A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the results below. From past studies, the publisher assumes $\sigma$ is 1.9 minutes and that the population of times is normally distributed.
12
12
$6 \quad 8$
8
10
$8 \quad 9 \quad 8$
12
$7 \quad 7$
$7 \quad 7$
10
Construct the $90 \%$ and $99 \%$ confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals.
The $90 \%$ confidence interval is (Round to one decimal place as needed.)

Step 1 ：The publisher wants to estimate the mean length of time all adults spend reading newspapers. A random sample of 15 people was taken with the results: 12, 12, 6, 8, 8, 10, 8, 9, 8, 12, 7, 7, 7, 7, 10. The standard deviation of the population is assumed to be 1.9 minutes and the population of times is normally distributed.

Step 2 ：To construct the confidence interval, we need to calculate the sample mean and then use the formula for the confidence interval which is: \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z\) is the Z-score corresponding to the desired confidence level, \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size.

Step 3 ：The sample mean is calculated as follows: \(\bar{x} = \frac{\sum x}{n} = \frac{12 + 12 + 6 + 8 + 8 + 10 + 8 + 9 + 8 + 12 + 7 + 7 + 7 + 7 + 10}{15} = 8.733333333333333\)

Step 4 ：The Z-scores for the 90% and 99% confidence levels are approximately 1.645 and 2.576 respectively. These can be found from a standard normal distribution table.

Step 5 ：The standard error is calculated as follows: \(SE = \frac{\sigma}{\sqrt{n}} = \frac{1.9}{\sqrt{15}} = 0.4905778905196061\)

Step 6 ：The 90% confidence interval is calculated as follows: \((\bar{x} - Z_{90}\times SE, \bar{x} + Z_{90}\times SE) = (8.733333333333333 - 1.645\times 0.4905778905196061, 8.733333333333333 + 1.645\times 0.4905778905196061) = (7.926404510809956, 9.54026215585671)\)

Step 7 ：The 99% confidence interval is calculated as follows: \((\bar{x} - Z_{99}\times SE, \bar{x} + Z_{99}\times SE) = (8.733333333333333 - 2.576\times 0.4905778905196061, 8.733333333333333 + 2.576\times 0.4905778905196061) = (7.469688427259727, 9.996978239406939)\)

Step 8 ：\(\boxed{\text{Final Answer: The 90% confidence interval is approximately (7.9, 9.5) and the 99% confidence interval is approximately (7.5, 10.0). The 99% confidence interval is wider than the 90% confidence interval.}}\)