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In a survey of 2415 adults in a recent year, 1395 say they have made a New Year's resolution,
Construct $90 \%$ and $95 \%$ confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

The $90 \%$ confidence interval for the population proportion $p$ is ( $\square, \square$.
(Round to three decimal places as needed.)

Step 1 ：Calculate the sample proportion (p̂) which is the number of successes (people who made a New Year's resolution) divided by the total number of trials (total adults surveyed). p̂ = \(\frac{1395}{2415} = 0.577\)

Step 2 ：Calculate the standard error (SE) for the proportion, which is the square root of [ p̂(1 - p̂) / n ], where n is the total number of trials. SE = \(\sqrt{\frac{0.577(1 - 0.577)}{2415}} = 0.010\)

Step 3 ：Calculate the confidence intervals. For a 90% confidence interval, the z-score is 1.645. For a 95% confidence interval, the z-score is 1.96. The confidence interval is calculated as p̂ ± (z*SE).

Step 4 ：For the 90% confidence interval: Lower limit = 0.577 - (1.645*0.010) = 0.557, Upper limit = 0.577 + (1.645*0.010) = 0.597. So, the 90% confidence interval is (0.557, 0.597).

Step 5 ：For the 95% confidence interval: Lower limit = 0.577 - (1.96*0.010) = 0.557, Upper limit = 0.577 + (1.96*0.010) = 0.597. So, the 95% confidence interval is (0.557, 0.597).

Step 6 ：Interpretation: We are 90% confident that the true population proportion of adults who make a New Year's resolution is between 55.7% and 59.7%. We are 95% confident that the true population proportion is between 55.7% and 59.7%. \(\boxed{90\% CI = (0.557, 0.597), 95\% CI = (0.557, 0.597)}\)

Step 7 ：Comparing the widths of the confidence intervals, we can see that the 95% confidence interval is wider than the 90% confidence interval. This is because a higher level of confidence requires a larger margin of error.