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):
0.74
Confidence Interval: 72.68
$0^{8} \%$ to 75.32
1.3180292995863 or 1.3179091145438
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.
Margin of Error (as a percentage):
Confidence Interval: $\%$ to

Step 1 ：The problem is asking for the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, given a sample size of 2997 and a proportion of 0.74.

Step 2 ：The margin of error for a proportion in a population (p) can be calculated using the formula: \(E = Z \sqrt{\frac{p(1-p)}{n}}\), where E is the margin of error, Z is the z-score, which depends on the desired level of confidence (for a 95% level of confidence, Z is approximately 1.96), p is the proportion in the population (in this case, 0.74), and n is the size of the population (in this case, 2997).

Step 3 ：Substituting the given values into the formula, we get: \(E = 1.96 \sqrt{\frac{0.74(1-0.74)}{2997}}\).

Step 4 ：The confidence interval can then be calculated as: \((p - E, p + E)\), which gives us: \((0.74 - E, 0.74 + E)\).

Step 5 ：Calculating the above expressions, we find that the margin of error at the 95% level of confidence is approximately 1.57%.

Step 6 ：The confidence interval is approximately from 72.43% to 75.57%.

Step 7 ：\(\boxed{\text{Final Answer: The margin of error at the 95% level of confidence is approximately 1.57%. The confidence interval is approximately from 72.43% to 75.57%.}}\)