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 44.1 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.)

lower limit
upper limit
margin of error

(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
upper limit
margin of error

(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.)

lower limit
upper limit
margin of error

(d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase?

As the confidence level increases, the margin of error decreases.As the confidence level increases, the margin of error remains the same.    As the confidence level increases, the margin of error increases.

(e) Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?

As the confidence level increases, the confidence interval increases in length.As the confidence level increases, the confidence interval remains the same length.    As the confidence level increases, the confidence interval decreases in length.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 138.5

Population standard deviation =    = 44.1

Sample size = n = 31

a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 * ( 44.1/  31 )

= 13.0

At 90% confidence interval estimate of the population mean is,

  ± E

138.5 ± 13.0

( 125.5, 151.5 )  

lower limit = 125.5

upper limit = 151.5

margin of error = 13.0

b) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

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

= 1.96 * ( 44.1/  31 )

= 15.5

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

  ± E

138.5 ± 15.5

( 123.0, 154.0 )  

lower limit = 123.0

upper limit = 154.0

margin of error = 15.5

c) At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_0.005 = 2.576

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

= 2.576 * ( 44.1/  31 )

= 20.4

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

  ± E

138.5 ± 20.4   

( 118.1, 158.9 )  

lower limit = 118.1

upper limit = 158.9

margin of error = 20.4

d) As the confidence level increases, the margin of error increases

e) As the confidence level increases, the confidence interval increases in length.As the confidence level increases,