Parametric and nonparametric statistical methods for the life sciences  Session I


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1 Why nonparametric methods What test to use? Rank Tests Parametric and nonparametric statistical methods for the life sciences  Session I Liesbeth Bruckers Geert Molenberghs Interuniversity Institute for Biostatistics and statistical Bioinformatics (IBiostat) Universiteit Hasselt June 7, 2011
2 Why nonparametric methods What test to use? Rank Tests Table of contents 1 Why nonparametric methods Introductory example Nonparametric test of hypotheses 2 What test to use? Two independent samples More then two independent samples Two dependent samples More then two dependent samples Ordered hypotheses 3 Rank Tests Wilcoxon Rank Sum Test KruskalWallis Test Friedmann Statistic Sign Test JonckheereTerpstra Test
3 Why nonparametric methods What test to use? Rank Tests Introductory example Nonparametric test of hypotheses Why nonparametric methods?
4 Why nonparametric methods What test to use? Rank Tests Introductory example Nonparametric test of hypotheses Introductory Example The paper Hypertension in Terminal Renal Failure, Observations Pre and Post Bilateral Nephrectomy (J. Chronic Diseases (1973): ) gave blood pressure readings for five terminal renal patients before and 2 months after surgery (removal of kidney). Patient Before surgery After surgery Question: Does the mean blood pressure before surgery exceed the mean blood pressure two months after surgery?
5 Why nonparametric methods What test to use? Rank Tests Introductory example Nonparametric test of hypotheses Classical Approach Paired ttest: Patient Before surgery After surgery Difference D i Hypotheses: H 0 : µ d = 0 versus H 1 : µ d > 0 µ d : mean difference in blood pressure TestStatistic : t = D 1 (Di D) n(n 1) 2 follows a t distribution with n 1 d.f.
6 Why nonparametric methods What test to use? Rank Tests Introductory example Nonparametric test of hypotheses Assumptions The statistic follows a tdistribution if the differences are normally distributed ttest = parametric method Observations are made independent: selection of a patient does not influence chance of any other patient for inclusion (Two sample t test): populations must have same variances Variables must be measured in an interval scale, to interpret the results These assumptions are often not tested, but accepted.
7 Why nonparametric methods What test to use? Rank Tests Introductory example Nonparametric test of hypotheses Normal probability plot Normality is questionable!
8 Why nonparametric methods What test to use? Rank Tests Introductory example Nonparametric test of hypotheses Nonparametric Test of Hypotheses Follow same general procedure as parametric tests: State null and alternative hypothesis Calculate the value of the appropriate test statistic (choice based on the design of the study) Decision rule: either reject or accept depending on the magnitude of the statistic P H0 (T c) =?? Exact distribution Approximation for the exact distribution
9 Why nonparametric methods What test to use? Rank Tests Two independent samples More then two independent samples When to use what test
10 Why nonparametric methods What test to use? Rank Tests Two independent samples More then two independent samples What test to use? Choice of appropriate test statistic depends on the design of the study: number of groups? independent of dependent samples? ordered alternative hypothesis?
11 Why nonparametric methods What test to use? Rank Tests Two independent samples More then two independent samples Two Independent Samples Permeability constants of the human chorioamnion (a placental membrane) for at term (x) and between 12 to 26 weeks gestational age (y) pregnancies are given in the table below. Investigate the alternative of interest that the permeability of the human chorioamnion for a term pregnancy is greater than for a 12 to 26 weeks of gestational age pregnancy. X (at term) Y (1226weeks) Statistical Methods: ttest Wilcoxon Rank Sum Test
12 Why nonparametric methods What test to use? Rank Tests Two independent samples More then two independent samples More Than Two Independent Samples Protoporphyrin levels were determined for three groups of people  a control group of normal workers, a group of alcoholics with sideroblasts in their bone marrow, and a group of alcoholics without sideroblasts. The data is shown below. Does the data suggest that normal workers and alcoholics with and without sideroblasts differ with respect to protoporphyrin level? Group Protoporphyrin level (mg) Normal Alcoholics with sideroblasts Alcoholics without sideroblasts Statistical Methods: ANOVA KruskalWallis Test
13 Why nonparametric methods What test to use? Rank Tests Two independent samples More then two independent samples Two Dependent Samples Twelve adult males were put on liquid diet in a weightreducing plan. Weights were recorded before and after the diet. The data are shown in the table below. Subject Before After Statistical Methods: Paired ttest Sign test; Signedrank test
14 Why nonparametric methods What test to use? Rank Tests Two independent samples More then two independent samples Randomized Blocked Design Effect of Hypnosis: Emotions of fear, happiness, depression and calmness were requested (in random order) from 8 subject during hypnosis Response: skin potential (in millivolts) Subject Fear Happiness Depression Calmness Statistical Methods: Mixed Models Friedmann test
15 Why nonparametric methods What test to use? Rank Tests Two independent samples More then two independent samples Ordered Treatments Patients were treated with a drug a four dose levels (100mg, 200mg, 300mg and 400mg) and then monitored for toxicity. Drug Toxicity Dose Mild Moderate Severe Drug Death 100mg mg mg mg Statistical Methods: Regression JonckheereTerpstra Test
16 Wilcoxon Rank Sum Test
17 Wilxocon Rank Sum Test Detailed Example: Data : GAF scores Control Treatment Does treatment improve the functioning?
18 Parametric Approach: ttest t = X 1 X 0 s S X1, where S X1 X X 0 = s2 0 n 0 1 n0 t test: means of two normally distributed populations are equal H 0 : µ 1 = µ 0 H 1 : µ 1 µ 0 (one sided test H 1 : µ 1 µ 0 equal sample sizes two distributions have the same variance X 1 = 34.00, X 0 = 23.33, S X1 = 7.21, S X0 = t = 1.27 P H0 (t 1.27) =
19 Wilxocon Rank Sum Test Detailed Example: Control Treatment Order data: Position of patients on treatment as compared with position of patients in control arm? Ranks
20 Treatment is effective if treated patients rank sufficiently high in the combined ranking of all patients Test statistic such that: treatment ranks are high value test statistic is high treatment ranks are low value test statistic is low W S = S 1 + S S n (n=3, number of patients in treatment arm) Ranks W S = =14 Control (25) (10) (35) Treatment (36) (26) (40)
21 Reject null hypothesis when W S is sufficiently large : W S c P H0 (W S c) = α (alpha=0.05) Distribution of W S under H 0? Suppose no treatment effect (H 0 ) rank is solely determined by patients health status rank is independent of receiving treatment or placebo rank is assigned to patient before randomisation Random selection of patients for treatment random selection of 3 ranks out of 6 Randomisation divides ranks (1,2,...6) into two groups! Number of possible combinations : ( ) N n = N! n!(n n)!
22 All posibilities: (each as a probability of 1/20 under H 0 ) treatment ranks (4,5,6) (3,5,6) (3,4,6) (3,4,5) (2,5,6) w s treatment ranks (2,4,6) (2,4,5) (2,3,6) (2,3,5) (2,3,4) w treatment ranks (1,5,6) (1,4,6) (1,4,5) (1,3,6) (1,3,5) w s treatment ranks (1,3,4) (1,2,6) (1,2,5) (1,2,4) (1,2,3) w s
23 Distribution of W S under the null hypothesis: w P H0 (W s = w)
24 P HO (W S 14) = 0.1 Do not reject H 0. Conclusion: Treatment does not increase the GAF scores. Power of this study???
25 Large Sample Sizecase ( N ) n increases rapidly with N and n ( 20 ( ) = ) = 924 Asymptotic Null Distribution: Central Limit Theorem Sum T of large number of independent random variables is approximately normally distributed. ( ) T E(T ) P a Φ(a) Var(T ) where Φ(a) is the area to the left of a under a standard normal curve
26 If both n and m are sufficiently large: W S N(E(W S ); Var(W S )) E(W S ) = 1 2n(N + 1) Var(W S ) = 1 12nm(N + 1)
27 KruskalWallis Test
28 Kruskal Wallis test Example: Kruskal Wallis test: The following data represent corn yields per acre from three different fields where different farming methods were used. Method 1 Method 2 Method Question: is the yields different for the 4 methods?
29 Parametric Approach Oneway ANOVA Statistical test of whether or not the means of several groups are all equal Assumptions: Independence of cases The distributions of the residuals are normal : ɛ i (0, σ 2 ). Homoscedasticity variance between groups F = = variance within groups MSTR MSE Statistic follows a F distribution with s 1, n s d.f.
30 Small F: Large F:
31 OneWay ANOVA results X 1 = 89, X 2 = 88.33, X 3 = 99 σ 1 = 3.56, σ 2 = 6.65, σ 3 = 4.08 MSTR= , MSE = F= 6.11 P H0 (F 6.11) =
32 Ranks: Method 1 Method 2 Method R i. :
33 Hypothesis : H 0 : No difference between the treatments H 1 : Any difference between the treatments If treatments do not differ widely (H 0 ): R i. are close to each other R i. close to R.. If treatments do differ (H 1 ): R i. differ substantial R i. not close to R..
34 Evaluate the null hypothesis by investigating: K = 12 N(N + 1) s n i (R i. R.. ) 2 i=1 P H0 (K c) =? Exact distribution of K under H 0 : ranks are determined before assignment to treatment random assignment all possibilities same chance of being observed Number ( of possible combinations: multinomial coefficient : 11 ( 4,3,4) = 11 )( )( 4) = ( ) ( N n 1,n 2,...,n s = N )( N n1 ) ( n 1 n 2... N n1... n s 1 ) n s
35 A few possible configurations: Method 1 Method 2 Method 3 K (1,2,3,4) (5,6,7) (8,9,10,11) 8.91 (1,2,3,5) (4,6,7) (8,9,10,11) 8.32 (1,2,3,6) (4,5,6) (8,9,10,11) 7.84 (1,2,3,7) (4,5,6) (8,9,10,11) 7,48... (1,3,5,6) (2,4,8) (7,9,10,11) Each configuration has a probability of to happen.
36 Exact Distribution of K: P H0 (K 6.16) = Conclusion: Reject H 0 : there is a difference between the farming methods Large sample size approximation χ 2 distribution with s 1 d.f.
37 Friedmann Test
38 Friedmann Statistic Setting 1: complete randomization: KruskalWallis test pvalue = Treatment effect is blurred by the variability between subjects Setting 2: randomisation within age groups: pvalue Conclusion reject H 0
39 Procedure Divide subjects in homogeneous subgroups (BLOCKS) Compare subjects within the blocks w.r.t. treatment effects (Generalisation of the paired comparison design)
40 Example Data Agegroup treatment y y y y A B C Rank subjects within a block: Agegroup treatment y y y y A B C
41 Mean of ranks for: treatment A = R A. = 10 4 = 2.5 treatment B = R B. = 6 4 = 1.5 treatment C = R C. = 9 4 = 2.25 If these mean ranks are different reject H 0 If these mean ranks are close accept H 0
42 Measure for closseness of the mean ranks: if the R i. are all close to each other then they are close to the overall mean R.. and (R i. R.. ) 2 will be close to zero Friedman Statistic Q = 12N s(s + 1) s (R i. R.. ) 2 i=1
43 P H0 (Q c) =? Exact distribution of Q under H 0 : A few possible configurations: Agegroup Q Treatment y y y y A B C A B C A B C A B C
44 Exact Distribution of Q: Q Pr E E E E E02
45 Number of possibilities for the rank combinations: agegroup year: 3! = 6 agegroups are independent total number of possible combinations: (3!) 4 = 1296 Under the null these are all equally likely : (s!) N, s= treatment groups, N = of blocks P H0 (Q 3.5) = Do not reject H 0
46 Sign Test
47 Sign Test Special case of Friedmann test: blocks of size 2 subjects matched on e.g. age, gender,... twins two eyes (hands) of a person subject serves as own control: e.g. blood pressure before and after treatment Example: Pain scores for lower back pain, before and after having acupuncture Pain score Pain score Sign Pain score Pain score Sign Patient Before After Patient Before After
48 9 pairs out 15 where treatment comes out ahead (reduction in pain scores) Sign Test: S N = 9 P H0 (S N 9) =??? Exact Distribution of S N under H 0 is binomial N trials, N = number of pairs Success probability: 1 2 P H0 (S N 9) = ( ( 15 9 P H0 (S N = a) = ) + ( ( ) N 1 a 2 N ) ( ) 15 ) 1 =
49 JonckheereTerpstra Test
50 JonckheereTerpstra Test To be used when the H 1 is ordered. Ordinal data for the responses and an ordering in the treatment/groups. Example: Data: Three diets for rats Response: growth H 1 : Growth rate decreases from A to C : A B C A B C
51 Parametric Approach : Regression Models the relationship between a dependent and independent variable y i = β 0 + β 1 x i + ɛ i Assumptions ɛ i N(0, σ 2 ), ɛ i are independent homoscedasticity x i is measured without error
52 β 0 = 169, pvalue = < β 1 = 16, pvalue = Rsquare =
53 JonckheereTerpstra Test Based on MannWhitney statistics for two treatments Comparing the treatment groups two by two if W BA is large: growth A > growth B : (W BA = 18 if W BC is large: growth B > growth C : (W BC = 18 if W CA is large: growth A > growth C : (W BA = 23 JT Statistic: W = i<j W ij Reject H 0 when W is sufficiently large W = 59 P H0 (W c) = Compare with the result of a KruskalWallis Test: pvalue = The distribution of W follows a normal distribution for large samples
54 Parametric versus nonparametric tests Parametric tests: Assumptions about the distribution in the population Conditions are often not tested Test depends on the validity of the assumptions Most powerful test if all assumptions are met Nonparametric tests: Fewer assumptions about the distribution in the population In case of small sample sizes often the only alternative (unless the nature of the population distribution is known exactly) Less sensitive for measurement error (uses ranks) Can be used for data which are inherently in ranks, even for data measured in a nominal scale Easier to learn
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