Question
Times between successive crashes of a computer system were generated for a 6-month period and are given in increasing order as follows (time in hours):
1 10 20 30 40 52 63 70 80 90 100 102
130 140 190 210 266 310 530 590 640 1340
The parameter , mean
.
Use the Kolmogorov-Smirnov test to examine the goodness of fit of exponential distribution.
Answer :
Word Count : 878
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We are asked to test the goodness of fit of an exponential distribution using the Kolmogorov-Smirnov (K–S) test. Let's solve this step by step. Given data (in hours): 1, 10, 20, 30, 40, 52, 63, 70, 80, 90, 100, 102, 130, 140, 190, 210, 266, 310, 530, 590, 640, 1340 Number of observations: ( n = 22 ) Parameter: ( a = 0.00435 ), mean ( 1/\alpha = 230 \text{ hrs} ) --- Step 1: Exponential CDF The exponential cumulative distribution function is: [ F(t) = 1 - e^{-\alpha t} ] Here, ( \alpha = 1/230 \approx 0.00435 ), which matches the given ( a ). --- Step 2: Compute empirical CDF values The empirical CDF at the ( i^\text{th} ) data point is: [ F_{\text{emp}}(t_i) = \frac{i}{n}, \quad i = 1, 2, \dots, 22 ] So: [ F_{\text{emp}} = \frac{1}{22}, \frac{2}{22}, \dots, \frac{22}{22} ] Numerical values: [ F_{\text{emp}} = 0.0455, 0.091, 0.136, 0.182, 0.227, 0.273, 0.318, 0.364, 0.409, 0.455, 0.500, 0.545, 0.591, 0.636, 0.682, 0.727, 0.773, 0.818, 0.864, 0.909, 0.955, 1.0 ] --- Step _______ ________ ____ ___ _________ ___ ______.
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