Question

Thirty-four 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 42.5 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 126.9 Incorrect: Your answer is incorrect. upper limit 150.1 Incorrect: Your answer is incorrect. margin of error 11.6 Incorrect: Your answer is incorrect. (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 124.6 Incorrect: Your answer is incorrect. upper limit 152.4 Incorrect: Your answer is incorrect. margin of error 13.9 Incorrect: Your answer is incorrect. (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 120.3 Incorrect: Your answer is incorrect. upper limit 156.7 Incorrect: Your answer is incorrect. margin of error 18.2 Incorrect: Your answer is incorrect.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 138.5

Population standard deviation =    = 42.5

Sample size = n = 34

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 * ( 42.5 /  34 )

= 12.0

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

  ± E

138.5 ± 12.0

( 126.5, 150.5 )  

margin of error = 12.0

lower limit = 126.5

upper limit = 150.5

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 * ( 42.5 /  34 )

= 14.3

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

  ± E

138.5 ± 14.3

( 124.2, 152.8 )  

margin of error = 14.3

lower limit = 124.2

upper limit = 152.8

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 * ( 42.5 /  34 )

= 18.8

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

  ± E

138.5 ± 18.8

( 119.7, 157.3 )  

margin of error = 18.8

lower limit = 119.7

upper limit = 157.3