The personnel department of a large corporation wants to est

The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees reveals the following family dental expenses (in dollars): 115, 370, 250, 593, 540, 225, 177, 425, 318, 182, 275, and 228. Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation. Place your LOWER limit, in dollars rounded to 1 decimal place, in the first blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 123.4 would be a legitimate entry. Place your UPPER limit, in dollars rounded to 1 decimal place, in the second blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 567.8 would be a legitimate entry.

Solution

standard deviation = 147.928 (based on the data)

The degree of freedom =n-1=12-1=11

Given a=0.1, chisquare with 0.05 and df=11 is 4.57 (from chisquare table)

chisquare with 0.95 and df=11 is 19.68 (from chisquare table)

So 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation is

(sqrt(n-1)*s/sqrt(19.68), sqrt(n-1)*s/sqrt(4.57))

--> (sqrt(11)*147.928/sqrt(19.68), sqrt(11)*147.928/sqrt(4.57))

--> (110.6, 229.5)

So LOWER limit = 110.6

UPPER limit = 229.5