Chapter 11, Problem 9CT

(a)

To determine

To identify: The population and sample.

Answer to Problem 9CT

The sample consists of $1296$ adults who said they purchased a product online and $304$ adults who said they did not.

Explanation of Solution

Given:

Out of $1600$ randomly selected U.S adults, $81\%$ said they have purchased product online.

  

Calculation:

The sample here consists of $1600$ U.S adults who were surveyed.

Now,

  $81\%of1600=\frac{81}{100}\times 1600=1296$

The sample consists of $1296$ adults who said they purchased a product online and $304$ adults who said they did not.

Conclusion:

Hence, the sample consists of $1296$ adults who said they purchased a product online and $304$ adults who said they did not.

(b)

To determine

To find: The margin of error for the survey.

Answer to Problem 9CT

The margin of error for the survey is about $\pm 2.5\%$ .

Explanation of Solution

Given:

Out of $1600$ randomly selected U.S adults, $81\%$ said they have purchased product online.

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 $1600$ . That is,

  $n=1600$

From the margin of error formula given above,

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

  $\begin{array}{l}=\pm \frac{1}{\sqrt{1600}}\\ =\pm 0.025\end{array}$

Thus, the margin of error for the survey is about $\pm 2.5\%$ .

Conclusion:

Hence, the margin of error for the survey is about $\pm 2.5\%$ .

(c)

To determine

To give: The interval of all the U.S adults purchasing the product online

Answer to Problem 9CT

The exact percent of all the U.S adults who have purchased a product online is between $78.5\%and83.5\%.$

Explanation of Solution

Given:

Out of $1600$ randomly selected U.S adults, $81\%$ said they have purchased product online.

Calculation:

To find the interval that is likely to contain the exact percent of all U.S adults who have purchased a product online.

To find this interval, subtract and add $2.5\%$ to the percent of people surveyed who have purchased a product online $(81\%).$

  $p-\frac{1}{\sqrt{n}}=81\%-2.5\%=78.5\%$

and

  $p+\frac{1}{\sqrt{n}}=81\%+2.5\%=83.5\%$

Thus, it is likely that the exact percent of all the U.S adults who have purchased a product online is between $78.5\%and83.5\%.$

Conclusion:

Hence, the exact percent of all the U.S adults who have purchased a product online is between $78.5\%and83.5\%.$

(d)

To determine

To explain: the survey is used

Answer to Problem 9CT

The sample will give the biased result.

Explanation of Solution

Given:

Out of $1600$ randomly selected U.S adults, $81\%$ said they have purchased product online.

  

Calculation:

You should use the recent survey because the teachers at the school are not representative of the population and such a sample will always give a biased result.

Conclusion:

Hence, the sample will always give a biased result.