Question 1 17 pts A sample of 41 recent graduates from a specific program yielde

Question

Question 1 17 pts A sample of 41 recent graduates from a specific program yielded an average starting salary of $43,720 and a standard deviation of $3853 (a) Find the 90% confidence interval for the mean starting salary based on this sample and interpret it in the context of the problem. (b) Find the 90% confidence interval for the standard deviation in the starting salary based on this sample and interpret it in the context of the problem. (c) For this problem only. Also indicate what "90% confidence means" in your interpretations

Explanation / Answer

a)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    43720          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    3853          
n = sample size =    41          
              
Thus,              
Margin of Error E =    989.7701168          
Lower bound =    42730.22988          
Upper bound =    44709.77012          
              
Thus, the confidence interval is              
              
(   42730.22988   ,   44709.77012   ) [ANSWER]

We are 90% confident that the true mean starting salary for graduates of a specific program is between 42730.22988 and 44709.77012. [ANSWER]

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b)

As              
              
df = n - 1 =    40          
alpha = (1 - confidence level)/2 =    0.05          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    55.75847928          
chi^2(alpha/2) =    26.5093032          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    10649938.23          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    22400602.37          
              
Thus, the confidence interval for the variance is              
              
(   10649938.23   ,   22400602.37   )
              
Also, for the standard deviation, getting the square root of the bounds,              
              
(   3263.42431   ,   4732.927463   ) [ANSWER]

We are 90% confident that the true standard deviation of the starting salary for graduates of a specific program is between 3263.42431 and 4732.927463. [ANSWER]

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c)

There is 90% probability that the true population parameter is inside our confidence interval.

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