In a survey of 2973 adults, 1412 say they have started paying bills online in the last year. Construct a $99 \%$ confidence interval for the population proportion. Interpret the results. A $99 \%$ confidence interval for the population proportion is $(\square, \square)$.

Step 1 ：Given that in a survey of 2973 adults, 1412 say they have started paying bills online in the last year. We are asked to construct a $99 \%$ confidence interval for the population proportion.

Step 2 ：To construct a confidence interval for a population proportion, we can use the formula: \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(\hat{p}\) is the sample proportion, \(n\) is the sample size, and \(Z\) is the Z-score corresponding to the desired level of confidence.

Step 3 ：In this case, \(\hat{p} = \frac{1412}{2973}\), \(n = 2973\), and \(Z\) is the Z-score for a $99\%$ confidence interval, which is approximately $2.576$.

Step 4 ：Substituting these values into the formula, we get the confidence interval for the population proportion.

Step 5 ：The lower limit of the confidence interval is calculated as: \(0.475 - 2.576 \times \sqrt{\frac{0.475 \times (1-0.475)}{2973}} \approx 0.451\)

Step 6 ：The upper limit of the confidence interval is calculated as: \(0.475 + 2.576 \times \sqrt{\frac{0.475 \times (1-0.475)}{2973}} \approx 0.499\)

Step 7 ：So, the confidence interval for the population proportion is approximately \((0.451, 0.499)\). This means that we are $99\%$ confident that the true population proportion of adults who have started paying bills online in the last year is between $45.1\%$ and $49.9\%$.

Step 8 ：Final Answer: The $99 \%$ confidence interval for the population proportion is approximately \(\boxed{(0.451, 0.499)}\).