Question

Thirty 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 41.5 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:

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 138.5

Population standard deviation =    = 41.5

Sample size = n = 30

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 * (41.5 /  30 )

= 12.5

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

  ± E

138.5 ± 12.5   

( 126.0, 151.0)  

lower limit = 126.0

upper limit = 151.0

margin of error = 12.5

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 * (41.5 /  30)

= 14.9

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

  ± E

138.5 ± 14.9   

( 123.6, 153.4)  

lower limit = 123.6

upper limit = 153.4

margin of error = 14.9

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 * (41.5 /  30 )

= 19.5

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

  ± E

138.5 ± 19.5

( 119.0, 158.0)  

lower limit = 119.0

upper limit = 158.0

margin of error = 19.5