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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]$\left(z^\ \textless \ em\ \textgreater \ \text{-score }=1.96\right)$[/tex] for the proportion of subscribers who would like more national news coverage. Complete the statements.

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

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].

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].

Sagot :

GinnyAnswer
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].
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