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A consumer advocacy group is concerned about the variability in the cost of prescription medication. The group surveys eight
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Answer #1

a )

Sample variance ( s2 ) is best point estimate of population variance ( \sigma ^2 )

We have to find sample variance.

Using Excel , =VAR( )

ww 24.5 22.5 36.5 10 =VAR(A1:A3) 11

Variance = s2 = 37.41071

b)

We have to find 90% confidence interval for population variance.

Formula :

(n-1)s 2 xổ 2,1-1 (n − 1) s- Xí-a/2,n=1

Where ,

n is sample size = 8 , Variance = s2 = 37.41071

Xa/2.n-1 is right tail chi-square critical value at given confidence level.

Confidence level = 90% = 0.9  

Significance level = \alpha = 1 - 0.9 = 0.1 , \alpha /2 = 0.05

So, right tail critical value is ,

Xa/2.n-1 = ΧΟ, 05,7 = 14.0671

{ Using Excel function , =CHIINV(0.05,7 ) = 14.0671 }

Xi-a/2.n-1 is left tail critical value.

Xi-a/2.n-1= xổ 45 = 2.16735

{ Using Excel function , =CHIINV(0.95,7 ) = 2.1673 }
  

So, 90% confidence interval of population variance is ,

(8 - 1)37.4107 14.0671 <o (8 - 137.4107 <! 2.1673

18.62 <σ < 120.83

Confidence interval = 18.62 to 120.83

c)

Te group decides to begin a lobbying effort on it's members' behalf if the variance in the price does not equal 6

From part b , it is observed that limits of  confidence interval is greater than 6

So, the group should begin on its lobbying effort.

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