difference. The poll is predicted to show a gap of 6 per cent.

(c)

To determine

To find: The standard deviation for the proportional difference.

Answer to Problem 7E

4 percent

Explanation of Solution

Given:

  $p_M=0.59$

  $p_F=0.53$

  $n_M=n_F=300$

Formula used:

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

Calculation:

Estimating the standard deviation of difference with

  $\begin{array}{c}SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=\sqrt{\frac{p_M{q}_M}{n_M}+\frac{p_F{q}_F}{n_F}}\\ =\sqrt{\frac{0.59\left(1-0.59\right)}{300}+\frac{0.53\left(1-0.53\right)}{300}}\\ =\sqrt{\frac{0.59\left(0.41\right)}{300}+\frac{0.53\left(0.47\right)}{300}}\\ =0.04\end{array}$

4% is the standard deviation for the difference in the proportion.

(d)

To determine

To Construct: the sampling model for the differences in size in the proportions of men and women with his candidate's favourable perceptions that may occur in a survey like this.

Explanation of Solution

The sampling distribution of ${\widehat{p}}_M-{\widehat{p}}_F$ is under acceptable assumptions, modelled by an mean normal model $\mu =p_M-p_F$ and standard deviation $SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=\sqrt{\frac{p_M{q}_M}{n_M}+\frac{p_F{q}_F}{n_F}}\cdot$

  $\begin{array}{c}\mu =p_M-p_F\\ =0.59-0.53\\ =0.06\end{array}$

And standard deviation $SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=0.04$ (from part (c)).

Using the 68-95-99.7 rule

  $\begin{array}{c}\mu -SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=0.06-0.04=-0.02\\ \mu +SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=0.06+0.04=0.10\\ \mu -2SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=0.06-2\left(0.04\right)=0.14\\ \mu +2SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=0.06+2\left(0.04\right)=0.14\\ \mu -3SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=0.06-3\left(0.04\right)=-0.06\\ \mu +3SE\left({\widehat{p}}_M-{\widehat{p}}_F\right)=0.06+3\left(0.04\right)=0.18\end{array}$

Graph:

Below, the distribution model is shown.

They indicate that nearly 68% of the observations would have a difference in proportion between 0.02 and 0.10, approximately 95% of the observations would have a difference in proportion between -0.02 and 0.14, and roughly 99.7% of the observations would have a difference in proportion between -0.06 and 0.18.

(e)

To determine

To explain: The poll misleads the campaign, concluding that there is actually no gender difference.

Answer to Problem 7E

Yes

Explanation of Solution

Yes, if the gap in the sample proportion is just one or two standard deviations below the mean, then the poll could mislead the campaign, concluding that there is actually no gender gap.