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Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Unformatted text preview: Stat 312: Lecture 24 Goodness-of-fit Test for Distributions Moo K. Chung mchung@stat.wisc.edu April 29, 2003 Concepts 1. Knowing the exact distribution of a sample is important for statistical analysis. We want to test if a sample x1 , · · · , xn follows a certain probability distribution P . We perform a χ 2 goodness-of-fit test on the null hypothesis H0 : P (X ∈ Ci ) = pi for all i = 1, · · · , k For a continuous distribution, Ci are intervals. In-class problems Example 1. Let us see if the distribution of output tuft weight (Example 14.11.) follows an exponential distribution with parameter λ. Interval 0-8 8-16 16-24 24-32 32-40 40-48 48-56 56-64 64-70 Observed # 20 8 7 1 2 1 0 1 0 1 Solution. The density of the exponential random variable X is given by f (x) = λ e−x/λ , x ≥ 0. Note . 8 1 ˆ that 0 λ e−x/λ = 0.5. Solving this, we estimate λ = 11.54. Based on this, we compute the expected number of elements in each category. For instance, the expected number of element in the interval [8, 16] 8 1 is 40 0 11.54 e−x/11.54 dx = 40(pexp(16, 1/11.54) − pexp(8, 1/11.54)). > interval<-c(8,16,24,32,40,48,56,64,70) > p<- pexp(interval,1/r) [1] 0.50 0.75 0.87 0.94 0.97 0.98 0.99 0.996 0.998 > p[2:9]-p[1:8] [1] 0.25 0.12 0.06 0.03 0.016 0.008 0.004 0.002 > 40*(p[2:9]-p[1:8]) 10 5 2.5 1.3 0.6 0.3 0.2 0.06 Expected # 20 10 5 2.5 1.3 0.6 0.3 0.2 0.06 Then we compute χ2 = 50 and the P -value is pchisq(50, 7) = 0. So we reject H 0 and conclude that the exponential distribution is not a good fit. Self-study problems Example 14.11. This is slightly different from the in-class example setting. Exercise 12.14. Assignment 8 Due May 8, 11:00am. 14.6., 14.16., 14.18., 14.26., 14.32. ...
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This note was uploaded on 01/31/2008 for the course STAT 312 taught by Professor Chung during the Spring '04 term at Wisconsin.

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