Thirty small communities in Connecticut (population near 10,000 each) gave an average of $\bar{x}=138.8$ reported cases of larceny per year. Assume that $\sigma$ is known to be 44.7 cases per year. (a) Find a $90 \%$ confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error (b) Find a 95\% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) \begin{tabular}{lll} lower limit & 1.78 & \\ & & \\ upper limit & 31.82 & \\ margin of error & & \\ & & \end{tabular} c) Find a $99 \%$ confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 138.8, the standard deviation (\(\sigma\)) is 44.7, and the sample size (\(n\)) is 30.

Step 2 ：The formula for the confidence interval is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\), where \(Z\) is the Z-score corresponding to the desired confidence level.

Step 3 ：The Z-scores for 90%, 95%, and 99% confidence levels are approximately 1.645, 1.96, and 2.576 respectively.

Step 4 ：For a 90% confidence level, the lower limit is \(\bar{x} - Z \frac{\sigma}{\sqrt{n}} = 138.8 - 1.645 \frac{44.7}{\sqrt{30}}\), the upper limit is \(\bar{x} + Z \frac{\sigma}{\sqrt{n}} = 138.8 + 1.645 \frac{44.7}{\sqrt{30}}\), and the margin of error is \(Z \frac{\sigma}{\sqrt{n}} = 1.645 \frac{44.7}{\sqrt{30}}\). After calculation, the lower limit is \(\boxed{130.6}\), the upper limit is \(\boxed{147.0}\), and the margin of error is \(\boxed{8.2}\).

Step 5 ：For a 95% confidence level, the lower limit is \(\bar{x} - Z \frac{\sigma}{\sqrt{n}} = 138.8 - 1.96 \frac{44.7}{\sqrt{30}}\), the upper limit is \(\bar{x} + Z \frac{\sigma}{\sqrt{n}} = 138.8 + 1.96 \frac{44.7}{\sqrt{30}}\), and the margin of error is \(Z \frac{\sigma}{\sqrt{n}} = 1.96 \frac{44.7}{\sqrt{30}}\). After calculation, the lower limit is \(\boxed{128.8}\), the upper limit is \(\boxed{148.8}\), and the margin of error is \(\boxed{10.0}\).

Step 6 ：For a 99% confidence level, the lower limit is \(\bar{x} - Z \frac{\sigma}{\sqrt{n}} = 138.8 - 2.576 \frac{44.7}{\sqrt{30}}\), the upper limit is \(\bar{x} + Z \frac{\sigma}{\sqrt{n}} = 138.8 + 2.576 \frac{44.7}{\sqrt{30}}\), and the margin of error is \(Z \frac{\sigma}{\sqrt{n}} = 2.576 \frac{44.7}{\sqrt{30}}\). After calculation, the lower limit is \(\boxed{125.3}\), the upper limit is \(\boxed{152.3}\), and the margin of error is \(\boxed{13.5}\).