Lecture 11 Nonparametric Statistics Introduction Outline of Today Rank Sum Test Order Statistics 11/20/2018 SA3202, Lecture 11

Introduction Problem of Interest Classical methods of statistical inference rely heavily on detailed distributional assumptions about the underlying population The validity of such procedures may be seriously affected when these assumptions are not satisfied. Nonparametric/Distribution-free Methods To overcome the above problem, methods of statistical analysis have been developed which do not require detailed assumptions about the underlying distribution of the data Distribution-free the methods do not require exact form of the population distribution Nonparametric the analysis is not in terms of the parameters of the underlying distribution. 11/20/2018 SA3202, Lecture 11

Example Suppose we want to compare the effectiveness of two teaching methods A and B. Suppose Use Method A, n1=3 children taught Method B, n2=4 children taught Their performances are assessed after a period of time. The null hypothesis is that H0: The performances of the children are not dependent on the teaching methods That is, essentially, there is no difference between these two methods. A parametric testing procedure: Two-sample t-test A nonparametric testing procedure: Wilcoxon-Mann-Whitney Test 11/20/2018 SA3202, Lecture 11

The Wilcoxon-Mann-Whitney Rank Sum Test Procedure We rank the performances of all the 7 children as 1 (the best) 2 3 …….. 7 (the worst) and record the ranks of the children in Group A. 11/20/2018 SA3202, Lecture 11

Key Idea Under H0, the ranks of the children in Group A are simply a random sample from the numbers 1,2,…7. Thus the rank sum statistic W=Sum of ranks of the children in Group A can be used to test H0 since when Method A is better than B, W will be “too small” (the smallest value is ) when Method A is worse than B, W will be “too large”. (the largest value is ) Thus Reject H0 when W is “too small” or “too large” To conduct the test, we need the null distribution of W. 11/20/2018 SA3202, Lecture 11

The Null Distribution of the Rank Sum Test Basic Fact Under H0, the ranks of the children in Group A form a random sample of size 3 from 1,2,…7 (without replacement). There are 35 such samples, all equally likely. The Sampling Distribution: W Ranks of Children in Group A Probability 6 (1,2,3) 1/35=.0286 7 (1,2,4) 1/35 8 (1,2,5), (1,3,4) 2/35 9 10 11 12 13 14 15 16 17 (4,6,7) 1/35 18 (5,6,7) 1/35=.0286 11/20/2018 SA3202, Lecture 11

Based on this table, we can conduct the Rank Sum Test Based on this table, we can conduct the Rank Sum Test. For example, for the 5% significance test, we use the critical region W<=6 or W>=18. The exact size of the test is .0286+.0286=.0572. 11/20/2018 SA3202, Lecture 11

Remarks The rank sum test is a typical of a wide class of distribution free tests. Some features to note are Simplicity of ideas and assumptions: only classical probability theory, based on the notion of equal-likelihood has been used. Simplicity of Calculations: the calculations needed to obtain the null distribution of the test statistic are very simple in nature. (For large sample sizes, calculation is done using a computer) Also tables of critical values are available for many tests. Applicability to rank data. Note that in the above example, all we needed was the ranks of the children. This is a useful feature of many distribution-free tests, that can be applied to ordinal (rank) data. Drawback: low power ( the capability to detect the violation of the null hypothesis is low) 11/20/2018 SA3202, Lecture 11

Order Statistics Definition Let X1, X2, … Xn be a sample from a continuous distribution with cdf F and pdf f. The order statistics X(1), X(2), …., X(n) are defined as follows: X(1)= the smallest of X1, X2, … Xn X(2)=the second smallest of X1, X2, … Xn ………………………………….. X(n) = the largest of X1, X2, …. Xn For example, for a sample of size 4 x1=9, x2=3, x3=1, x4=5 The order statistics are 11/20/2018 SA3202, Lecture 11

Distribution of the Order Statistics Let Gr(x) and gr(x) denote the cdf and pdf of X(r ) respectively. Then Gn(x)=Pr(X(n)<x) G1(x)=Pr(X(1)<x) 11/20/2018 SA3202, Lecture 11

General Cases Note that we have the following important relationship: X(r ) <=x equivalent to #{Xi| Xi<=x}>= r That is, X(r )<=x if and only if at least r observations are less than or equal to x. Thus Gr(x)=Pr(X(r )<=x)=Pr(U>= r), U=# {Xi | Xi<= x} 11/20/2018 SA3202, Lecture 11

Now, regarding the event “ the observation is less than or equal to x” as “Success”, it follows that U has a Binomial distribution with success probability p=Pr(Xi<=x)=F(x) Thus Gr(x)=Pr(U>=r)= 11/20/2018 SA3202, Lecture 11

The pdf gr(x) of X(r ) is then obtained by differentiating the cdf Gr(x): 11/20/2018 SA3202, Lecture 11