1. Lecture9 non- parametric methods
Xiaojin YU
Department of Epi. And Biostatistics, School of public health，Southeast university

2. Review: Type of data qualitative data (categorical data)
(1) binary (dichotomous, binomial)
(2) multinomial (polytomous)
(3) ordinal
quantitative data
Firstly, we have a review of last class, in that class, we have introduced type of data,
There are two type, one is

3. Measures of central tendency- quantitative data
Mean: Normal distribution
Geometric mean: positively skew and data can be transferred into normal distribution by log scale.
Median: used by all data, in general, is often used to abnormal data.

4. Measures of Dispersion- quantitative data
Range,
Interquartile range ,
Variance and standard deviation ,
coefficient of variation

5. Compare means by t-testν
type
conditions H0
Single sample t-test
,S,n,μ0
n-1
μ=μ0
Paired t-test
,Sd,n
μd=0
np-1
Two group t-test
, s1,n1, ,
s2,n2
Assumption:
normality
Equality of variance
n1+n2-2

6. Comparison of Means between two groups
2018/11/18

8. Compare proportion by Chi-square test
Drugs
Effect of drug
total
Sample rates
effective
Not effective
Drug A 41 4 45
91.1
Drug B 24 11 35
68.6 65 15 70
Are the 2 population proportions equal or not?
How categorical variables are distributed among 2 population?

9. solution H0： πA = πB H1: πA≠πB, α=0.05
Calculate T and Test Statistic, Chi-square
A ( T )
41 (36.56)
4 (8.44)
24 (28.44)
11 (6.56)
Drug
positive
negative
R total A T T n1 B T T n2
C total m1 m2 n

10. Test Statistic
A ( T )
41 (36.56)
4 (8.44)
24 (28.44)
11 (6.56)

11. Conclusion
Since 6.573>3.84, P<0.05, we reject H0,accept H1 at 0.05 level, so We conclude that the two populations are not homogeneous with respected to effect of drug. The effects of drug A and drug B are not equivalent.

12. Compare Ordinal data

13. OUTLINE Basic logic of rank based methods
Rank sum test for 2 independent group (Completely random design)
Sign rank test for Paired design
Rank sum test for 3 or more independent group (Completely random design)
Multiple Comparison

14. Rank & Rank Sum Review of median Example of duration in Hospital:
month
rank

15. Task to you
How to compare boys are taller or girl are taller no measuring is allowed?

16. Solution to the task
Blue-male
Red- female 16

17. The locations of small value are in front(small rank), and great value are in the post(great rank).

18. Part I: Wilcoxon Rank Sum Test
Rank Sum Test for Comparing the Locations of Two Populations
Mann-Whitney test
review t-test for comparing 2 population means
Normality and homogeneity

19. The height of 3 and 4 is same, so their rank will be 3.5
2018/11/18

20. EXAMPLE 1: Table 9.1 Survival Times of Cats & Rabbits without oxygen14 35 12 30 11 28
9.5 25 8 23 20 50
6.5 21 19 49 18 48 5 17 46 4 16 3 15 44 2 13 34 1
rank
minutes
rabbits
Cats

21. STEP I: Test Hypothesis and sig. level
H0：M1=M2 population locations of survival time of both cat and rabbit are equal
H1： M1 ≠ M2 population locations of survival time of both cat and rabbit are not equal ；
a = 0.05

22. STEP II: Statistic Assign Ranks
9.5 25 19 49
T2=82.5
n2=12
T1=127.5
n1=8 20 50 8 23 18 48
6.5 21 17 46 16 5 15 44 4 14 35 3 13 34 2 12 30 1 11 28
rank
time
Pooled sample
To pool n1 +n2 observations to form a single sample
rank all observations of the pooled sample from smallest to largest in column 2 and 4
Mid-ranks are used by tied values

23. STEP II: Statistic Test statistic T
Calculate the rank sums for the two samples respectively, denotes by T1 and T2.
Take the Ti with small n as T.
n1=8<n2=12, so T= T1 =127.5.
Sum(T1 ,T2)=N(N+1)/2=210

24. STEP III: Determine P Value, conclusion
From table in appendix E， by n1=8,n2-n1=4， we have the critical interval of Tα (58-110)
Since T=127.5, is beyond of Tα, so,P≤α。Given α=0.05, P<0.05;
H0 is rejected, it concludes that the survival times of cats and rabbits in the environment without oxygen might be different.
Cat will survive for longer time without oxygen.

25. BASIC LOGIC
N=N1+N2
GIVEN N, the total rank sum is fixed and can be calculated .
If H0 is true, the total rank sum should be assigned between 2 groups with weight of ni.

26. Normal Approximation
n1>10 or n2-n1 >10
Correction of ties

27. EXAMPLE 2: Table 9-2 Results From a Clinic Trial for Hypertension
10292
7663 ~
189
120 69
total
5827.5
4252.5
157.5 64 37 27
2(healed)
1384.5
2662.5
106.5
88-125 38 13 25
1 effect.
3080
748 44
1-87 87 70 17
0 ineffect.
DrugB
DrugA
Average rank
Range of rank
Drug B
Rank sum
effect
TA=7663 n=69;
TB=10293,n=120

28. Part II: Wilcoxon’s Signed Rank Test
Wilcoxon(1945) H0: Md=0
Example:
A test procedure the data on 28 patients data(14 pairs) from a sequential analysis double blind clinical trial for cancer of the head and neck will be used. (Bakowski MT, etc. Int. J. Radiation Oncology Biology Physics 1978 ,4 : )

29. Wilcoxon’s Signed Rank Test often used
1) quantitative data---t-test for pairs design
the difference of pairs must be normal, if its distribution is skew then must used Signed Rank Test.
Qualitative data--- pairs design
ordinal

30. Example 9.3 2 treatment groups : radiotherapy + drug (B)
radiotherapy + placebo (A)
The tumor response within three months of completion of treatment was assessed for each patient in terms of complete regression (CR), partial regression (PR), no change (NC) and progression of the disease (P).
Scored from 1 to 5 as follows:
5 = CR with no recurrence subsequently up to 6 months ore, 4 = CR initially but with a subsequent recurrence within 6 months,
3 = PR,
2 = NC,
1 = P.

31. 2)

32. STEP I: Test Hypothesis
H0 ： Md=0 population Median of differences is equal to zero;
H1 ： Md≠0 population Median of differences is equal to zero;
α=0.05

33. STEP II: statistic Assigning Rank
1) Calculate the difference di=xi-yi, and ignore all the pairs with zero differences.
2) Rank the absolute values of non-zero dis from the smallest to the largest such that each di gets a rank; if there is a tie, what will we do?

34. Ties: These six patients all have differences of 1 and therefore the rank numbers 1, 2, 3, 4, 5 and 6 must be divided amongst them.
That is, they all have a rank of
( )/6 =3.5
3) Assign the initial signs of dis to their ranks

35. Test Statistics T valid number of pairs n=10;
Find the sum of the ranks with positive signs and denote by T+;
Find the sum of the ranks with negative signs and denote by T-;
Sum(T+ ,T-)=n(n+1)/2=55
Let T=min(T+ ,T-) or anyone。
T-=48, T+=7

36. Step 3) Determine the P value & Conclude A Conclusion
n<25，find the critical value range Tα in table 10.3 (P184) .
n=10，T=48 or 7，in this example, given the value of α=0.05, find the critical value T0.05 is (8~47)，T is not in the interval，P<0.05, H0 is not rejected。It can not conclude that the results from two different between 2 treatments.

37. Normal Approximation
When n> 25，the table 10.3 can’t help. Then we turn to the normal approximation.
In fact it can be proved that if H0 is true, when n is large enough, the distribution of statistic T will close to a normal distribution with

38. Correction of Continuity
If there is tie, the statistic is

39. Part III: Kruskal-Wallis Test
Similar to one-way ANOVA /chi-square test
Used to test location of more than 2 populations

40. Example 9.4
Allocate 24 person randomly to 1 of 3 groups: no exercise; 20 minutes of jogging per day; or 60 minutes of jogging per day. At the end of a month, ask each participant to rate how depressed they now feel, on a Likert scale that runs from 1 ("totally miserable") through to 100 (ecstatically happy"). Question:Does physical exercise alleviate depression?

41. Report on depression from 3 groups and ranksRi

42. Test Hypothesis H0: M1=M2=…=Mk
3 populations have the same population location
H1: M1,M2,…Mk are not all equal : 3 populations have different population location , At least one of the populations has a median different from the others.
a = 0.05

43. Test Statistic -H Let N=n1+n2+n3
Ri the sum of the ranks associated with the ith sample, like 76.5,79.5,144
The average rank is (N+1)/2
The sample average rank for ith sample is Ri/ni
12/{N(N+1)}standard the test statistic in terms of the overall sample size N.

44. Solution to Example There are k-1=2 degree of freedom in this example.
R1=76.5 n1=8 R2=79.5 n2=8 R3=144 n3=8
There are k-1=2 degree of freedom in this example.

45. Adjusted Formulae for Tied
the number of individuals within the j-th tied subgroup

46. CRITICAL VALUE Table 11 H-critical values C2 –Critical Values
when n is big enough， H is distributed as c2 distribution approximately with n = k – 1

47. Conclusion k=3，，the critical value is 5.99 .
Since 7.27>5.99,the P<0.05. we reject H0. that is, there is evidence that at least one of the groups is different from others.

48. NONPARAMETRIC test
Nonparametric: That are not focused on testing hypothesis about the parameters of the population.
Distribution-free: make no assumptions about the distribution of the data; and are suitable for small sample sizes or large samples where parametric assumptions are violated
– Use ranks of the data values rather than actual data values themselves
– Loss of power when parametric test is appropriate 48 48

49. Parametric and non-parametric equivalents

50. Loss of power when parametric test is appropriate
NONPARAMETRIC test
Advantages
More different types of data
Numerical Data with unknown distribution or skewed distribution
Ordinal variable or the measurement data that are given with rank only
Disadvantage
A waste of data
Loss of power when parametric test is appropriate 50 50

51. Learning Objectives
1. Understand when nonparametric statistical methods are appropriate.
2. Know how to perform the Wilcoxon Signed-Rank Test and when it should be used.
3. Know how to perform the Wilcoxon Rank-Sum Test and when it should be used.

52. THANK YOU FOR YOUR ATTENTION!