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Thirty-three small communities in Connecticut (population near 10,000 each) gave an average of x = 138.5...

Thirty-three 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 41.5 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    

Solutions

Expert Solution

Solution :

Given that,

= 138.5

= 41.5

n = 33

(a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 * (41.5 / 33)

= 11.9

Margin of error = E = 11.9

At 90% confidence interval estimate of the population mean is,

- E < < + E

138.5 - 11.9 < < 138.5 + 11.9

126.6 < < 150.4

Lower limit = 126.6

Upper limit = 150.4

(b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * (41.5 / 33)

= 14.2

Margin of error = E = 14.2

At 90% confidence interval estimate of the population mean is,

- E < < + E

138.5 - 14.2 < < 138.5 + 14.2

124.3 < < 152.7

Lower limit = 124.3

Upper limit = 152.7

(C)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * (41.5 / 33)

= 18.6

Margin of error = E = 18.6

At 99% confidence interval estimate of the population mean is,

- E < < + E

138.5 - 18.6 < < 138.5 + 18.6

119.9 < < 157.1

Lower limit = 119.9

Upper limit = 157.1


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