Some neighborhoods in a particular city are known for high rate of car theft. Th
ID: 3397152 • Letter: S
Question
Some neighborhoods in a particular city are known for high rate of car theft. The monthly averages of the stolen cars in a given year are shown below. Determine the 99% confidence interval on the mean monthly stolen cars. (Assume Normal population) (6.74, 9.91) (6.08, 9.68) (6.23,8.82) (7.88, 11 99) Refer to question 3. Determine a 95% confidence interval for the population standard deviation of stolen cars in the city. Assuming all measurements are normally distributed. (1.51, 3.52) (2.28, 12.39) (1.50, 3.12) (2.26, 9.71)Explanation / Answer
3.
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 7.875
t(alpha/2) = critical t for the confidence interval = 3.105806516
s = sample standard deviation = 2.012291774
n = sample size = 12
df = n - 1 = 11
Thus,
Lower bound = 6.070841348
Upper bound = 9.679158652
Thus, the confidence interval is
( 6.08 , 9.68 ) [ANSWER, B]
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4.
As
df = n - 1 = 11
alpha = (1 - confidence level)/2 = 0.05
Then the critical values for chi^2 are
chi^2(alpha/2) = 19.67513757
chi^2(alpha/2) = 4.574813079
Thus, as
lower bound = (n - 1) s^2 / chi^2(alpha/2) = 2.263897767
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) = 9.736463381
Thus, the confidence interval for the variance is
( 2.263897767 , 9.736463381 )
Also, for the standard deviation, getting the square root of the bounds,
( 1.50 , 3.12 ) [ANSWER, C]
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