1. NONPARAMETRIC STATISTICS
CHAPTER 15
NONPARAMETRIC STATISTICS

2. Learning Objectives
Determine situations where nonparametric procedures are better alternatives to the parametric tests
Understand the assumptions of nonparametric tests
Use one- and two-sample nonparametric tests
Use nonparametric alternatives to the single-factor ANOVA

3. Nonparametric vs. Parametric
Used an assumption that we are working with random samples from normal populations
Called parametric methods
Based on a particular parametric family of distributions
Describe procedures called nonparametric methods
Make no assumptions about the population distribution other than that it is continuous

4. Why Nonparametric Procedures
Distributions are not close to normal
Data need not be quantitative but can be categorical (such as yes or no, defective or non defective) or rank data
Are usually very quick and easy to perform
Provides considerable improvement over the normal-theory parametric methods
Not utilize all the information provided by the sample
Requirement of a larger sample size

5. Which One? Which one to choose?
If both methods are applicable to a particular problem
Use the more efficient parametric procedure
Otherwise, use the non parametric procedure

6. SIGN TEST
Used to test hypotheses about the median of a continuous distribution
Mean of a normal distribution equals the median
Sign test can be used to test hypotheses about the mean of a normal distribution
Used the t-test in Chapter 9
Sign test is appropriate for samples from any continuous distribution
Counterpart of the t-test

7. Description of the Test
Use the following differences
Xi is ith the sample observation and is the specified median value
Number of plus signs is a value of a binomial random variable that has the parameter p=1/2
Reject the if the proportion of plus signs is significantly different from 1/2

8. Using P-value Use the P-value If r+ < n/2 the P-value
If the P-value is less than the significance level , we will reject H0 and conclude that H1 is true

9. The Normal Approximation
Binomial distribution has well approximately a normal distribution when n >10 and p=0.5
Mean=np and the variance=np(1-p)
Test statistics
Critical region can be chosen from the table of the standard normal distribution

10. Sign Test for Paired Samples
Applied to paired observations drawn from two continuous populations
Define the paired difference as
Test the hypothesis that the two populations have a common median
Equivalent to
Done by applying the sign test to the n observed differences

11. Example
Ten samples were taken from a plating bath used in an electronics manufacturing process, and the bath pH was determined.
The sample pH values are 7.91, 7.85, 6.82, 8.01, 7.46, 6.95, 7.05, 7.35, 7.25, 7.42
Manufacturing engineering believes that pH has a median value of 7.0. Do the sample data indicate that this statement is correct? Use the sign test with =0.05 to investigate this hypothesis. Find the P-value for this test

12. Calculate the differences
Use the general procedure covered in Chapter 8
Parameter of interest is the median of the distribution of pH
The
=0.05

13. Data and the observed plus signs
Solution - Cont
Data and the observed plus signs i xi
xi-7
Sign 1
7.91
+ 0.91 + 2
7.85
+ 0.85 3
6.82
- 0.18 - 4
8.01
+ 1.01 5
7.46
+ 0.46 6
6.95
- 0.05 7
7.05
+ 0.05 8
7.35
+ 0.35 9
7.25
+ 0.25 10
7.42
+ 0.42
5. Test statistic is the observed number of plus differences r+=8
6. Reject H0 if the P-value corresponding to r=8 is less than or equal to = 0.05

14. Solution-Cont.
7. Since r >n/2=5, we calculate the P-value by using the binomial formula with n=10 and p=0.5
Hence, the P-value = 2P(R+8|p=0.5)
Since P=0.109 is not less than = 0.05, we cannot reject the null hypothesis
8. Observed number of plus signs r = 8 was not large or enough to indicate that median pH is different from 7.0

15. Using Table Table of critical values for the sign test
Appendix Table VII is for two-sided and one-sided alternative hypothesis
Let R=min (R+, R-)
Reject H0
If r-≤ critical value; if (>) used for H1
If r+≤ critical value; if (<) used for H1
If r≤ critical value; if (≠) used for H1

16. Wilcoxon Signed-rank Test
Sign test uses only the plus and minus signs of the differences
Does not take into consideration the size or magnitude of these differences
Uses both direction (sign) and magnitude
In case of symmetric and continuous distributions
Test H0 as µ=µ0

17. Description of the Test
Compute the following quantities
Xi- 0
Xi is ith the sample observation i and 0 is the specified median or mean value
Rank the absolute differences in ascending order
Give the ranks the signs
W+ be the sum of the positive ranks and W- be the sum of the negative ranks, and let W min(W+,W- )
Table VIII contains critical values of W
Reject H0
If w-≤ critical value; if (>) used for H1
If w+≤ critical value; if (<) used for H1
If w≤ critical value; if (≠) used for H1

18. Large-Sample Approximation
Large sample size (n>20)
has approximately a normal distribution
Mean and variance
Test statistics
Appropriate critical region can be chosen from a table of the standard normal distribution

19. Paired Observations
Applied to paired observations drawn from two continuous and symmetric populations
Define the paired difference as
Test the hypothesis that the two populations have a common mean
Equivalent to testing that the mean of the differences

20. Description of the Test
Differences are first ranked in ascending order of their absolute values
Ranks are given the signs of the differences
Ties are assigned average ranks
W+ be the sum of the positive ranks and W- be the sum of the negative ranks, and let W min(W+,W- )
Table VIII contains critical values of W
Reject H0
If w-≤ critical value; if (>) used for H1
If w+≤ critical value; if (<) used for H1
If w≤ critical value; if (≠) used for H1

21. Example
Consider the data in the previous example and assume that the distribution of pH is symmetric and continuous.
Use the Wilcoxon signed-rank test with =0.05 to test the following hypothesis H0: µ=7 vs. H1: µ≠7

22. Solution 1. Parameter of interest is the mean of the pH 2. H0: µ=7
4. α=0.05
5. Test statistic
w=min (w+, w-)
6. Reject H0 if w<w*0.05=8 from Table VIII

23. Solution – Cont. 7. Signed rankxi
xi-7
Signed Rank 1
7.05
+ 0.05
+ 1.5 2
6.95
-0.05
- 1.5 3
6.82
- 0.18 4
7.25
+ 0.25
+ 4 5
7.35
+ 0.35
+ 5 6
7.42
+ 0.42
+ 6 7
7.46
+ 0.46
+ 7 8
7.85
+ 0.85
+ 8 9
7.91
+ 0.91
+ 9 10
8.01
+1.01
+ 10
Determine the minimum value of the following
w+ = ( )= 50.5
w – = ( ) = 4.5
Test statistic is w = min (50.5,4.5)

24. Solution-Cont. 8. Since w=4.5 is less than the critical value w0.05 =8
Reject the null hypothesis

25. WILCOXON RANK-SUM TEST
Statistical inference for two samples
Wilcox on rank-sum test is a non parametric
alternative
Two independent continuous populations X1 and X2 with means 1 and 2
Wish to test the following hypotheses
n1 and n2 are sample size

26. Description of the Test
Arrange all n1+n2 observations in ascending order of magnitude and assign ranks to them
Ties are assigned average rank
W1 be the sum of the ranks in the smaller sample (1), and define W2 to be the sum of the ranks in the other sample
Also can be found
Table IX contains the critical value of the rank sums for two significance levels
Reject H0
If w2 ≤ critical value; if (>) used for H1
If w1 ≤ critical value; if (<) used for H1
If either w1 or w2 ≤ critical value; if (≠) used for H1

27. Large-Sample Approximation
When both n1 and n2 are moderately large
Distribution of w1 can be well approximated by the normal distribution with the following mean and variance
Test statistic
Appropriate critical region can be chosen from the table

28. Kruskal-Wallis Test
Recall the single-factor analysis of variance model
Error terms ij were with mean zero and variance
Kruskal-Wallis test is a nonparametric alternative
Error terms ij are assumed to be from the same continuous distribution

29. Description of the Test
Compute the total number of observations
Rank all N observations from smallest to largest
Assign the smallest observation rank 1, the next smallest rank 2, , and the largest observation rank N
Rij be the rank of observation Yij
Ri. denote the total and the. average of the ni ranks

30. Test Statistic Calculate
H has approximately a chi-square distribution with a-1 degrees of freedom
Reject H0 if the observed value h is greater than the critical value, or
Critical region can be chosen from the Chi-square distribution table depending on whether the test is a two-tailed, upper-tail, or lower-tail test

31. Ties in the Kruskal-Wallis Test
Observations are tied, assign an average rank
use the following test statistic
ni is the number of observations in the ith treatment
N is the total number of observations
S2 is just the variance of the ranks

32. Example 15-7
Montgomery (2001) presented data from an experiment in which five different levels of cotton content in a synthetic fiber were tested to determine whether cotton content has any effect on fiber tensile strength. The sample data and ranks from this experiment are shown in following Table
Does cotton percentage affect breaking strength? Use α=0.01

33. Rank all observations from smallest to largest
Solution
Rank all observations from smallest to largest
Cotton % 7 9 10
Rank 1 2 3 4 5 11 12 6 8 14 15 17 18 13 19 16 20 22 23 25 21 24
Assign average rank ( )/3 = 2
Perform the same calculations for the other tied observations

34. Solution-Cont. Data and Ranks for the Tensile Testing Experiment
There is a fairly large number of ties
Use the equation that was defined for the tied observations

35. Solution-Cont. Thus Test statistic
Since h> 13.28, we would reject the null hypothesis
Conclude that treatments differ
Same conclusion is given by the usual analysis of variance

36. Next Agenda Introduces statistical quality control
Fundamentals of statistical process control