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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.)

llaven  Nov 25, 2018
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\(\text{An }\alpha \text{ confidence interval means that }\alpha \text{ of the probability mass lies within the interval}\\ \text{This means that }\dfrac{1-\alpha}{2} \text{ of the probability mass lie in each of the two tails}\\ \text{a 90% confidence interval has 2.5% probability in the lower tail and }\\ \text{we can look up the corresponding z-score}\\ \text{it turns out that the z-score for 2.5% is }z=-1.645\\ \text{thus our 90% confidence interval is given by}\\ [\mu - 1.645 \sigma,~\mu + 1.645\sigma] \\ \mu \text{ is given to be }\mu = 138.5\\ \text{the standard deviation of the mean is given by }\sigma = \dfrac{\sigma_p}{\sqrt{N}} = \dfrac{44.5}{100} = 0.445\\ \text{and thus our 90% confidence interval is } [137.768, 139.232]\)

 

\(\text{The z-scores for 95% and 99% are }\\ z_{95\%} = -1.96,~z_{99\%} = -2.58\\ \text{see if you can find the intervals for those confidence levels}\)

Rom  Nov 26, 2018

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