In a survey of 2465 adults in a recent year 1482 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 width of the confidence intervals. the 90% confidence interval for the population proportion P equals equals? The 95% equals?

To find the confidence intervals for the population proportion, we will use the formula:
CI = \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Given:
n = 2465, \hat{p} = \frac{1482}{2465} = 0.6012
For 90% Confidence Interval:
z_{90\%} = 1.645
CI_{90\%} = 0.6012 \pm 1.645 \times \sqrt{\frac{0.6012 \times 0.3988}{2465}}
CI_{90\%} = 0.6012 \pm 0.0156
CI_{90\%} = (0.5856, 0.6168)

For 95% Confidence Interval:
z_{95\%} = 1.96
CI_{95\%} = 0.6012 \pm 1.96 \times \sqrt{\frac{0.6012 \times 0.3988}{2465}}
CI_{95\%} = 0.6012 \pm 0.0183
CI_{95\%} = (0.5829, 0.6195)

Therefore, the 90% confidence interval for the population proportion P is (0.5856, 0.6168) and the 95% confidence interval is (0.5829, 0.6195) . The 95% confidence interval is wider than the 90% confidence interval due to the higher critical value associated with a higher confidence level.

\textbf{Answer:}
The 90% confidence interval for the population proportion P is (0.5856, 0.6168) .
The 95% confidence interval for the population proportion P is (0.5829, 0.6195) .