Question 924701
Thirty-five 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.
 (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.)
x-bar = 138.5
ME = 1.645*44.5/sqrt(35) = 12.38
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 lower limit  = 138.5-12.38 = 126.1
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 upper limit = 138.5+12.38 = 150.0
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(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.)
 lower limit:: 138.5-1.96*44.5/sqrt(35) = 123.8 
upper limit:: 138.5+1.96*44.5/sqrt(35) = 153.2
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(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.)
Comment:: Change the "z" value to 2.5758 and do the arithmetic
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 lower limit 
upper limit
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margin of error
Cheers,
Stan H.
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