Question

Practice: Compute the Kolmogorov-Smirnov Test Statistic 1 point possible (graded) Let X1, ..., Xn be iid samples with cdf F, and let F° denote the cdf of Unif(0,1). Recall that F° (t) = t. 1(t € [0,1]) +1.1(t > 1). We want to use goodness of fit testing to determine whether or not X1,...,x, iid Unif(0,1). To do so, we will test between the hypotheses = 70 H : F(t) H :F(t) + F. To make computation of the test statistic easier, let us first reorder the samples from smallest to largest, so that X(1) X(2) <... <X(n) is the reordered sample. In this set-up, the Kolmogorov-Smirnov test statistic is given by the formula i - 1 TR == n max max X01 (X4) € (0,1)||* – X91(X(c) € (0,1))}. i=1,..., n You observe the data set x consisting of 5 samples: x=0.8,0.7,0.4, 0.7,0.2 Using the formula above, what is the value of T5 for this data set? (You are encouraged to use computational tools.)

Answer 1

Let X2, X2, --- Xn be lid samples with cof F, and let F° denote the c.dif. of unif (0.I). Recall that Fº(t) = t:i(de(0,13 + 1-1 (71) We want to use goodness of fit testing to determine whether X, X2, ..., Xn a unif(0, 1). To do so, we will test between the hypothesis or hot Ho: FCT) – FOS H: F(t) & Fo first reorder To make Computation of the test Statistic easier, let us the samples from smallest to largest, so that X , ² X cest... LXcno is the reordered somble. In this set-up, the Kolmogorov - Smirnor test statistic is given by the formula Tr In max 1=1,2,.in { marec 1971. - Yes I CXjelo_1]], [4 - Xer, t (%, C019))} You observe the data set X consisting of samples X = 0.8, 0.7.0.4, 0.7.0.2 we reorder using the formula above, what is the value of To for this data set first the data set in the smallest to largest order, then, R 0.220.420.750.720.8 Xex) = 0.2 , X (2) = 0.4, X (3) = 0.7, X14, = 0.7, X 6,70.8 55 max (1 2 - 1 - Xw (Xoj € (0,1)|, 17 To max jin 4(3,101) d = 12,5 Ce Scanned with CamScanner

Now, we Consider max 1-1 (1 (Xa je 10,1), | 5 - You) 1. (Xevine € (0.1))) 1 ** Put i= l in equation (X* -0.2 max Xul & Xces (Xuve Lo,1)) (141 - Xw t (Xu, € (0,17)],11 max(10-0121.15-0.41)= max (1-0-21, lol) - max (0-2,0) = 0.2 Put i= 2 in equation (** & X (2) X (2) = 0.4 max max mand 1 2 1 16x9 € (0,1]][, I X62, 1. ( Yenje [0,171) 1x(13-0-4), 13-0-41) = max(1-0-2), 101) – max (0-2,0) = 0.2 Put i=3 in equation (** (1334 - Xco) 1 (Xcg, €60,17)), 113. - Yo, HlXcg, f[0,17)|). & Xc3, = 0.7 - marl 19-0-41.13 -0-71)= max(1-0-3),|-0-41)= max(0-3,0-1) = 0.3 - (1 - X (Xa, La -17) x Ca, Co,1721) mox(13-0-71,14-0-71)= mox ( 1-0-10,10-11) - max(0-1,0-1) = 0-1 – 0 i= 5 in equation (xx max( 15.1 - You, I Crow, € (0,1)|| 5 - Xog) ? (X 5,9€ (0,1]) 1 ) & Xoy = 0.8 3 Put (= 4 in equation (**) Max & X(a) = 0.7 = max Put & XD mood19-08 , 11-0-01) – mor(101, 10-21)= mor(0, 0-2) = 0.8 - 6 Scanned with CamScanner

Therefore, Ty=55 max max CU I - Xan 1 (Xaj C10,1)), ) - You (Xerj€(0,1)))} L-1.93...,5 = 15 max i=1,2,3,45 { 0.2, 0.1, 0.3, 0.1, 0.2 a} (using equation & ho J5 x 0.3 11 > Т. 5 0.355 T5 = 0.6708 Round off 4 decimal place Therefore Kolmogorov - Smistnou test statistic data 18 for the given T 5 0.355 0.6708 Ang Scanned with CamScanner