Chapter 11.5, Problem 17E

a.

To determine

To calculate:

Margin of error for a survey.

Answer to Problem 17E

Margin of error for a survey that has $1020$ sample size is $3.13\%$ .

Explanation of Solution

Given information:

Sample size: $1020$

Percentage: $41\%$

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 $=1020$

Substitute the values in above formula,

  $\begin{array}{l}=\pm \frac{1}{\sqrt{n}}\\ =\pm \frac{1}{\sqrt{1020}}\\ =0.0313\\ =3.13\%\end{array}$

Hence, Margin of error for a survey that has $1020$ sample size is $3.13\%$ .

b.

To determine

To calculate:

Interval for a survey.

Answer to Problem 17E

Interval for a survey is $37.87\%$ .

Explanation of Solution

Given information:

Sample size: $1020$

Percentage: $41\%$

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 $=1020$ and $41\%$ .

So,Substitute the values in above formula,

  $41\%-3.13\%=37.87\%$

Hence, Interval for a survey is $37.87\%$ .