s340text-KStestOnly

Exact magnitude if we were interested in further

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Unformatted text preview: ere interested in further improving our process since a number closer to 0:10 may result in the need for additional sampling/experimentation. 3.2 Kolmogorov-Smirnov Test If we wanted to compare a theoretical distribution F to the empirical CDF from ^ ^ the data F , then we will use this test. If F and F “appear” close, then we will assume our data follows from F . For this to work we will need to have a working knowledge of how several CDFs. ^ Step 1 We determine how far F and F are from each other. We call this, d, distance the K-S Distance. Step 2 We ask if d is “large” We compare d to the distances we’ get by com. d paring F to the empirical CDF and if the distance is relatively small then we would expect that we got data that did have distribution F . If our p-value is small, we reject F as the distribution. 3.2.1 Building an Empirical CDF 1 First we assume that every observation has weight n where n is the number of observations. The the empirical CDF is de…ned as: n 1X ^ F (x) = I (xi n i=1 6 x) This translates to the number of observations observed at values less than or equal to x. Take the example of a random set of data: f1; 1; 2; 5g. This has the ^ following ECDF. [If the picture doesn’ work it’ …le: ksexample1ecdf] F can t s be built for discrete and continuous data. We assume for the continuous data that P(xi = a; xj = a)=0, for i 6= j . Put simply, to create an Empirical CDF: 1. Order the data. 2. Assume we point is equiprobable. 3. Make a step function or equal to x. 3.2.2 Ix n where Ix is the number of data points less than The K-S Distance At every step of our ECDF we will calculate a two distances: lim F (a) and lim F (a) a%x a&x ^ F (x) ^ F (x). The is the “up”and “down”distances. The K-S distance d the maxfdi;j g. Example. We have received the data set f1; 4; 9; 16g, and we suspect that it follows an exponential distribution. 7 8 Step 1 Build the empirical cdf Empirical CDF 0 . 1 8 . 0 6 . 0 x) ( F 4 . 0 2 . 0 0 . 0 -5 0 10 5 15 20 X Figure 3.1: Theoretical vs. Empirical cdfs Since we believe that this follows an exponential distribution, we can estimate its parameter: ^= 1 = 2 x 15 Now, we can look at the comparison between the theoretical and empirical CDF probabilities: F (1) = 1 e ^ F (1) = 1 4 2 15 F (4) = 1 e ^ F (4) = 1 2 8 15 F (9) = 1 e 3 ^ F (9) = 4 18 5 F (16) = 1 e ^ F (16) = 1 32 15 Let di;U represent the “up” distance and di;D the “down” distance for observation i. Then, for this set of data, we have the following: d1;U = 0:1252 d2;U = 0:0866 d3;U = 0:0512 d4;U = 0:1184 d1;D = 0:1248 d2;D = 0:1634 d3;D = 0:1980 d4;D = 0:1316 Thus, our KS Distance is d = 0:1980. Step 2 Create data n data sets derived from F : At this point, we will create n sets of ECDFs from F and denote them as ^ ^ Fi (i = 1; ::; n). Then, we calculate the K-S Distance between F and each Fi . 9 Empirical CDF 0 . 1 d4,U d4,D 8 . 0 d3,U 6 . 0 d3,D x) ( F d2,U 4 . 0 d2,D 2 . 0 d1,U d1,D 0 . 0 -5 0 5 10 15 20 X Figure 3.2: Theoretical vs. Empirical cdfs with K-S Distances Finally, we consider the fr...
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