Question

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.

a.Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation.

b.
Construct a 95% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation.

c. Construct a 99% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation

Answer 1

a). Given :

a =
`$0.1$`,

chisquare with the
`$0.05$`,

`$df=11$` is
`$4.57$` (from the chisquare table)

Chisquare with the
`$0.95$`, df =
`$11$` is
`$19.68$` (from chisquare table)

So 90% Cl is

`$\left((v(n-1)* s)/(v19.68), (v(n-1)* s)/(v4.57)\right)$`

`$=(√(11)* 147.928)/(√(19.68)),(√(11)* 147.928)/(√(4.57))$`

`$=(110.5947,229.5031)$`

b). Given :

a =
`$0.05$`

Chisquare with
`$0.025$`

df =
`$n-1=11$` is A,
`$3.82$`

Chisquare with
`$0.975$`, df = 11 is
`$21.92$`

So, 95% Cl is

`$\left((v(n-1)* s)/(v21.92), (v(n-1)* s)/(v3.82)\right)$`

`$=(√(11)* 147.928)/(√(21.92)),(√(11)* 147.928)/(√(3.82))$`

`$=(104.7916,251.0239)$`

c). Given a =
`$0.01$`, chisquare with
`$0.005$`, df =
`$11$` is
`$2.6$`

Chisquare with
`$0.995$`, df =
`$11$` is
`$26.76$`

So 99% CI is

`$\left((v(n-1)* s)/(v26.76), (v(n-1)* s)/(v2.6)\right)$`

`$=(√(11)* 147.928)/(√(26.76)),(√(11)* 147.928)/(√(2.6))$`

`$=(94.84265,304.2706)$`