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 σ is known to be 40.1 cases per year.

(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.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 138.5

Population standard deviation =    = 40.1

Sample size = n =31

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (40.1 /  31 )

E= 14.1
At 95% confidence interval estimate of the population mean
is,

- E < < + E

138.5 -14.1 <   <138.5 +14.1

124.4<   < 152.6

( 124.4, 152.6)