Question

Suppose you have the following aircraft part failure time data (in thousands of hours).

Part XRR-012 failure times:

10.03 11.61 6.13 12.24 42.10 6.66 38.59 22.09 33.34 23.56 17.82 10.47 20.80 16.51 4.86 23.64* 40.50* 45.07* 41.57* 80.48* 12.48* 21.93* 45.03* 10.75* 7.76* 8.18* 38.20* 28.28* 18.55* 24.44* Where the asterisk (*) indicates the part is still active on an aircraft (that is, the part is in current use and has not failed yet).

Describe how you would fit a gamma distribution to the first 15 data points (first 3 rows of data).

Mean of data set Var of data set Mean divided with var gives one of the parameter β β obtain the other parameter α

The sum of the first 15 obs is 276.75

Mean 276.75/15 = 18.45 Var 1931.1/15-1 = 137.94

Parameter 18.45/137.94 which is roughly 0.1338

Mean = α/ β 18.45= α/0.1338 α = 2.46861 α = 2.4686

Probability density function of Gamma distribution is as follows f(x)=〖0.1338〗^2.4686/(⎾2.4686) x^((2.4686-1) ) e^((-0.1338 x)); α>0,β>0

Your manager asks you to fit a gamma distribution to all of the data. Describe how you might do this.

The mean for 30 obs is 723.6.3/30 = 24.1223 Var 8125.726/30-1 = 280.1974

Parameter 24.1223/280.1974 which is roughly 0.0861

Mean is also 24.1233= α/0.08609 = 2.0767

Gamma distribution: f(x)=〖0.0861〗^2.0767/(⎾2.0767) x^((2.0767-1) ) e^((-0.0861x)); α>0,β>0

Perform task (b) and estimate the mean part lifetime using a gamma distribution. Show your work and state any assumptions.

The mean for 30 obs is 723.6.3/30 = 24.1223

Suppose a colleague has fit a Weibull distribution to all of the data. Your manager asks you to determine which fit is better, your gamma fit from part (b) or the Weibull fit of your colleague. Explain how you might make this decision.(HELP WITH THIS QUESTION)

Answer 1