Thirty-four small communities in Connecticut (population near 10,000 each) gave

Question

Thirty-four 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 43.3 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     (d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase? As the confidence level increases, the margin of error decreases. As the confidence level increases, the margin of error remains the same.     As the confidence level increases, the margin of error increases. (e) Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length? As the confidence level increases, the confidence interval increases in length. As the confidence level increases, the confidence interval decreases in length.     As the confidence level increases, the confidence interval remains the same length.

Explanation / Answer

a) here std error =std deviation/(n)1/2 =43.3/(34)1/2 =7.4259

for 90% CI; z=1.645

therefore 90% confidence interval =sample mean -/+ z*std error =126.3 ; 150.7

margin of error =z*std error =12.2

b) for 95% CI ; z=1.96

therefore 95% confidence interval =sample mean -/+ z*std error =123.9 ; 153.1

margin of error =z*std error =14.6

c)

for 99% CI ; z=2.5758

therefore 95% confidence interval =sample mean -/+ z*std error =119.4 ; 157.6

margin of error =z*std error =19.1

d)As the confidence level increases, the margin of error increases.

e)As the confidence level increases, the confidence interval increases in length.

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