Thirty-one 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 43.5cases per year.
(a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the upper, and lower limit? (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 upper, and lower limit? (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 upper, and lower limit? (Round your answers to one decimal place.)
Solution :
Given that,
Point estimate = sample mean =
= 138.5
Population standard deviation =
= 43.5
Sample size = n =31
a)At 90% confidence level
= 1 - 90%
= 1 - 0.90 =0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645
Margin of error = E = Z/2
* (
/
n)
= 1.645 * ( 43.5 / 31
)
= 12.9
At 90% confidence interval estimate of the population mean is,
±
E
138.5 ± 12.9
( 125.6, 151.4 )
b)At 95% confidence level
= 1 - 95%
= 1 - 0.95 =0.05
/2
= 0.025
Z/2
= Z0.025 = 1.96
Margin of error = E = Z/2
* (
/
n)
= 1.96 * ( 43.5 / 31
)
= 15.3
At 95% confidence interval estimate of the population mean is,
±
E
138.5 ± 15.3
( 123.2, 153.8 )
c)At 99% confidence level
= 1 - 99%
= 1 - 0.99 =0.01
/2
= 0.005
Z/2
= Z0.005 = 2.576
Margin of error = E = Z/2
* (
/
n)
= 2.576 * ( 43.5 / 31
)
= 20.1
At 99% confidence interval estimate of the population mean is,
±
E
138.5 ± 20.1
( 118.4, 158.6 )
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