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In: Statistics and Probability

The personnel department of a large corporation wants to estimate the family dental expenses of its...

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 97% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation

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Expert Solution

Solution:

Given a sample of size n = 12

115, 370, 250, 593, 540, 225, 177, 425, 318, 182, 275,228

x x2
115 13225
370 136900
250 62500
593 351649
540 291600
225 50625
177 31329
425 180625
318 101124
182 33124
275 75625
228 51984
Sum 3698 1380310

Sample variance s2 =

= [1/(12 - 1)][1380310 - (36982/12) ]

= 21882.696969697

s2 = 21882.696969697

Our aim is to construct 97% confidence interval for the population standard deviation

c = 97% = 0.97

= 1 - 0.97 = 0.03

/ 2 = 0.015 and 1 - ( / 2) = 0.985

Now , n = 12

d.f. = n - 1 = 12 - 1 = 11

Now , using chi square table ,

/2,df = 0.015,11 = 23.503

1- /2,df = 0.985,11 = 3.363

Now ,

The 97% confidence interval for 2 is,

( 12 - 1) *21882.696969697/ 23.503< 2 < ( 12 -1 ) * 21882.696969697/ 3.363

10241.6570934 < 2 <  71575.874715

Now , the confidence interval for the population standard deviation is given by taking square root of the confidence interval for 2

10241.6570934 < <  71575.874715

101.20 < <  267.54

orchestra answered 1 year ago
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