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 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.) (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.)(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.)
Accepted Solution
A:
Answer:Part A: Error Margin= 11.9Interval is 126.6 to 150.4Part B:Error Margin=14.1Interval is 124.4 to 152.6Part C:Error Margin=18.6Interval is 119.9 to 157.1Step-by-step explanation:Confidence Interval Z90% 1.64595% 1.9699% 2.58Formula used in all three parts:Error Margin=Z*σ/[tex]\sqrt{n}[/tex]Upper limit of Interval=x+ Error MarginLower limit of Interval=x- Error MarginPart A:Error Margin=1.645*42.7/[tex]\sqrt{35}[/tex]Error Margin= 11.9Upper limit of Interval=138.5+ 11.9Upper limit of Interval=150.4Lower limit of Interval=138.5 - 11.9Lower limit of Interval=126.6Interval is 126.6 to 150.4Part B:Error Margin=1.96*42.7/[tex]\sqrt{35}[/tex]Error Margin=14.1Upper limit of Interval=138.5+ 14.1Upper limit of Interval=152.6Lower limit of Interval=138.5 - 14.1Lower limit of Interval=124.4Interval is 124.4 to 152.6Part C:Error Margin=2.58*42.7/[tex]\sqrt{35}[/tex]Error Margin=18.6Upper limit of Interval=138.5+ 18.6Upper limit of Interval=157.1Lower limit of Interval=138.5 - 18.6Lower limit of Interval=119.9Interval is 119.9 to 157.1