Chapter 11.5, Problem 18E

a.

To determine

To calculate:

Margin of error for a survey.

Answer to Problem 18E

Margin of error for a survey that has $1022$ sample size is $3.13\%$ .

Explanation of Solution

Given information:

Sample size: $1022$

Percentage: $76\%$

Calculation:

In order to find the margin of error formula used is,

  $z.\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

Where,

  $z=$$z-$ Score.

  $\widehat{p}=$ Sample proportion.

  $n=$ sample size.

Here given is sample size $=1022$

Substitute the values in above formula,

  $\begin{array}{l}=\pm \frac{1}{\sqrt{n}}\\ =\pm \frac{1}{\sqrt{1022}}\\ =0.031280\\ =3.13\%\end{array}$

Hence, Margin of error for a survey that has $1022$ sample size is $3.13\%$ .

b.

To determine

To calculate:

Interval for a survey.

Answer to Problem 18E

Interval for a survey is $72.87\%$ .

Explanation of Solution

Given information:

Sample size: $1022$

Percentage: $76\%$

Calculation:

In order to find the interval of the survey,

  $I=P-M_e$

Where,

  $I=$ Interval of the survey.

  $P=$ Percentage.

  $M_e=$ Margin of error.

Here given is sample size $=1022$ and $76\%$ .

So,Substitute the values in above formula,

  $76\%-3.13\%=72.87\%$

Hence, Interval for a survey is $72.87\%$ .