# Distribution and confidence interval

Please see attached

(The material is from Confidence Intervals For Variances. Please show each step of your solution. Thank you.)

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attached file.

(The material is from Confidence Intervals For Variances. Please show each step of your solution. Thank you.)

Proof.

(a) By the hint, we are going to find the Moment generating function of M. By the definition, we have

Note that are mutually independent (as it is a sample ). So,

.................(1)

Now we are going to compute . It is

where f(x) is the p.d.f. of X, which follows exponential distribution with mean . So,

Now we use substitution y= . Then

, when t<1/2.

Hence, by (1), we have

Note that the Moment generating function of the chi-squared distribution with degree of freedom 2n is . So, we have therefore proved that W follows

(b) By part a), we know that W follows . Hence,

i.e.,

i.e.,

Hence, a confidence interval for is

(c) Given n=7, and . We have

, 6.571 and

. So, a 90% confidence interval for is

=[55.326, 199.422]

So, the endpoints of a 90% confidence interval for are 55.326 and 199.422.

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