+7 Thirty-three small communities in Connecticut (population near 10,000 each) gave an average of x = 138.5 reported cases of larceny per year. Assume that σ is known to be 44.5 cases per year. Each answer must have the upper limit, lower limit, and margin or error.
(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.)
(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.)
(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.)
+6248 \(\text{An }\alpha \text{ confidence interval means that }\alpha \text{ of the probability mass lies within the interval}\\ \text{This means that }\dfrac{1-\alpha}{2} \text{ of the probability mass lie in each of the two tails}\\ \text{a 90% confidence interval has 2.5% probability in the lower tail and }\\ \text{we can look up the corresponding z-score}\\ \text{it turns out that the z-score for 2.5% is }z=-1.645\\ \text{thus our 90% confidence interval is given by}\\ [\mu - 1.645 \sigma,~\mu + 1.645\sigma] \\ \mu \text{ is given to be }\mu = 138.5\\ \text{the standard deviation of the mean is given by }\sigma = \dfrac{\sigma_p}{\sqrt{N}} = \dfrac{44.5}{100} = 0.445\\ \text{and thus our 90% confidence interval is } [137.768, 139.232]\)
\(\text{The z-scores for 95% and 99% are }\\ z_{95\%} = -1.96,~z_{99\%} = -2.58\\ \text{see if you can find the intervals for those confidence levels}\)
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