Chapter PV, Problem 9RE

(a)

To determine

To explain what the margin of error means.

Explanation of Solution

It is given in the question that the Pew research foundation sampled $1593$ respondents and asked how they obtained news. In Pew’s report, $33\%$ of respondents say that they now obtain news from the internet at least once a week and a margin of error of $\pm 3\%$ for this result was found. Thus, the margin of error in this context is the amount by which the statistics varies assuming a normal sampling distribution, that varies by $33\%\pm 3\%$ .

(b)

To determine

To explain how would you expect the margin of error for this statistic to change in comparison with the margin of error for the percentage of all respondents.

Explanation of Solution

It is given in the question that the Pew research foundation sampled $1593$ respondents and asked how they obtained news. In Pew’s report, $33\%$ of respondents say that they now obtain news from the internet at least once a week and a margin of error of $\pm 3\%$ for this result was found. Thus, we would expect the margin of error for this statistic to change in comparison with the margin of error for the percentage of all respondents since margin of error ME and the standard error SE are related then we expect a higher margin of error for a smaller pool of respondents if $SE\propto \frac{1}{\sqrt{n}}$ .

(c)

To determine

To find a confidence interval for this statistics.

Answer to Problem 9RE

  $(0.387,0.513)$ .

Explanation of Solution

It is given in the question that the Pew research foundation sampled $1593$ respondents and asked how they obtained news. In Pew’s report, $33\%$ of respondents say that they now obtain news from the internet at least once a week and a margin of error of $\pm 3\%$ for this result was found. Now, let us calculate the standard error as:

  $\begin{array}{l}S{E}_{\text{pooled}}=\sqrt{\frac{0.45(1-0.45)}{239}}\\ =0.032\end{array}$

Thus, the $95\%$ confidence interval is then calculated as:

  $\begin{array}{l}0.45\pm (1.96\times 0.032)\\ =(0.387,0.513)\end{array}$

(d)

To determine

To explain how does the margin of error for your confidence interval compare with the values in parts (a) and (b) and why.

Explanation of Solution

It is given in the question that the Pew research foundation sampled $1593$ respondents and asked how they obtained news. In Pew’s report, $33\%$ of respondents say that they now obtain news from the internet at least once a week and a margin of error of $\pm 3\%$ for this result was found. Thus, the margin of error for your confidence interval compare with the values in parts (a) and (b) as for the same reason as in part (b) we expect it to be larger.