STAT4290 April 18th 2011 Notes

STAT4290 April 18th 2011 Notes - ≤ =-limn Gxn limn PD xn...

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Testing the form of the distribution Two-sided Ho: F(x) =  Fo(x)     for all x  R Ha:  F(x)   Fo(x)    for at least one x D =  sup |fo(x) – S(x)| (greatest vertical distance between Fo(x) and S(x)) One-sided Ho: F(x) = >  Fo(x)   for all x  R Ha:  F(x)  < Fo(x)                  for at least one x D + = sup |fo(x) – S(x)| (greatest vertical distance by Fo(x) above   S(x)) Or Ho: F(x) <= Fo(x)     for all x  R Ha:  F(x)  >  Fo(x)                 for at least one x D - = sup | S(x) – fo(x)  | <greatest vertical distance by  S(x) above Fo(x)> When F(x) is continuous, and under H0, the exact distribution of D+ and D- is give by                   Greatest integer less than or equal to n (1-x) = - = [ ( - )] ( - - ) - ( + ) - Gx 1 xj 0 n 1 x nj 1 x jn n j x jn j 1 As  → ∞, n →∞ = →∞
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Unformatted text preview: +≤ = - -limn Gxn limn PD xn 1 e 2x2 , ≤ = +≤ -≤ ≈[ ( )] So PD x PD x and D x G x 2 Exact quantiles – Table A13 Notes: they are exact only if F(x) is continuous. If it’s discrete, the quantiles lead to a conservative test – it’s going to be harder to reject Ho than it should be. But – do we want to reject Ho. For discrete F(x), we can calculate exact p-values One-sided test I. Let + Dobs denote the observed value of + D-Compute the probabilities , ≤ ≤ ( -+ fj 0 j n 1 Dobs ), by drawing a horizontal line directly on a graph of Fo(x), with ordinate -1 +- : Dobs jn-={ -+- fj 1 Dobs jn heigh to Foxat botton of jumpif the horizontal like doesn'tintersect Foxat a jumpif the horizontal line intersect Foxat a jump-Compute the constants e0, e1,…. As e0 = 1-<Look at the book…please>-II. ffff--...
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