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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 in 2004 reveals the following dental expenses (in dollars): 115, 370, 250, 93, 540, 225, 177, 425, 318, 182, 275, and 228. The sample mean is ___________. Construct a 95% confidence interval estimate of the mean family dental expenses for all employees of this corporation. The upper boundary/limit is _________ and the lower boundary/limit is ________. (keep two decimal points).

Solutions

Expert Solution

Solution:

x x2
115 13225
370 136900
250 62500
93 8649
540 291600
225 50625
177 31329
425 180625
318 101124
182 33124
275 75625
228 51984
x=3198 x2=1037310

The sample mean is

Mean     = (x / n) )

= (115,+370+ 250+ 93,+540+ 225+ 177+ 425+ 318+ 182+ 275+228 / 12)

= 3198 / 12

= 266.5

Mean = 266.5

The sample standard is S

  S = ( x2 ) - (( x)2 / n ) n -1

= (1037310( (3198 )2 / 12 ) 11

   = (1037310 - 852267 / 11)

= (185043/ 11)

= 16822.0909

= 129.7

The sample standard is = 129.7

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,11 = 2.201

Margin of error = E = t/2,df * (s /n)

= 2.201 * (129.7 / 12)

= 82.41

Margin of error = 82.41

The 95% confidence interval estimate of the population mean is,

- E < < + E

266.5 - 82.4 1< <266.5 + 82.41

184.09 < < 348.91

The upper boundary/limit = 348.91

The lower boundary/limit =184.09

milcah answered 2 years ago
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