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    Distribution and confidence interval

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

    © BrainMass Inc. brainmass.com October 7, 2022, 7:03 pm ad1c9bdddf
    https://brainmass.com/statistics/confidence-interval/distribution-confidence-interval-166236

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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.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 7, 2022, 7:03 pm ad1c9bdddf>
    https://brainmass.com/statistics/confidence-interval/distribution-confidence-interval-166236

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