Monotonic relationship of two variables, X and Y.

Post on 17-Jan-2016

232 views 2 download

Tags:

Transcript of Monotonic relationship of two variables, X and Y.

Monotonic relationship of two variables, X and Y

0

4

8

12

16

0 1 2 3 4

Y

X

Deterministic monotonicity

If X growsthen

Ygrows

too

0

4

8

12

16

0 1 2 3 4

Y

X

Stochastic monotonicity

***

*

*

*

*

*

**

*

* *

*

*

*

*

If X growsthenlikely

Ygrows

too

Ss X Y 1 1 35 2 1.5 34 3 2 36 4 3 37 5 7 38 6 10 39

An example

Ss X rank Y rank 1 1 1 35 2 2 1.5 2 34 1 3 2 3 36 3 4 3 4 37 4 5 7 5 38 5 6 10 6 39 6

Rank data separately for X and Y

Spearman-s rank correlation (rS):

Correlation between ranksIn the above example:

r = 0.91, rS = 0.94

DiscordancyConcordancy

+

A

B

C

D X

Y

Concordancy and discordancy

pp

Kendall-s tau

p+: Proportion of concordantpairs in the population

p-: Proportion of discordantpairs in the population

1 +1 If X and Y are independent:

= 0: no stochastic monotonicity = deterministic

monotone decreasing (inreasing) relationship

Features of Kendall’s

p p

p p

A Kendall’s gamma

For discrete X and Y variables

1 +1 If X and Y are independent: = 0 = 0: no stochastic monotonicuty If = 1: p+ = 0

If = +1: p = 0

Features of Kendall’s

Testing the H0: = 0null hypothesis

Sample tau: Kendall’s rank correlation coefficient (r)

Testing stochastic monotonicity = testing the significancy of r

+

A

B

C

D X

Y

Computation of sample tau

++

C+

c = n = 4d = n= 2

r = (4-2) /(4+2)

= 2/6 = 0.33

c = # of concordanciesd = # of discordanciesT = # of total couples

= n(n-1)/2

r = (c - d)/T, = (c - d)/(c+d)

In which cases will r = ?

Formulea of r and

Ss X Y 1 1 35 2 1.5 34 3 2 36 4 3 37 5 7 38 6 10 39

An example

r(p < 0.02);

rS(p < 0.02);

r(p < 0.10);

Comparison of several Comparison of several independent samplesindependent samples

-60

-40

-20

0

20

40

60

80G

SR

-dec

reas

e

Agr1 Agr2 Agr3 Light Verbal

Groups

Normal Person. disorder

Holocaustgroup

0

0.5

1

1.5

2

2.5

Average Rorschach time (min)

Comparison of population means

H0: E(X1) = E(X2) = ... = E(XI)

H0: 1 = 2 = ... = I

One way independent sample ANOVA

SStotal = SSb + SSw

SStotal: Total variability

SSb: Between sample variability

SSw: Within sample variability

Basic identity

Varb = SSb/(I - 1) = SSb/dfb

- Treatment variance

Varw = SSw/(N - I) = SSw/dfw

- Error variance

One-way ANOVA

Test statistic: F = Varb/Varw

Treatment variance

1

)(

1

2

1

I

xxn

I

SSVar

I

iii

bb

Error variance

I

ii

I

iii

ww

df

Vardf

IN

SSVar

1

1

H0: 1 = 2 = ... = I

F = Varb/Varw ~ F-distribution

Assumptions of ANOVA

F F: reject H0 at level

+

Independent samplesNormality of the dependent variable

Variance homogeneity (identical population variances)

Assumptions of ANOVA

Welch test James test Brown-Forsythe test

Robust ANOVA’s

Levene test

O’Brien test

Testing variance homogeneity

Var1 Var2 ... VarI

or (and)

n1 n2 ... nI

Trust in the result of ANOVA

Different sample sizes

Substantially different sample variances

When to apply a robust ANOVA?

Conventional test: Tukey-Kramer test (Tukey’s HSD test)

Robust test: Games-Howell test

Post hoc analyses

Hij: i = j

Nonlinear coefficientof determination

Explained variance: eta2 = SSb/SStotal

Nonlinear correlationcoefficient: eta

SStotal = SSb + SSw

An exampleAn example

Agr1 Agr2 Agr3 Light Verb.

n i 5 4 6 4 4

xi 14.506.75 5.20 -13.45-30.08

s i 29.609.15 6.96 13.11 14.57

Levene test:

F(4, 7) = 0.784 (p > 0.10, n.s.)

O’Brien test:

F(4, 8) = 1.318 (p > 0.10, n.s.)

Testing variance homogeneity

Treatment var.: Varb = 1413.9 Error variance: Varw = 286.2

F(4, 18) = 1413.9/286.2= 4.940**

Nonlinear coeff. of determin.:eta2 = SSb/SStotal = 0.523

Conventional ANOVA

Welch test:W(4, 8) = 5.544*

James test:U = 27.851+

Brown-Forsythe test:BF(4, 9) = 5.103*

Robust ANOVA’s

Tukey-Kramer test: T12= 0.97 T13= 1.28T14= 3.48 T15= 5.55**T23= 0.20 T24= 2.39T25= 4.35* T34= 2.42T35= 4.57* T45= 1.97

Pairwise comparison of means