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: $\%$ to $\%$ 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 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 The more error we allow, the less precise our estimate. Therefore, as the confidence level increases, the precision of our estimate stays roughly the same decreases

Step 1 ：We are given that the proportion of the sample that regularly uses Facebook as a news source is 74%, or 0.74, and the sample size is 2997. We are asked to find the margin of error and confidence interval at the 90% level of confidence. The z-score for a 90% level of confidence is approximately 1.645.

Step 2 ：We can use the formula for the margin of error for a proportion, which is given by: \[E = Z \sqrt{\frac{{p(1-p)}}{n}}\] where E is the margin of error, Z is the z-score, p is the proportion of the sample with the characteristic of interest, and n is the sample size.

Step 3 ：Substituting the given values into the formula, we get: \[E = 1.645 \sqrt{\frac{{0.74(1-0.74)}}{2997}}\]

Step 4 ：Solving the above expression, we find that the margin of error, E, is approximately 0.013180292995862807, or 1.32% when expressed as a percentage.

Step 5 ：To calculate the confidence interval, we subtract and add the margin of error from/to the sample proportion. So, the lower limit of the confidence interval is \(0.74 - E = 0.7268197070041372\), and the upper limit of the confidence interval is \(0.74 + E = 0.7531802929958628\).

Step 6 ：Thus, the margin of error at the 90% level of confidence is approximately 1.32%, and the confidence interval is from 72.68% to 75.32%.