(a) Find a 90% confidence interval for the pepulation mean annual number of repo
ID: 3358319 • Letter: #
Question
(a) Find a 90% confidence interval for the pepulation mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place) upper limit margin of erree (b) Find a 95w% 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 it margin of error 12.0 lower limit upper imt margin of error (d) Compare the margins of error for parts (a) through (e). As the confidence sevels increase, do the margins of error increase C As the confidence level increases, the margin of error remains the same C As the confidence level increases, the margin of error decreases C As the confidence level increases, the margin of error increases As the confidence leve·eoeases, the con dence interval increases in length. O As the confidence level increases, the confidence interval remains the same length As the contidence level increases, the con idence interval decreases in length Noed Help?Explanation / Answer
Ans:
a)n=31
90% confidence interval for mean
=138.5+/-1.645*40.5/sqrt(31)
=138.5+/-12
=(126.5, 150.5)
Margin of error=12
b)
95% confidence interval for mean
=138.5+/-1.96*40.5/sqrt(31)
=138.5+/-14.3
=(124.2,152.8)
Margin of error=14.3
c)
99% confidence interval for mean
=138.5+/-2.58*40.5/sqrt(31)
=138.5+/-18.8
=(119.7,157.3)
Margin of error=18.8
d)As the confidence level increase ,margin of error increases.
e)As,the confidence level increasses,length of Confidence interval increases(i.e. become wider)
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