Chapter 11, Problem 12CR

To determine

To find: The interval of having exact person of US adults.

Answer to Problem 12CR

The exact percent of all the U.S adults who prefer saving money over spending it is between $58.9\%and65.1\%.$

Explanation of Solution

Given:

Out of $1017$ U.S adults, $62\%$ said they prefer saving money over spending it.

Calculation:

For a random sample of size $n$ , taken from a large population, the margin of error is approximated by

Margin of error $=\pm \frac{1}{\sqrt{n}}$ .So, if is the percent of the sample responding in a certain way, then the percent of the population who would respond the same way is likely to be between $p-\frac{1}{\sqrt{n}}andp+\frac{1}{\sqrt{n}}$

To find the margin of error for the survey.

Given that sample size is $1017$ . That is,

  $n=1017$

From the margin of error formula given above,

Margin of error $=\pm \frac{1}{\sqrt{n}}$

  $\begin{array}{l}=\pm \frac{1}{\sqrt{1017}}\\ =\pm 0.031\end{array}$

Thus, the margin of error for the survey is about $\pm 3.1\%$ .

To find the interval that is likely to contain the exact percent of all the U.S adults who prefer saving money over spending it.

To find the interval, subtract and add $3.1\%$ to the percent of people surveyed who prefer saving money over-spending it $(62\%).$

  $p-\frac{1}{\sqrt{n}}=62\%-3.1\%=58.9\%$

and

  $p+\frac{1}{\sqrt{n}}=62\%+3.1\%=65.1\%$

Thus, it is likely that the exact percent of all the U.S adults who prefer saving money over spending it is between $58.9\%and65.1\%.$

Conclusion:

Hence, the exact percent of all the U.S adults who prefer saving money over spending it is between $58.9\%and65.1\%.$