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 42.7 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 means 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

solution:-
given that mean = 138.5 , standard deviation = 42.7
sample n = 30

(a)90% confidence for z is 1.645
confidence interval formula
mean +/- margin of error
mean +/- z * s/sqrt(n)
138.5 +/- 1.645 * 42.7/sqrt(30)
138.5 +/- 12.8
=> (125.7 , 151.3)
lower limit = 125.7
upper limit = 151.3
margin of error = 12.8

(b)95% confidence for z is 1.96
confidence interval formula
mean +/- margin of error
mean +/- z * s/sqrt(n)
138.5 +/- 1.96 * 42.7/sqrt(30)
138.5 +/- 15.3
(123.2 , 153.8)
lower limit = 123.2
upper limit = 153.8
margin of error = 15.3

(c)99% confidence for z is 2.58
confidence interval formula
mean +/- margin of error
mean +/- z * s/sqrt(n)
138.5 +/- 2.58 * 42.7/sqrt(30)
138.5 +/- 20.1
(118.4 , 158.6)
lower limit = 118.4
upper limit = 158.6
margin of error = 20.1