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

2. 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

3. 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

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

5. 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

6. 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 such samples, all equally likely.
The Sampling Distribution:
W Ranks of Children in Group A Probability
(1,2,3) /35=.0286
(1,2,4) /35
(1,2,5), (1,3,4) /35 9 10 11 12 13 14 15 16
(4,6,7) /35
(5,6,7) /35=.0286
11/20/2018
SA3202, Lecture 11

7. 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 =.0572.
11/20/2018
SA3202, Lecture 11

8. 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

9. 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

10. 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

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

12. 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

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