Question

Suppose twenty-two communities have an average of  = 139.6 reported cases of larceny per year. Assume that σ is known to be 45.4 cases per year. Find a 90%, 95%, and 98% 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?

Group of answer choices

The 90% confidence level has a margin of error of 74.7; the 95% confidence level has a margin of error of 89.0; and the 98% confidence level has a margin of error of 105.8. As the confidence level increases, the margins of error increase.

The 90% confidence level has a margin of error of 3.4; the 95% confidence level has a margin of error of 4.0; and the 98% confidence level has a margin of error of 4.8. As the confidence level increases, the margins of error increase.

The 90% confidence level has a margin of error of 105.8; the 95% confidence level has a margin of error of 89.0; and the 98% confidence level has a margin of error of 74.7. As the confidence level increases, the margins of error decrease.

The 90% confidence level has a margin of error of 22.6; the 95% confidence level has a margin of error of 19.0; and the 98% confidence level has a margin of error of 15.9. As the confidence level increases, the margins of error decrease.

The 90% confidence level has a margin of error of 15.9; the 95% confidence level has a margin of error of 19.0; and the 98% confidence level has a margin of error of 22.6. As the confidence level increases, the margins of error increase

Answer 1

sample mean, xbar = 139.6
sample standard deviation, σ = 45.4
sample size, n = 22

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 45.4/sqrt(22)
ME = 15.9

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 45.4/sqrt(22)
ME = 19

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

ME = zc * σ/sqrt(n)
ME = 2.33 * 45.4/sqrt(22)
ME = 22.6

The 90% confidence level has a margin of error of 15.9; the 95% confidence level has a margin of error of 19.0; and the 98% confidence level has a margin of error of 22.6. As the confidence level increases, the margins of error increase