Question

Thirty-two 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 44.1 cases per year.

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.) lower limit = upper limit = margin of error =

Answer 1

Solution :

Given that,

Sample size = n = 32

Z/2 = 2.576

Margin of error = E = Z/2* ( /n)

= 2.576 * (44.1 / 32)

Margin of error = E = 20.1

At 99% confidence interval estimate of the population mean is,

- E < < + E

138.5 - 20.1 < < 138.5 + 20.1

118.4 < < 158.6

lower limit = 118.4 , upper limit = 158.6