1. Nonparametrics and goodness-of-fit
Tron Anders Moger

2. Nonparametric statistics
In tests we have done so far, the null hypothesis has always been a stochastic model with a few parameters.
T tests
Tests for regression coefficients
Test for autocorrelation …
In nonparametric tests, the null hypothesis is not a parametric distribution, rather a much larger class of possible distributions

3. Parametric methods we’ve seen:
Estimation
Confidence interval for µ
Confidence interval for µ1 -µ2
Testing
One sample T test
Two sample T test
The methods are based on the assumption of normally distributed data (or normally distributed mean)

4. Nonparametric statistics
The null hypothesis is for example that the median of the distribution is zero
A test statistic can be formulated, so that
it has a known distribution under this hypothesis
it has more extreme values under alternative hypotheses

5. Non-parametric methods:
Estimation
Confidence interval for the median
Testing
Paired data: Sign test and Wilcoxon signed rank test
Two independent samples: Mann-Whitney test/Wilcoxon rank-sum test
Make (almost) no assumptions regarding distribution

6. Confidence interval for the median, example
Betaendorphine concentrations (pmol/l) in 11 individuals who collapsed during a half-marathon (in increasing order):
Find that median is 107.6
What is the 95% confidence interval?

7. Confidence interval for the median in SPSS:
Use ratio statistics, which is meant for the ratio between two variables. Make a variable that has value 1 for all data (unit)
Analyze->Descriptive statistics->Ratios
Numerator: betae Denominator: unit
Click Statistics and under Central tendency check Median and Confidence Intervals

8. SPSS-output:
95% confidence interval for the median is (71.2, 177.0):

9. The sign test
Assume the null hypothesis is that the median of the distribution is zero.
Given a sample from the distribution, there should be roughly the same number of positive and negative values.
More precisely, number of positive values should follow a binomial distribution with probability 0.5.
When the sample is large, the binomial distribution can be approximated with a normal distribution
Sign test assumes independent observations, no assumptions about the distribution
SPSS: Analyze->nonparametric tests->two related samples. Choose sign under test type

10. Example from lecture 5: Energy intake kJ
SUBJECT PREMENST POSTMENS
Number of cases read: Number of cases listed: 11
Want to test if
energy intake
is different
before and after
menstruation.
H0: Median difference=0
H1: Median difference≠0 +

11. Using the sign test:
The number of positive differences (+’s) should follow a binomial distribution with P=0.5 if H0: median difference between premenst and postmenst is 0
What is the probability of observing 11 +’s?
Let P(x) count the number of + signs
P(x) is Bin(n=11,P=0.5)
The p-value of the test is the probability of observing 11 +’s or something more extreme if H0 is true, hence
However, because of two-sided test, the p-value is *2=0.001
Clear rejection of H0 on significance-level 0.05

12. In SPSS:
Recall: Found p-value=0.000 using t-test, so p-value from sign-test is a bit larger
This will always be the case
P-value

13. Wilcoxon signed rank test
Takes into account the strength of the differences, not just if they are positive or negative
Here, the null hypothesis is a symmetric distribution with zero median. Do as follows:
Rank all values by their absolute values, from smallest to largest
Let T+ be the sum of ranks of the positive values, and T-corresponding for negative values
Let T be the minimum of T+ and T-
Under the null hypothesis, T has a known distribution (p.874).
For large samples, the distribution can be approximated with a normal distribution
Assumes independent observations and symmetrical distributions
SPSS: Analyze->nonparametric tests->two related samples. Choose Wilcoxon under test type

14. The energy intake example:
Difference Rank (+) Rank(-)
premenst-postmenst
1350, No
1250, negative
1755, differences!
1020,00 3
745,00 2
1835,00 11
1540,
725,00 1
1330,00 5
1435,00 7
Rank sum: T+:66 T-:0
Min(66,0)=0
Test H0: Median difference =0 vs H1: Median difference≠0 with α=0.05
From p.874, find T=11 for n=11 and α/2=0.025
Reject H0
Also, see that p-value is smaller than 0.01

15. In SPSS: Sum of ranks P-value based on normal approximation
P-value based on Wilcoxon
distribution

16. Sign test and Wilcoxon signed rank test
Often used on paired data.
E.g. we want to compare primary health care costs for the patient in two countries: A number of people having lived in both countries are asked about the difference in costs per year. Use this data in test.
In the previous example, if we assume all patients attach the same meaning to the valuations, we could use Wilcoxon signed rank test on the differences in valuations

17. Wilcoxon rank sum test (or the Mann-Whitney U test)
Here, we do NOT have paired data, but rather n1 values from group 1 and n2 values from group 2.
Assumptions: Independence within and between groups, same distributions in both groups
We want to test whether the values in the groups are samples from different distributions:
Rank all values as if they were from the same group
Let T be the sum of the ranks of the values from group 1.
Under the assumption that the values come from the same distribution, the distribution of T is known.
The expectation and variance under the null hypothesis are simple functions of n1 and n2.

18. Wilcoxon rank sum test (or the Mann-Whitney U test)
For large samples, we can use a normal approximation for the distribution of T.
The Mann-Whitney U test gives exactly the same results, but uses slightly different test statistic.
In SPSS:
Analyze->nonparametric tests-> two independent samples tests
Test type: Mann-Whitney U

19. Example We have observed values are the groups different?
If we assume that the values in the groups are normally distributed, we can solve this using the T-test.
Otherwise we can try the normal approximation to the Wilcoxon rank sum test
Test H0: Groups come from same distribution vs
H1: Groups come from different distributions
on 5% significance level

20. Example (cont.)
Rank
Group 1
Group 2
Wilcoxon T: 16 1
1.3 2
1.5 3
2.1 4
3.2 5
3.4 6
4.3 7
4.9 8
6.3 9
7.1
E(T)=
n1(n1+n2+1)/2
=25
SE(T)=
√n1n2(n1+n2+1)/12
=4.08
Z=(T-E(T))/SE(T)
=(16-25)/4.08
=-2.20
p-value: 0.032
Reject H0
Rank sum:

21. Number of cases read: 22 Number of cases listed: 22
ID GROUP ENERGY
....
Number of cases read: Number of cases listed: 22
Example from lecture 4:
Energy expenditure
in two groups,
lean and obese.
Want to test if they
come from different distributions.

22. Wilcoxon Rank Sum Test in SPSS:
Sum of ranks
Z-value
P-value based on normal approx.
Exact P-value
Exactly the same result as we got when using the T distribution!

23. Spearman rank correlation
This can be applied when you have two observations per item, and you want to test whether the observations are related.
Computing the sample correlation (Pearson’s r) gives an indication.
We can test whether the Pearson correlation could be zero but test needs assumption of normality.

24. Spearman rank correlation
The Spearman rank correlation tests for association without any assumption on the association:
Rank the X-values, and rank the Y-values.
Compute ordinary sample correlation of the ranks: This is called the Spearman rank correlation.
Under the null hypothesis that X values and Y values are independent, it has a fixed, tabulated distribution (depending on number of observations)

25. Concluding remarks on nonparametric statistics
Tests with much more general null hypotheses, and so fewer assumptions
Often a good choice when normality of the data cannot be assumed
If you reject the null hypothesis with a nonparametric test, it is a robust conclusion
However, with small amounts of data, you can often not get significant conclusions
In SPSS, you only get p-values from non-parametric tests, more interesting to know the mean difference and confidence intervals

26. Some useful comments:
Always start by checking whether it is appropriate to use T tests (normally distributed data)
If in doubt, a useful tip is to compare p-values from t-tests with p-values from non-parametric tests
If very different results, stay with the p-values from the non-parametric tests

27. Contingency tables
The following data type is frequent: Each object (person, case,…) can be in one of two or more categories. The data is the count of number of objects in each category.
Often, you measure several categories for each object. The resulting counts can then be put in a contingency table.

28. Testing if probabilities are as specified
Example: Have n objects been placed in K groups, each with probability 1/K?
Expected count in group i:
Observed count in group i:
Test statistic:
Test statistic has approx. distribution with K-1 degrees of freedom under null hypothesis.

29. The Chi-square distribution
The Chi-square distribution with n degrees of freedom is denoted
It is equal to the sum of the squares of n independent random variables with standard normal distributions.

30. Testing association in a contingency table
Treatment A B C
TOTAL 1 23 14 19
R1=56 2 7 10
R2=31 3 9 5 54
R3=68
C1=46
C2=26
C3=83
155 G R O U P
n values in total

31. Testing association in a contingency table
If the assignment of observations to the two categories is independent, the expected count in a cell can be computed from the marginal counts:
Actual observed count:
Test statistic:
Under null hypothesis, it has distribution with (r-1)(c-1) degrees of freedom
In SPSS: Analyze->Descriptives->Crosstabs

32. Example from Lecture 4:
Spontanous abortions among nurses helping with operations and other nurses
Want to test if there is any association between abortions and type of nurse
Op.nurses
Other nurses
No. interviewed 67 92
No. births 36 34
No. abortions 10 3
Percent abortions
27.8
8.8

33. Example: Abortions No abortions Total Op.nurses 10 26 36 Other nurses31 34 13 57 70
Expected cell values: Abortion/op.nurses: 13*36/70=6.7
Abortion/other nurses: 13*34/70=6.3
No abortion/op.nurses: 57*36/70=29.3
No abortion/other nurses: 57*34/70=27.7

34. We get: This has a chi-square distribution with (2-1)*(2-1)=1 d.f.
Want to test H0: No association between abortions and type of nurse at 5%-level
Find from table 7, p. 869, that the 95%-percentile is 3.84
This gives you a two-sided test!
Reject H0: No association
Same result as the test for different proportions in Lecture 4!

35. In SPSS: Check Expected under Cells, Chi-square under statistics, and
Display clustered bar charts!

36. Goodness-of-fit tests
Sometimes, the null hypothesis is that the data comes from some parametric distribution, values are then categorized into K groups, and we have the counts from the categories.
To test this hypothesis:
Estimate the m parameters in the distribution from data.
Compute expected counts.
Compute the test statistic used for contingency tables.
This will now have a chi-squared distribution with K-m-1 d.f. under the null hypothesis.

37. Poisson example from lecture 3:
No. of No. Expected no. observed
cases observed (from Poisson dist.)
Calculation: P(X=2)=0.9362*e-0.936/2!=0.17
Multiply by the number of weeks: 0.17*47=8.1
Poisson distribution fits data fairly well!

38. Test if the data can come from a Poisson distribution:
Test statistic: approx. with 6-1-1=4 d.f.
Calculate:
From table 7, p. 869, see that 95%-percentile of chi-square ditribution with 4 d.f. is 9.49
Reject H0: The data come from a Poisson distribution

39. Tests for normality
Visual methods, like histograms and normality plots, are very useful.
In addition, several statistical tests for normality exist:
Kolmogorov-Smirnov test (which I have showed you before)
Bowman-Shelton test (tests skewness and kurtosis)
However, with enough data, visual methods are more useful

40. Next time: Seen tests for comparing means in two independent groups
What if you have more than two groups?
Analysis of variance