Question

Thirty-five 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 41.1 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

a)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.1 /2) = 1.645
138.5 ± Z (0.1/2 ) * 41.1/√(35)
Lower Limit = 138.5 - Z(0.1/2) 41.1/√(35)
Lower Limit = 127.1
Upper Limit = 138.5 + Z(0.1/2) 41.1/√(35)
Upper Limit = 149.9

Margin of Error = Z (0.1/2 ) * 41.1/√(35) = 11.4

b)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
138.5 ± Z (0.05/2 ) * 41.1/√(35)
Lower Limit = 138.5 - Z(0.05/2) 41.1/√(35)
Lower Limit = 124.9
Upper Limit = 138.5 + Z(0.05/2) 41.1/√(35)
Upper Limit = 152.1

Margin of Error = Z (0.05/2 ) * 41.1/√(35) = 13.6   

c)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
138.5 ± Z (0.01/2 ) * 41.1/√(35)
Lower Limit = 138.5 - Z(0.01/2) 41.1/√(35)
Lower Limit = 120.6
Upper Limit = 138.5 + Z(0.01/2) 41.1/√(35)
Upper Limit = 156.4

Margin of Error = Z (0.01/2 ) * 41.1/√(35) = 17.9