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 40.3 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

Formula

Confidence Interval for population Mean When Population Standard deviation is known

Margin of error:

Given,

Sample average reported cases of larceny per year : Sample average : = 138.5

Population Standard deviation : = 40.3 cases per year

Sample Size : n= 30

(a)

for 90% confidence level = (100-90)/100 = 0.1

/2 = 0.05

Z_/2 = Z_0.05 = 1.6449

90% confidence interval for the population mean annual number of reported larceny cases in such communities

Margin of error

lower limit	126.4
upper limit	150.6
margin of error	12.1

(b)

for 95% confidence level = (100-95)/100 = 0.05

/2 = 0.025

Z_/2 = Z_0.025 = 1.96

95% confidence interval for the population mean annual number of reported larceny cases in such communities

Margin of error :

lower limit	124.1
upper limit	152.9
margin of error	14.4

(c)

for 99% confidence level = (100-99)/100 = 0.01

/2 = 0.005

Z_/2 = Z_0.005 = 2.5758

99% confidence interval for the population mean annual number of reported larceny cases in such communities

Margin of error:

lower limit	119.5
upper limit	157.5
margin of error	19.0