Lab Number Six: Non-Parametric Tests :

The Mann-Whitney U-Test, Kruskal Wallis Test, Kolmogorov-Smirnov goodness of fit test, Kolmogorov-Smirnov 2 sample test

The Mann-Whitney U-test is a popular non-parametric test that allows for the statistical comparison of differences between two independent sets of data that are, at the very least, strongly ordinal. Rather than assuming that both samples are drawn from populations with normal distributions, it simply tests the null hypothesis that these two groups are drawn from populations with identical distributions. In other words, no assumptions are made about the nature of the distributions.

Computational Formulae:

U₁ = n₁.n₂ + n₁(n₁ + 1)/2 -GR₁

U₂ = n₁.n₂ + n₂(n₂ + 1)/2 -GR₂

where U is the test statistic,

where n₁ and n₂ are sample sizes,

where R₁ and R₂ are calculated ranks.

If ranks are tied use the average of the ranks.

Example

A map reading test, which yields only ordinal level data is given to 8 randomly selected men and eight randomly selected women. Do men differ from women in terms of map reading? Test with  a= 0.05.

Find the null and the alternate hypothesis

H₀: there is no difference in male and female map reading

H₁: there is a difference in male and female map reading

Calculate U_A and U_B

U_A= N1 N2+ N1(N1+1)/2 - R1

U_A= 8x8+ 8(8+1)/2 - 62

U_A= 64+ 36 - 62

U_A=38

U_B = N₁ N₂ - U_A

U_B = 64-38= 26

 U_B is the lower value and is therefore compared to the critical value from the tables which for N1=8 and N2=8 is 13. As the calculated value is greater than this H₀ can not be rejected.

adapted from: 

http://www.unn.ac.uk/academic/ss/psychology/resource/py071/ass2/stats/ExampleU/ExampleofMannWhitneyUtest.htm

INSTRUCTIONS (10 marks)

1. To be calculated manually:

The 1986 asset levels of the top nine banks from the United Kingdom, and the top eight banks from France, are listed below.

Please note that the values of the figures presented are rounded to the nearest billion and are expressed in US dollars.

British Banks (Top 9)

70 24 123 13 47 14 117 78 39

French Banks (Top 8)

41 116 40 26 156 133 42 143

You are asked to compare these two groups, by employing the Mann-Whitney U-test. Please answer the following (remember to show all work):

a. State the null (H_o) and the research (H₁) hypotheses.

b. Using both the formal and informal methods, calculate BY HAND the value of U. (Hint: compare the Test Statistic to the computational value of U).

c. At the 95% confidence level, what is the critical value of U?

d. Are we dealing with a one- or two-tailed test? Why? Do we accept or reject the null hypothesis?

2. To be calculated using SPSS for Windows: (10 marks)

Similarly, these are the assets of the top 20 Japanese and German banks, in $US billions. Again, compare these two groups and answer the following questions:

German Banks

19 73 133 20 58 76 18 49

32 102 77 15 13 12 10 63

56 17 16 14

Japanese Banks

162 240 191 131 212 205 116 138

78 111 204 81 161 114 73 125

101 124 115 100

a. State the null and the research hypotheses.

b. Using the informal method, calculate the value of U. How does your answer compare to the SPSS for Windows output? Explain any differences.

c. What is the critical value of U? Assume " = 0.05.

d. Are we dealing with a one- or two-tailed test? Why? Do we accept or reject the null hypothesis?

Non-Parametric Tests Part II The Kruskal-Wallis H-Test for K-Samples

Kruskal Wallis Test

Unlike the Mann-Whitney U-Test, the Kruskal-Wallis H-test is designed to provide a non-parametric alternative to one-way analysis of variance for more than two sample sets. The assumptions underlying the Kruskal-Wallis H test are similar to those of the Mann-Whitney U-test. Thus, it is normally applied to rank-order (ordinal) data, and it does not assume that the samples being tested are derived from normal distributions. It is important to remember that when samples of greater than 5 individuals are examined, the Kruskal-Wallis H-test uses the same critical values as the chi-square distribution. In determining the degrees of freedom for a given test, simply employ the k-1 correction; i.e. the number of samples minus 1.

Computational Formulae:

H = 12/N(N+1) . E(R_i²/n_i) - 3(N+1)

1 - E(t³-t)/(n³-n) Tie Correction (for the purposes of this lab use the average of the tied ranks if required)

Example

Four different methods of growing corn were randomly assigned to a large number of plots of land, and the yield per acre was computed for each plot, as given in the table.

 

 Yield per acre X_ij for four treatments

Method 1 Method 2 Method 3 Method 4  83  91 101  78  91  90 100  82  94  81  91  81  89  83  93  77  89  84  96  79  96  83  95  81  91  88  94  80  92  91    81  90  89        84

The hypotheses may be stated as follows:

• H_0: The four methods are equivalent.
• H₁: Some methods of growing corn tend to give higher yields than others.

The observations are ranked from the smallest (77) of rank 1 to the largest (101) of rank N=34. 

Tied values are given average ranks. 

The ranks of the observations, and the sums R_i are given below

Yields X_ij for and assigned ranks R_ij

Method 1 Ranks 1 Method 2 Ranks 2 Method 3 Ranks 3 Method 4 Ranks 4  83.0  11.0  91.0  23.0 101.0  34.0  78.0   2.0  91.0  23.0  90.0  19.5 100.0  33.0  82.0   9.0  94.0  28.5  81.0   6.5  91.0  23.0  81.0   6.5  89.0  17.0  83.0  11.0  93.0  27.0  77.0   1.0  89.0  17.0  84.0  13.5  96.0  31.5  79.0   3.0  96.0  31.5  83.0  11.0  95.0  30.0  81.0   6.5  91.0  23.0  88.0  15.0  94.0  28.5  80.0   4.0  92.0  26.0  91.0  23.0      81.0   6.5 90.0  19.5  89.0  17.0              84.0  13.5                         ni   9    10     7     8 G(Ri)2 38612.3   23409.0   42849   1482.3 G/ni 4290.3   2340.9   6121.3   185.3

H=12/34(35)*12937.7 - 3(35) = 21.7

making use of the fact that the test is P² distributed with df=number of samples-1 when there are more than 5 items per sample the critical value is 7.815

we accept H₁

For two samples, the Kruskal-Wallis test is equivalent to the Mann-Whitney test.
data table and top of second table from: http://www.stams.strath.ac.uk/~bob/classes/323/notes/whole/node91.html

INSTRUCTIONS  (10 marks)

1. To be calculated manually (remember to show all work):

In 1988, Statistics Canada recognized 27 census metropolitan areas across the nation. Among other criterion, a minimum population of 100,000 is required for inclusion as a Canadian CMA. The following data lists the 27 CMA's in no particular order, and provides both the population of the CMA as of June 1st, 1988, and the per capita income (PCI) of each CMA in Canadian dollars.

1. Organize the 27 CMA's into three distinctive groups, based on the following criteria:

a. CMA's with populations of between 100,000 and 249,999.

b. CMA's with populations of between 250,000 and 499,999.

c. CMA's with populations in excess of 500,000 inhabitants.

2. State the null and research hypotheses.

3. Determine the ranks of the CMA's.

4. Calculate the value of H by using the provided formula.

5. Do we accept or reject the null hypothesis? Why?

II. To be completed using SPSS for Windows: (10 marks)

1. Input the same data in SPSS for Windows and account for the differences (if any) between the hand calculated H-value and the one provided by SPSS for windows.

2. Based on the answer provided by SPSS for Windows, would you accept or reject the null hypothesis at alpha = .05 ?

3. Be sure to hand in your output from SPSS for Windows write your name on your copy.

Kolmogorov-Smirnov One-Sample Test

It is a more powerful alternative to chi-square goodness-of-fit tests when its assumptions are met. Whereas the chi-square test of goodness-of-fit tests whether in general the observed distribution is not significantly different from the hypothesized one, the K-S test tests whether this is so even for the most deviant values of the criterion variable. Thus it is a more stringent test.

There is some disagreement among statisticians whether the K-S one sample test applies only to continuous distributions, not discrete ones such as Poisson. Most sources I consulted say a comparison with Poisson is allowed.

The assumptions of the K-S test are as follows:

1. Randomness - sample members must be randomly drawn from the population.
   
2. Independence - Sample scores must be independent of each other.
   
3. Measurement Scale - the categories must be ordinal or higher.

Using K-S to compare to a poisson distribution

Example

H₀: the observed distribution is poisson distributed

H₁: the observed distribution in not poisson distributed

Number of collisions between passenger trains on British rail, 1970 to 1983

Problem (20 marks)

1. To be calculated manually (remember to show all work):

You may wish to use your calculations from lab 5 for the poisson probabilities.

Lightening strikes per day were recorded in Alberta for 6 summers (July and August).

Calculate using a weighted mean

Calculate the probability P(X) for the number of strikes per day using the Poisson formula:

Now determine if the pattern of lightning strikes is consistent with a Poisson distribution.

Introduction:  Kolmogorov-Smirnov Two-Sample Test

When can you use this test?

2) Whenever Mann-Whitney U, or the Kruskal-Wallis test could be used.

Its different because Mann-Whitney and Kruskal-Wallis use ranks, Kolmogorov-Smirnov-type tests are based on cumulative distributions.

Example

In the table below we have the average air temperatures in F at Nottingham Castle for 1920 and 1921

H₀:  there is no difference in mean monthly temperature between 1920 and 1921

H₁: there is a difference in mean month temperature between 1920 and 1921

we accept H₀

Problem (20 marks)

Below is a comparison of murder rates for 10 cities in the US for 1960 and 1970

Determine if there is any difference between 1960 and 1970.

Step 1: H_o: There is no difference in the mean murder rate for 1960 and 1970.

H₁: There is a difference in the mean murder rate for 1960 and 1970.

Now complete the test.

Problem: (20 marks)

Collect some data of interest and perform one of the test covered in this exercise.