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Question 6

According to a recent survey of 1417 voters, $46 \%$ feel that the president is doing an acceptable job. Construct a $98 \%$ confidence interval to estimate the proportion of voters who feel the president is doing an acceptable job.
\[
\widehat{p}=
\]
\[
\frac{\alpha}{2}=
\]
\[
z_{\frac{a}{2}}=
\]

Margin of Error: $E=$
We are $98 \%$ confident that the proportion of voters who feel the president is doing an acceptable job is between (round to 4 decimal places) and
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Step 1 ：First, we need to calculate the standard error. The sample proportion \(\widehat{p}\) is 0.46 and the sample size \(n\) is 1417. We can plug these values into the formula for the standard error: \(\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\).

Step 2 ：Next, we need to find the z-score that corresponds to the desired confidence level. For a 98% confidence level, the alpha level is 0.02 (1 - 0.98), and \(\frac{\alpha}{2}\) is 0.01. The z-score that corresponds to an \(\frac{\alpha}{2}\) of 0.01 is approximately 2.33.

Step 3 ：Then, we calculate the margin of error, which is the z-score multiplied by the standard error: \(E = z \times SE\).

Step 4 ：Finally, we calculate the confidence interval, which is the sample proportion plus or minus the margin of error: \(\widehat{p} \pm E\).

Step 5 ：Substituting the values, we get the standard error \(SE = 0.01324\), the margin of error \(E = 0.03085\), and the confidence interval \(\widehat{p} \pm E = 0.4292, 0.4908\).

Step 6 ：Final Answer: We are 98% confident that the proportion of voters who feel the president is doing an acceptable job is between \(\boxed{0.4292}\) and \(\boxed{0.4908}\).