A poll of 2997 U.S. adults found that $74 \%$ regularly used Facebook as a news source.
Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the $90 \%$ level of confidence. Report answers to at least 2 decimal places.
Margin of Error (as a percentage):
Confidence Interval: 72.68
Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the $95 \%$ level of confidence. Report answers to at least 2 decimal places.
Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the $99 \%$ level of confidence. Report answers to at least 2 decimal places.
Margin of Error (as a percentage):
Confidence Interval: $\%$ to
Answer
The confidence interval is approximately \(\boxed{72.68\%}\) to \(\boxed{75.32\%}\).
Step 1 ：First, we calculate the standard error using the formula \(\sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the sample proportion and \(n\) is the sample size. In this case, \(p = 0.74\) and \(n = 2997\).
Step 2 ：Next, we find the z-score for a 90% confidence level. The z-score is the number of standard deviations a given proportion is away from the mean. For a 90% confidence level, the z-score is approximately 1.645.
Step 3 ：We then calculate the margin of error by multiplying the standard error by the z-score.
Step 4 ：Finally, we calculate the confidence interval by subtracting and adding the margin of error from the sample proportion.
Step 5 ：Using these calculations, we find that the margin of error for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence, is approximately \(\boxed{1.32\%}\).
Step 6 ：The confidence interval is approximately \(\boxed{72.68\%}\) to \(\boxed{75.32\%}\).
