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Chi-squared analysis and budgeting weekly repair costs

by J Flanders
Tags: analysis, budgeting, chisquared, costs, repair, weekly
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J Flanders
#1
Jan15-08, 04:24 PM
P: 6
* I posted this in the Coursework section, but I wasn't sure if it would be answered there. *
Here's my question:

A plant supervisor is interested in budgeting weekly repair costs for a certain type of machine. Records over the past years indicate that these repair costs have an exponential distribution with mean 20 for each machine studied. Let Y1, Y2, Y3, Y4, Y5 denote the repair costs for five of these machines for the next week. Find a number c such that P(Y1 + Y2 + Y3 + Y4 + Y5 > c) = 0.05, assuming that the machines operate independently.

I was given in the previous problem that if Y has an exponential distribution with mean X, U = 2Y/X has a chi-squared distribution with 2 degrees of freedom.

I'm not quite sure what to do here. I think I solve for Y to get Y = UX/2, which means Y1, Y2, Y3, Y4, and Y5 each are independent chi-squared distributed random variables, each with 20 degrees of freedom. Then Y1 + Y2 + Y3 + Y4 + Y5 has a chi-squared distribution with (20)(5) = 100 degrees of freedom. Then I look at a chi-squared table for 100 d.f. and alpha = 0.05. Let c = 124.342.

Is this right and/or make sense?
Thanks for any help.
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EnumaElish
#2
Jan15-08, 05:09 PM
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You know each Y is distributed exponentially; you don't need to solve for it.

The questions are:

1. whether Y1 + Y2 + Y3 + Y4 + Y5 = 5Y

2. whether you can extrapolate the scaling factor given in your book from k = 2 to k = 5.

And the answers are:

1. Each of Yi (i = 1, ..., 5) represents a distinct random draw from a distribution. However, 5Y represents a single draw multiplied by 5. 5Y is scaling a random variable with a factor of 5, and does not represent 5 distinct draws made from the distribution.

2. Let m be the mean of each Y. Exponential distribution is a special case of the gamma distribution. Exp(1/m) = Gamma(1, m). Gamma is scalable: if X is Gamma(1, m) then kX is Gamma(1, km). So (2/m)X is Gamma(1, (2/m)m) = Gamma(1, 2) = Exp(1/2). And it so happens that Exp(1/2) is ChiSq(2).

So there is a very specific chain of relationships that connect U = 2Y/m to Chi-Squared, and I am not sure that it will hold for V = 5Y/m.

Your best bet is to use the property: "if Z = Y1 + Y2 + Y3 + Y4 + Y5 then Z is Gamma(5, m)" to solve for c in P(Z > c) = 0.05 using the Gamma distribution.
J Flanders
#3
Jan16-08, 01:08 PM
P: 6
I cannot find any tables of Gamma distributions. How would I find Gamma(5, 20)? Is there a way to turn it into a chi-squared distribution?

Thanks for the help from before.

EnumaElish
#4
Jan17-08, 11:32 AM
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Chi-squared analysis and budgeting weekly repair costs

ChiSq[k] happens to be equal to Gamma[k/2](2), where [] is the degrees of freedom and () is the remaining set of parameters. So Gamma[5](2) = ChiSq[10]. Moreover if Z ~ Gamma[5](20) then Z/10 ~ Gamma[5](2) = ChiSq[10]. If you calculate the sum Y1 + ... + Y5 then divide it by 10, you can use the ChiSq[10] table to calculate the applicable probability.

To verify this, you can numerically calculate the probabilities using the formulas here: http://met-www.cit.cornell.edu/reports/RR_91-2.html or purchase their book.


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