Question

Thirty-two small communities in Connecticut (population near 10,000 each) gave an average of x = 138.5 reported cases of larceny per year. Assume that ? is known to be 41.3 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

Solution :

Given that,

= 138.5

= 41.3

n = 32

(a)

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 * (41.3 / 32)

=12.01

At 90% confidence interval is,

- E < < + E

138.5-12.01 < < 138.5+12.01

126.49< < 150.51

lower limit:126.49

upper limit:150.51

(b)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 = 1.96

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

=1.96 * (41.3 / 32)

= 14.3097

At 99% confidence interval is,

- E < < + E

138.5-14.3097 < < 138.5+14.3097

124.1903< < 152.8097

lower limit:124.1903

upper limit:152.1903