Thirty-one small communities in Connecticut (population near 10,000 each) gave a

Question

Thirty-one small communities in Connecticut (population near 10,000 each) gave an average of x = 138.5 reported cases of larceny per year. Assume that is known to be 42.7 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.)

(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.)

Explanation / Answer

n =31 , mean = 138.5 , s = 42.7
a) z value for 90%CI = 1.645

CI = mean + / - z * ( s / sqrt(n))
= 138.5 + / - 1.645 * (42.7 / sqrt(31))
= (125.88 , 151.12)

Lower limit = 125.88
Upper limit = 151.12

margin of error = z * (s/sqrt(n))
= 1.645 * (42.7 / sqrt(31))
= 12.616

b)
z value for 95% CI = 1.96

CI = mean + / - z * ( s / sqrt(n))
= 138.5 + / - 1.96 * (42.7 / sqrt(31))
= (123.47 , 153.53)

Lower limit = 123.47
Upper limit = 153.53

margin of error = z * (s/sqrt(n))
= 1.96 * (42.7 / sqrt(31))
= 15.032
c)
z value for 99% CI = 2.576

CI = mean + / - z * ( s / sqrt(n))
= 138.5 + / - 2.576 * (42.7 / sqrt(31))
= (118.74 , 158.26)

Lower limit = 118.74
Upper limit = 158.26

margin of error = z * (s/sqrt(n))
= 2.576 * (42.7 / sqrt(31))
= 19.756

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