Chapter 22, Problem 8E

(a)

To determine

To find: what sampling style is expected to be used by the magazine.

Explanation of Solution

Here, the magazine will split the entire population into two strata, American car owners and Japanese car owners, and then pick 450 American car owners and 450 Japanese car owners. So, the magazine aims to use the design of stratified sampling.

(b)

To determine

To find: what difference their poll would expect to reveal.

Answer to Problem 8E

2 percent

Explanation of Solution

78%- 76%= 2%.

Since there is a 2% gap, it is predicted the poll to display a gap of 2 percent.

(c)

To determine

To find: the standard deviation for the proportion difference.

Answer to Problem 8E

3 percent

Explanation of Solution

Given:

  $\begin{array}{l}{p}_j=0.78\\ p_A=0.76\\ n_j=n_A=450\end{array}$

Formula used:

  $SE\left({\widehat{p}}_j-{\widehat{p}}_A\right)=\sqrt{\frac{p_j{q}_j}{n_j}+\frac{p_A{q}_A}{n_A}}$

Calculation:

Estimated the standard deviation of the gap with

  $\begin{array}{c}SE\left({\widehat{p}}_j-{\widehat{p}}_A\right)=\sqrt{\frac{p_j{q}_j}{n_j}+\frac{p_A{q}_A}{n_A}}\\ =\sqrt{\frac{0.78\left(1-0.78\right)}{450}+\frac{0.76\left(1-0.76\right)}{450}}\\ =\sqrt{\frac{0.78\left(0.22\right)}{450}+\frac{0.76\left(0.24\right)}{450}}\\ =0.03\end{array}$

For the difference in the proportions, the standard deviation is 3%.

(d)

To determine

To construct: a model of sampling for the difference in proportions that may occur in such a poll.

Explanation of Solution

The sampling distribution of ${\widehat{p}}_j-{\widehat{p}}_A$ is, under acceptable assumptions, modelled by an mean normal model $\mu =p_j-p_A$ and standard deviation $SE\left({\widehat{p}}_j-{\widehat{p}}_A\right)=\sqrt{\frac{p_j{q}_j}{n_j}+\frac{p_A{q}_A}{n_A}}\cdot$

  $\begin{array}{c}\mu =p_J-p_A\\ =0.78-0.76\\ =0.02\end{array}$

Using the 68-95-99.7 rule

  $\mu -SE\left({\widehat{p}}_J-{\widehat{p}}_A\right)=0.02-0.03=-0.01$

  $\mu +SE\left({\widehat{p}}_J-{\widehat{p}}_A\right)=0.02+0.03=0.05$

  $\mu -2SE\left({\widehat{p}}_J-{\widehat{p}}_A\right)=0.02-2\left(0.03\right)=-0.04$

  $\mu +2SE\left({\widehat{p}}_J-{\widehat{p}}_A\right)=0.02+2\left(0.03\right)=0.08$

  $\mu -3SE\left({\widehat{p}}_J-{\widehat{p}}_A\right)=0.02-3\left(0.03\right)=-0.06$

  $\mu +3SE\left({\widehat{p}}_J-{\widehat{p}}_A\right)=0.02+3\left(0.03\right)=0.10$

Graph:

The distribution model graph:

They indicate that nearly 68% of the observations will have a difference in proportion between -0.01 and 0.05, nearly 85% of the observations will have a difference in proportion between -0.04 and 0.08, and nearly 99.7% of the observations will have a difference in proportion between -0.06 and 0.10.

(e)

To determine

To explain: that the poll misled the paper, concluding that American car owners are much happier with their cars than Japanese car owners.

Explanation of Solution

Yes, if the sample ratio gap is just one-two standard deviations underneath the mean, then the survey could mislead the campaign, indicating that American car owners are much satisfied with their cars than Japanese car owners.