Suppose twenty-two communities have an average of $\bar{x} = 139.6$ reported cases of larceny per year. Assume that $\sigma$ is known to be $45.4$ cases per year. Find a $90\%$, $95\%$, and $98\%$ confidence interval for the population mean annual number of reported larceny cases in such communities. Compare the margins of error. As the confidence level increase, do the margins of error increase?
The $90\%$ confidence level has a margin of error of $74.7$; the $95\%$ confidence level has a margin of error of $89.0$; and the $98\%$ confidence level has a margin of error of $105.8$. As the confidence level increases, the margins of error increase.
The $90\%$ confidence level has a margin of error of $105.8$; the $95\%$ confidence level has a margin of error of $89.0$; and the $98\%$ confidence level has a margin of error of $74.7$. As the confidence level increases, the margins of error decrease.
The $90\%$ confidence level has a margin of error of $15.9$; the $95\%$ confidence level has a margin of error of $19.0$; and the $98\%$ confidence level has a margin of error of $22.6$. As the confidence level increases, the margins of error increase.
The $90\%$ confidence level has a margin of error of $3.4$; the $95\%$ confidence level has a margin of error of $4.0$; and the $98\%$ confidence level has a margin of error of $4.8$. As the confidence level increases, the margins of error increase.
The $90\%$ confidence level has a margin of error of $22.6$; the $95\%$ confidence level has a margin of error of $19.0$; and the $98\%$ confidence level has a margin of error of $15.9$. As the confidence level increases, the margins of error decrease.
