Question

Suppose 22 communities have an average of reported cases of car theft per year. Assume that σ is known to be 37.3 cases per year. Find an 85%, 90%, and 95% confidence interval for the population mean annual number of reported larceny cases in such communities. Compare the margins of error. As the confidence level increase, do the margins of error increase?

Answer 1

Solution :

Given that,

= 37.3

n = 22

A ) At 80% confidence level the z is ,

  = 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z_/2 = Z_0.10 = 1.282

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

= 1.282 * (37.3 / 22 )

= 10.19

Margin of error = E = 10.19

B ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (37.3 / 22 )

= 13.08

Margin of error = E = 13.08

C ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (37.3 / 22 )

= 15.59

Margin of error = E = 15.59

As the confidence level increase, the margins of error increase