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 45.1 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

Answer 1

a) At 90% confidence interval the critical value is z_0.05 = 1.645

Margin of error = z_0.05 *

= 1.645 * 45.1/

= 13.3

The 90% confidence interval is

+/- E

= 138.5 +/- 13.3

= 125.2, 151.8

Lower limit = 125.2

Upper limit = 151.8

b) At 95% confidence interval the critical value is z_0.025 = 1.96

Margin of error = z_0.025 *

= 1.96 * 45.1/

= 15.9

The 95% confidence interval is

+/- E

= 138.5 +/- 15.9

= 122.6, 154.4

Lower limit = 122.6

Upper limit = 154.4

c) At 99% confidence interval the critical value is z_0.005 = 2.575

Margin of error = z_0.005 *

= 2.575 * 45.1/

= 20.9

The 99% confidence interval is

+/- E

= 138.5 +/- 20.9

= 117.6, 159.4

Lower limit = 117.6

Upper limit = 159.4