Some neighborhoods in a particular city are known for high r

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 arc shown below. Determine the 90% confidence interval on the mean monthly stolen cars. (Assume Normal population)ever to question 3. Determine a 95% confidence interval for the population standard deviation of stolen cars in the city. Assuming all measurements arc normally distributed. (1.38, 3.32) (1.90, 11.0) (2.03, 11.65) (1.42, 3.41)

Solution

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 =    0.05          
X = sample mean =    7.875          
t(alpha/2) = critical t for the confidence interval =    1.795884819          
s = sample standard deviation =    2.012291774          
n = sample size =    12          
df = n - 1 =    11          
Thus,              
              
Lower bound =    6.831773026          
Upper bound =    8.918226974          
              
Thus, the confidence interval is              
              
(   6.83   ,   8.92   ) [answer, C]

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4.

As              
              
df = n - 1 =    11          
alpha = (1 - confidence level)/2 =    0.025          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    21.92004926          
chi^2(alpha/2) =    3.815748252          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    2.032043792          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    11.67333301          
              
Thus, the confidence interval for the variance is              
              
(   2.032043792   ,   11.67333301   )
              
Also, for the standard deviation, getting the square root of the bounds,              
              
(   1.42   ,   3.41   ) [ANSWER, D]