Question

Suppose thirty-nine communities have an average of xbar=123.6 reported cases of larceny per year. Assume that o is known to be 40.3 cases per year. Find an 85%, 95%, and 98% confidence interval for the population mean annual number of reported larceny cases in such communities. Compare the lengths of the confidence intervals. As the confidence levels increase, do the confidence intervals increase in length? Round your answers to one decimal place.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 123.6

Population standard deviation =    = 40.3

Sample size = n = 39

a) At 85% confidence level

= 1 - 85%  

= 1 - 0.85 =0.15

/2 = 0.075

Z/2 = Z0.075  = 1.44

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

= 1.44 * (40.3 /  39)

= 9.29

At 85% confidence interval estimate of the population mean is,

  ± E

123.6 ± 9.29

( 114.31, 132.89)  

b) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

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

= 1.96 * (40.3 /  39)

= 12.6

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

  ± E

123.6 ± 12.6

( 111.0, 136.2 )  

c) At 98% confidence level

= 1 - 98%  

= 1 - 0.98 =0.02

/2 = 0.01

Z/2 = Z0.01  = 2.326

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

= 2.326 * (40.3 /  39 )

= 15.0

At 98% confidence interval estimate of the population mean is,

  ± E

123.6 ± 15.0

( 108.6, 138.6 )  

As the confidence level increases, the confidence interval increases in length