Chapter 6.4, Problem 27PS

Solution

(a)

To find: The margin of error for the survey.

The margin of error for the survey is $\pm 4.8\%$ .

Given information: A survey of 439 teenagers are given in the United States where 14% said that they worked during their summer vacation.

Formula used: The formula that is used for evaluating the margin of error is $\pm \frac{1}{\sqrt{n}}$ .

Calculation: The number of teenagers is 439 so one has $n=439$, thus the margin of error is evaluated as:

  $\begin{array}{c}\pm E=\pm \frac{1}{\sqrt{n}}\\ =\pm \frac{1}{\sqrt{439}}\\ =\pm 0.048\\ =\pm 4.8\%\end{array}$

Hence the margin of error for the survey is $\pm 4.8\%$ .

(b)

To find: Interval that is likely to contain the exact percent of all U.S. teenagers who worked during their summer vacation.

The interval will be $\pm 9.2\%$ to $\pm 18.8\%$ .

Given information: A survey of 439 teenagers are given in the United States where 14% said that they worked during their summer vacation.

Explanation:

As 14% said they worked during summer vacation so the interval is evaluated as:

  $14\%\pm 4.8\%=9.2\%\text{ to 18}\text{.8\%}$ .

Hence the interval will be $\pm 9.2\%$ to $\pm 18.8\%$ .