Chapter 6, Problem 19TP

Solution

a.

To State:

The number of people surveyed.

  $500$

Given:

The results of a survey of the newsgathering habits of adults aged 18-29 are displayed in the table.

What is your primary source of news?
TV	$32\%$
Internet	$41\%$
Newspaper	$12\%$
Radio	$15\%$
(Margin of error: $\pm 4.47\%$)

Concepts Used:

The margin of error is $\pm \frac{1}{\sqrt{n}}$ where $n$ is the number of people surveyed.

Calculations:

Determine the number of people surveyed.

  $\begin{array}{c}\pm 4.47\%=\pm \frac{1}{\sqrt{n}}\\ {0.0447}^2=\frac{1}{n}\\ n\approx 500\end{array}$

Conclusion:

  $500$ people were surveyed.

b.

To State:

Why might the conclusion “Young adults don’t generally get news from the TV” be inaccurate to draw from this data.

The margin of error is too large to make such a conclusion on the basis of given data.

Given:

The results of a survey of the newsgathering habits of adults aged 18-29 are displayed in the table.

What is your primary source of news?
TV	$32\%$
Internet	$41\%$
Newspaper	$12\%$
Radio	$15\%$
(Margin of error: $\pm 4.47\%$)

Concepts Used:

The population proportion may lie anywhere between $p-\frac{1}{\sqrt{n}}$ and $p+\frac{1}{\sqrt{n}}$ . Where $p$ is the estimate of population proportion and $\pm \frac{1}{\sqrt{n}}$ is the margin for error.

Calculations:

The actual percentage of young adults that get news from TV can lie anywhere between $32-4.47=27.53\%$ and $32+4.47=36.47\%$ . Also, The actual percentage of young adults that get news from Internet can lie anywhere between $41-4.47=36.53\%$ and $41+4.47=45.47\%$ . Note that there can be a situation where TV is a close runner up to Internet as the news source with $36.47\%$ and $36.53\%$ of the population getting news from these media respectively.

Conclusion:

The margin of error is too large to make such a conclusion on the basis of given data.

c.

To State:

The probability, to the nearest tenth of a percent, that the Internet is the primary source of news for at least three out of the six people Aleida chooses to survey at random.

  $47.6\%$

Given:

The results of a survey of the newsgathering habits of adults aged 18-29 are displayed in the table.

What is your primary source of news?
TV	$32\%$
Internet	$41\%$
Newspaper	$12\%$
Radio	$15\%$
(Margin of error: $\pm 4.47\%$)

Concepts Used:

In a binomial experiment with $n$ independent trials each having probability of success $p$ . The probability of $r$ successes is given by ${}_n{C}_r{p}^r{\left(1-p\right)}^{n-r}$ .

The probability of occurrence of the union of many events that are mutually exclusive is the sum of their individual probabilities.

The definition ${}_n{C}_r=\frac{n!}{r!\left(n-r\right)!}$ .

The probability of occurrence of an event and that of its complement add up to $1$ .

Calculations:

The probability that Internet is the news source is $41\%$ . The preferred news source of each person is independent. Calculate the probability that at least three out of the six people prefer Internet. The experiment is equivalent to a binomial experiment with $6$ independent trials each having probability of success $0.41$ .

  $\begin{array}{c}P\left(\text{at least 3 prefer Internet }\right)=1-P\left(\text{less than 3 prefer Internet }\right)\\ =1-P\left(\text{none prefer Internet}\right)-P\left(\text{one prefers Internet}\right)-P\left(\text{2 prefer Internet }\right)\\ =1{-}_6{C}_0{\left(0.41\right)}^0{\left(0.59\right)}^6{-}_6{C}_1{\left(0.41\right)}^1{\left(0.59\right)}^5{-}_6{C}_2{\left(0.41\right)}^2{\left(0.59\right)}^4\\ =1-{\left(0.59\right)}^4\left({0.59}^2+6\times 0.41\times 0.59+15\times {0.41}^2\right)\\ \approx \text{0}\text{.4764}\\ \approx 47.6\%\end{array}$

Conclusion:

  $47.6\%$ is the probability, to the nearest tenth of a percent, that the Internet is the primary source of news for at least three out of the six people Aleida chooses to survey at random.

d.

To State:

Why Aleida’s conclusion that the poll was inaccurate might be wrong.

The sample size of Aleida is too small to reliably refute the poll.

Given:

The results of a survey of the newsgathering habits of adults aged 18-29 are displayed in the table.

What is your primary source of news?
TV	$32\%$
Internet	$41\%$
Newspaper	$12\%$
Radio	$15\%$
(Margin of error: $\pm 4.47\%$)

Aleida finds that four of the six respondents (or $66.7\%$ ) to her poll said that the Internet was the primary source of news.

Concepts Used:

The margin of error is $\pm \frac{1}{\sqrt{n}}$ where $n$ is the number of people surveyed.

Calculations:

Determine the margin for error in her poll’s results.

  $\begin{array}{c}\pm \frac{1}{\sqrt{n}}=\pm \frac{1}{\sqrt{6}}\\ \approx 40.8\%\end{array}$

Conclusion:

Since the actual percentage can lie between $66.7\pm 40.8$ , the result is practically meaningless. The sample size of Aleida is too small to reliably refute the poll as inaccurate.

e.

To State:

To the nearest tenth of a percent, what is the margin of error in Aleida’s survey.

  $\pm 40.8\%$

Given:

The results of a survey of the newsgathering habits of adults aged 18-29 are displayed in the table.

What is your primary source of news?
TV	$32\%$
Internet	$41\%$
Newspaper	$12\%$
Radio	$15\%$
(Margin of error: $\pm 4.47\%$)

Known from Part d:

The margin of error in Aleida’s Poll is $\pm 40.8\%$ .