Lecture06 - Applied Nonparametrics STA 4502/5507 Yiyuan...

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Applied Nonparametrics STA 4502/5507 Yiyuan She Department of Statistics, Florida State University Fall 2009
Possible Course Flow I Estimating success probabilities I Single location: estimates, tests, intervals I Two locations: testing, estimating differences between locations I Scale comparisons I Multiple locations and factors I Independence I Nonparametric regression I Other topics ...
Multiple Location I Compared two population centers via locations (medians) in chapter 4 I Now, compare multiple locations I Parametric equivalent is analysis of variance (ANOVA) I ANOVA assumes means exist, variances exist, data follows particular distribution
Problem Setup I k populations (treatments) I k - 1 treatments and 1 control I In first case, looking for differences in locations I Second case compares treatments to control I Also want to compare all k treatments to each other
Assumptions I N = k j =1 n j , n j observations from j th treatment I All N observations are independent I X 1 , j , X 2 , j , . . . , X n j , j are data from treatment j , following continuous distribution F j I F j ( t ) = F ( t - τ j ) , t ( -∞ , ) , j = 1 , 2 , . . . , k where F is a continuous distribution function with unknown median θ I τ j is the treatment effect for population j
X 1 , 1 X 1 , 2 X 1 , 3 · · · X 1 , k X 2 , 1 X 2 , 2 X 2 , 3 · · · X 2 , k X 3 , 1 X 3 , 2 X 3 , 3 · · · X 3 , k X 4 , 2 X 4 , 3 · · · X 4 , k X 5 , 2 · · · X 5 , k X 6 , 2 · · · X 6 , k . . . . . . X n 2 , 2 . . . X n k , k
Assumptions I This corresponds to the parametric one-way ANOVA X i , j = θ + τ j + ε i , j , i = 1 , 2 , . . . , n j , j = 1 , 2 , . . . , k I θ is overall median I τ j is the treatment j effect I Errors ε i , j are iid with median 0 from continuous distribution I If errors are normally distributed, then medians = means = 0, constant variance
Kruskal - Wallis Test I No difference among treatment effects τ j H 0 : τ 1 = τ 2 = · · · = τ k or, H 0 : F 1 = F 2 = · · · = F k = F I Only difference is in medians, all have same scale
Kruskal - Wallis Test I Alternative H 1 : Not H 0 I At least one treatment effect is different, τ 1 , · · · , τ k not all equal
Kruskal - Wallis Test I Order the N combined sample values X i , j I Get ranks r i , j I For each j , set R j = n j X i =1 r i , j and R · j = R j / n j I R · j is the average rank for sample from treatment j
Kruskal - Wallis Test I Test statistic: H = 12 N ( N + 1) k X j =1 n j R · j - N + 1 2 2 or, H = 12 N ( N + 1) k X j =1 R 2 j n j - 3( N + 1) I Second slightly easier if doing this by hand I Reject H 0 if H h α I Exact null distribution available. See Table A.12 ( k = 3 , 4 , 5)
Kruskal - Wallis Test I Why N +1

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