Answered

A major newspaper conducted a random survey of 1,800 of its 75,000 subscribers. The subscribers were asked if the paper should increase its coverage of national news. Out of the 1,800 in the sample, 594 responded that they want more national news.

Construct a [tex]$95 \%$[/tex] confidence interval ( [tex]$z^\ \textless \ em\ \textgreater \ $[/tex]-score [tex]$=1.96$[/tex] ) for the proportion of subscribers who would like more national news coverage.

Complete the statements:

1. The estimated population proportion ( [tex]$\hat{p}$[/tex] ) is [tex]$\square \%$[/tex].

2. When the margin of error is calculated using the formula [tex]$E=z^\ \textless \ /em\ \textgreater \ \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex], to the nearest tenth of a percent, the result is [tex]$\square \%$[/tex].

3. With [tex]$95 \%$[/tex] confidence, it can be said that the proportion of subscribers who would like more coverage of national news is between [tex]$\square \%$[/tex] and [tex]$\square \%$[/tex].



Answer :

GinnyAnswer
Let's walk through the steps to construct a [tex]$95\%$[/tex] confidence interval for the proportion of subscribers who would like more national news coverage.

1. Calculate the sample proportion ([tex]\(\hat{p}\)[/tex]):

The sample proportion is calculated as the number of subscribers who responded positively divided by the total number of subscribers surveyed.
[tex]\[ \hat{p} = \frac{594}{1800} = 0.33 \][/tex]

2. Calculate the margin of error (E):

The margin of error at a [tex]$95\%$[/tex] confidence level with a [tex]\(z^*\)[/tex]-score of 1.96 is calculated using the formula:
[tex]\[ E = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]
Substituting the values:
[tex]\[ E = 1.96 \cdot \sqrt{\frac{0.33 \cdot (1 - 0.33)}{1800}} \approx 0.02172271928956716 \][/tex]

To convert the margin of error to a percentage:
[tex]\[ \text{Margin of Error} \approx 2.172271928956716\% \][/tex]
Rounding to the nearest tenth of a percent gives:
[tex]\[ \text{Margin of Error} \approx 2.2\% \][/tex]

3. Construct the confidence interval:

The confidence interval is calculated as:
[tex]\[ \hat{p} \pm E \][/tex]
For the lower bound:
[tex]\[ \text{Lower Bound} = 0.33 - 0.02172271928956716 \approx 0.30827728071043285 \][/tex]
For the upper bound:
[tex]\[ \text{Upper Bound} = 0.33 + 0.02172271928956716 \approx 0.3517227192895672 \][/tex]

To convert these bounds to percentages:
[tex]\[ \text{Lower Bound} \approx 30.827728071043285\% \][/tex]
[tex]\[ \text{Upper Bound} \approx 35.17227192895672\% \][/tex]

Rounding to the nearest tenth of a percent:
[tex]\[ \text{Lower Bound} \approx 30.8\% \][/tex]
[tex]\[ \text{Upper Bound} \approx 35.2\% \][/tex]

So, to complete the statements:

- The estimated population proportion is [tex]\(33\%\)[/tex].
- When the margin of error is calculated using the formula [tex]\(E = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)[/tex], to the nearest tenth of a percent, the result is [tex]\(2.2\%\)[/tex].
- With [tex]$95\%$[/tex] confidence, it can be said that the proportion of subscribers who would like more coverage of national news is between [tex]\(30.8\%\)[/tex] and [tex]\(35.2\%\)[/tex].