Chapter 19, Problem 13E

(a)

To determine

To find the margin of error for the $2007$ poll if we want $90\%$ confidence in our estimate of the percent of national adults who are baseball fans.

Answer to Problem 13E

  $0.025$ .

Explanation of Solution

It is given in the question that in a poll taken in March , Gallup asked $\text{1006}$ national adults whether they were baseball fans or not. Thus, to find the margin of error for the $2007$ poll if we want $90\%$ confidence in our estimate of the percent of national adults who are baseball fans, we have,

First find the critical value as:

  $\begin{array}{l}{z}^{*}=invNorm(0.95,0,1)\\ =1.645\end{array}$

Then, the margin of error will be:

  $\begin{array}{l}ME=1.645\times \sqrt{\frac{0.36\times 0.64}{1006}}\\ =0.025\end{array}$

Thus, $0.025$ margin of error for the $2007$ poll is needed if we want $90\%$ confidence in our estimate of the percent of national adults who are baseball fans.

(b)

To determine

To explain what that margin of error mean.

Explanation of Solution

It is given in the question that in a poll taken in March , Gallup asked $\text{1006}$ national adults whether they were baseball fans or not. Then, the margin of error will be:

  $\begin{array}{l}ME=1.645\times \sqrt{\frac{0.36\times 0.64}{1006}}\\ =0.025\end{array}$

Thus, $0.025$ margin of error for the $2007$ poll is needed if we want $90\%$ confidence in our estimate of the percent of national adults who are baseball fans. Thus, the margin of error is the error region about the point estimate.

(c)

To determine

To explain would the margin of error be larger or smaller.

Answer to Problem 13E

The margin of error will be larger.

Explanation of Solution

It is given in the question that in a poll taken in March , Gallup asked $\text{1006}$ national adults whether they were baseball fans or not. Then, the margin of error will be:

  $\begin{array}{l}ME=1.645\times \sqrt{\frac{0.36\times 0.64}{1006}}\\ =0.025\end{array}$

Thus, $0.025$ margin of error for the $2007$ poll is needed if we want $90\%$ confidence in our estimate of the percent of national adults who are baseball fans. Thus, a $99\%$ confidence level contain a larger range which means that the margin of error must be larger.

(d)

To determine

To find that margin of error.

Answer to Problem 13E

  $0.039$ .

Explanation of Solution

It is given in the question that in a poll taken in March , Gallup asked $\text{1006}$ national adults whether they were baseball fans or not. Thus, to find the margin of error be larger or smaller, we have,

First find the critical value as:

  $\begin{array}{l}{z}^{*}=invNorm(0.995,0,1)\\ =2.576\end{array}$

Then, the margin of error will be:

  $\begin{array}{l}ME=2.576\times \sqrt{\frac{0.36\times 0.64}{1006}}\\ =0.039\end{array}$

(e)

To determine

To explain will smaller margins of error produce greater or less confidence in the interval.

Answer to Problem 13E

It will produce less confidence in the interval.

Explanation of Solution

It is given in the question that in a poll taken in March , Gallup asked $\text{1006}$ national adults whether they were baseball fans or not. Then, the margin of error will be:

  $\begin{array}{l}ME=2.576\times \sqrt{\frac{0.36\times 0.64}{1006}}\\ =0.039\end{array}$

Thus, smaller margins of error produce less confidence in the interval because smaller margins of error will mean smaller confidence intervals which occur with decreasing confidence levels.

(f)

To determine

To explain do you think there is been a change from $\text{2006 to 2007}$ in the real proportion of national adults who are baseball fans.

Answer to Problem 13E

No, there is no change.

Explanation of Solution

It is given in the question that in a poll taken in March , Gallup asked $\text{1006}$ national adults whether they were baseball fans or not. Then, the margin of error will be:

  $\begin{array}{l}ME=2.576\times \sqrt{\frac{0.36\times 0.64}{1006}}\\ =0.039\end{array}$

Thus, we think that there is no change from $\text{2006 to 2007}$ in the real proportion of national adults who are baseball fans because both proportions belong in the confidence interval $\left(0.321,0.399\right)$ .