To construct a [tex]$95 \%$[/tex] confidence interval for the proportion of subscribers who would like more national news coverage, we need to follow these steps:
1. Calculate the sample proportion [tex]$\hat{p}$[/tex]:
[tex]\[
\hat{p} = \frac{\text{number of favorable responses}}{\text{sample size}} = \frac{594}{1800}
\][/tex]
[tex]\[
\hat{p} = 0.33 \Rightarrow \hat{p} = 33.0\%
\][/tex]
So, the estimated population proportion, [tex]$\hat{p}$[/tex], is [tex]\(33.0 \%\)[/tex].
2. Calculate the margin of error (E):
[tex]\[
E = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\][/tex]
Using the values:
[tex]\[
z^* = 1.96
\][/tex]
[tex]\[
\hat{p} = 0.33
\][/tex]
[tex]\[
n = 1800
\][/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 approximately:
[tex]\[
E = 2.172271928956716 \% \Rightarrow E \approx 2.2 \%
\][/tex]
3. Construct the confidence interval:
[tex]\[
\text{Lower bound} = \hat{p} - E = 33.0\% - 2.2\% = 30.8\%
\][/tex]
[tex]\[
\text{Upper bound} = \hat{p} + E = 33.0\% + 2.2\% = 35.2\%
\][/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].
