In a survey of 2268 adults in a recent year, 1242 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 (Round to three decimal places as needed.)

Step 1 ：First, calculate the sample proportion (p̂), which is the number of successes (people who have made a New Year's resolution) divided by the total number of trials (total adults surveyed).

Step 2 ：Then, use the formula for the confidence interval for a population proportion, which is p̂ ± Z*√((p̂(1-p̂))/n), where Z is the Z-score corresponding to the desired level of confidence, p̂ is the sample proportion, and n is the sample size.

Step 3 ：For a 90% confidence interval, the Z-score is approximately 1.645. For a 95% confidence interval, the Z-score is approximately 1.96.

Step 4 ：Calculate these confidence intervals.

Step 5 ：The 90% confidence interval for the population proportion is approximately (0.530, 0.565), and the 95% confidence interval is approximately (0.527, 0.568). This means that we are 90% confident that the true population proportion of adults who have made a New Year's resolution is between 53.0% and 56.5%, and we are 95% confident that the true population proportion is between 52.7% and 56.8%.

Step 6 ：The width of the 95% confidence interval is larger than the width of the 90% confidence interval, which makes sense because a higher level of confidence requires a larger interval.

Step 7 ：Final Answer: The 90% confidence interval for the population proportion p is approximately \(\boxed{(0.530, 0.565)}\), and the 95% confidence interval is approximately \(\boxed{(0.527, 0.568)}\).