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Two sample tests

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Two sample tests. Study designs. Single sample, compare two sub-samples (cross-sectional survey) Compare samples from 2 different populations (2 cross sectional surveys, case-control study) Single sample; subjects randomly allocated to different interventions (experiment, clinical trial). - PowerPoint PPT Presentation
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Two sample tests
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Page 1: Two sample tests

Two sample tests

Page 2: Two sample tests

Study designs

• Single sample, compare two sub-samples (cross-sectional survey)

• Compare samples from 2 different populations (2 cross sectional surveys, case-control study)

• Single sample; subjects randomly allocated to different interventions (experiment, clinical trial)

Page 3: Two sample tests

Simple randomization

1. Generate n uniform (0,1) random deviates.2. If ui<0.5 assign to intervention A to unit i; if ui> 0.5 assign B.3. Note nA is a random variable with E(nA)=0.5.

Page 4: Two sample tests

Restricted randomization

1. Generate a U(0,1) deviate, ui, for each unit in the sample.2. Sort the deviates from smallest to largest.3. Assign intervention A to the units with the n/2 smallest ui’s.4. Note this results in half of the sample assigned to each intervention;i.e. nA is fixed.

Page 5: Two sample tests

1 2

2 21 2

1 21 1

Suppose we have N subjects. How

many should be allocated to each group?

Let N=n n .

V(Y Y ) .n N n

Page 6: Two sample tests

1 1 2

2 21 22 2

1 1 1

1 1

2 2

Find n that minimizes V(Y Y ).

dV0.

dn n (N n )

n.

n

Page 7: Two sample tests

2 21 2

1 2

If we randomly assign subjects to

interventions, it is reasonable to assume

. Then the optimum allocation

is n n .

Page 8: Two sample tests

1i 1 2j 2

2k k k k

2 21 2

1 2 1 21 2

Two independent samples, normally

distributed data

Y i 1,2,3,...,n ; Y j 1,2,3,...,n .

Y ~ N( , / n ).

Y Y ~ N( , )n n

Page 9: Two sample tests

2 2

If X and Y are 2 r.v. and a and b are

constants, then

E(aX+bY)=aE(X)+bE(Y), and

V(aX+bY)=a V(X)+b V(Y)+2abcov(X,Y)

Page 10: Two sample tests

1 2

2 21 2

1 21 2

Therefore if a=1 and b=-1:

Y Y is approximately distributed as

N( , ).n n

Page 11: Two sample tests

1 2 1 22 21 2

1 2

1 2 1 22 21 2

1 2

(Y Y ) ( )~ N(0,1).

n n

but

(Y Y ) ( ) does not follow t-distn.

s sn n

Page 12: Two sample tests

.2nn

s)1n(s)1n(s

:average) (weighted estimate Pooled

. estimates s and estimates s

.:assumption Additional

21

222

2112

p

222

221

222

21

Page 13: Two sample tests

2nn

21p

212121

t~

n1

n1

s

)()YY(

Page 14: Two sample tests

Example:An experiment was conducted to see if a drug could prevent premature birth. 30 women atrisk of premature birth were assigned to take the drug or a placebo (15 in each group).Outcome: birthweight.

Page 15: Two sample tests

0 1 2

A 1 2

28,

.05

H : (1=drug; 2=placebo)

H :

C {t t }

C {t 1.7}

Page 16: Two sample tests

BirthweightsDrug Placebo

6.9 6.4

7.6 6.7

7.3 5.4

7.6 8.2

6.8 5.3

7.2 6.6

8.0 5.8

5.5 5.7

5.8 6.2

7.3 7.1

8.2 7.0

6.9 6.9

6.8 5.6

5.7 4.2

8.6 6.8

Page 17: Two sample tests

1 1

2 2

2 22p

.o5 0

y 7.08; s 0.899

y 6.26; s 0.961

14(.899) 14(.961)s 0.8659

287.08 6.26

t 2.412

.93115

t C , reject H ; p .01.

Page 18: Two sample tests

k

2 20 1 2

2 2 2k k k n 1

2

Testing H : :

We know that

(n 1)s / ~ k=1,2.

It can be shown that the distribution of the

ratio of two independent distributed

random variables divided by their dfs is

the F distribution

1 2 with df and df degrees

of freedom.

Page 19: Two sample tests

k

1 2

2k

2 1,

1 2

In general:

If X ~ ; k=1,2,

XY= ~ F

X

Page 20: Two sample tests

1 2

2 2k k k k

2 22 1 1 1

2 21 2 2 2

2 21 1

n 1,n 12 22 2

It follows that letting

X (n 1)s / ; k=1,2,

(n 1)(n 1)s /Y=

(n 1)(n 1)s /

s / ~ F

s /

Page 21: Two sample tests

1 2

1 2 1 2

1 2 1 2

22 2 1

0 1 2 n 1,n 122

0

n 1,n 1, / 2 n 1,n 1,1 / 2

2 21 1

n 1,n 1, / 2 n 1,n 1,1 / 22 22 2

0

sUnder H : , ~ F .

s

So, to test H , we find the critical values

F and F .

s sIf F or F ,

s s

we reject H .

Page 22: Two sample tests

1 22 1

, ,, ,1

Note :

1F .

F

Page 23: Two sample tests

2 2 2 20 1 2 A 1

.025,14,14 .025,14,14

2

2

0

Example birthweights:

H : ; H : .

C {F 2.79 F 1/ 2.79 0.36}

.899F 0.88

.961do not reject H .

Page 24: Two sample tests

2 21 2What do we do if ?

Fisher Behrens problem.

1. Satterwaite approximation.

2. Transform the data.

3. Nonparametric methods.

Page 25: Two sample tests

2 2 21 1 2 2

2 2 2 21 1 2 2

1 2

Satterwaite Approximation :

Statistic is usual t statistic.

(s / n s / n )df= 2

(s / n ) (s / n )n 1 n 1

Page 26: Two sample tests

2 21 2

2

4

If , we can consider a variance

stabilizing transformation.

Some examples:

If , W= Y.

If , W=ln(Y).

If , W=1/Y.

Page 27: Two sample tests

1 2

Notes:

(1) We transform both Y and Y .

(2) I rarely use transformations.

Page 28: Two sample tests

1 2

j j

2j j j

If n and n are large, the homogeniety

of variance assumption is not important.

Recall if n is large, Y is approximately

distributed as N( , / n ).

Page 29: Two sample tests

0 1 2

1 22 21 2

1 2

To test H : , we use

(Y Y )z= .

s sn n

Page 30: Two sample tests

0

1 2

Under H , z is approximately distributed

as a N(0,1) variate. The approximation

gets better as n ,n .

Page 31: Two sample tests

j

This approximation is good enough for

practical purposes if n 25; j=1,2.

Note also that the assumption that the Y's

are normally distributed is not needed

for this statistic (Central Limit Theorm).

Page 32: Two sample tests

A study was done to compare the percent body fat of 3rd gradersAt schools on 2 Native AmericanReservations: Gila River (TohonaO’odham) and White River (Apache).

Page 33: Two sample tests

0 T A A T A

T A

.05

H : ; H :

n 63; n 35.

C {z 1.96}.

Page 34: Two sample tests

T T

A A

2 2

0

y 37.9%; s 8.66

y 32.8% s 6.88

37.9 32.8z 3.20

8.66 6.8863 35

reject H ; p=0.0014

Page 35: Two sample tests

If the sample sizes are small and the Y's

are not normally distributed:

1. Transform the Y's.

2. Nonparametric method

Page 36: Two sample tests

1

Wilcoxon-Mann Whitney rank sum test:

1. Pool the two samples and rank them from

smallest to largest.

2. Replace the observations with their ranks.

3. Compute the sum of the ranks, W ,in

group 1.

Page 37: Two sample tests

j j

0 1 2

A 1 2

What hypothesis does the Wilcoxon

procedure test?

Assume Y ~ F (y); j=1,2.

H : F (y) F (y)

H : F (y) F (y ),

where is a constant.

Page 38: Two sample tests

1 2

1

0

1

There are N=n n subjects in our study.

NThus there are possible outcomes.

n

Under H , each is equally likely. We compute

the distribution of W by enumeration.

Page 39: Two sample tests

1 2

Example : 7 students are taking a series

of exams. They are randomly divided into

2 groups: n 3, n =4. After the first exam,

group 1 is told they scored badly on exam 1

regardless of their score; group

2 was

told nothing.

Page 40: Two sample tests

0

A

The null hypothesis is that telling the

students that they are doing poorly

will have no effect on their

subsequent grade.

H : 0

H : 0

Page 41: Two sample tests

Group1 Group2

65 89

73 70

69 92

88

7There are 35 possible outcomes

3

of the study.

Grades on second exam

Page 42: Two sample tests

Ranks W1 Ranks W1 Ranks W1

1,2,3 6 1,5,6 12 2,6,7 15

1,2,4 7 1,5,7 13 3,4,5 12

1,2,5 8 1,6,7 14 3,4,6 13

1,2,6 9 2,3,4 9 3,4,7 14

1,2,7 10 2,3,5 10 3,5,6 14

1,3,4 8 2,3,6 11 3,5,7 15

1,3,5 9 2,3,7 12 3,6,7 16

1,3,6 10 2,4,5 11 4,5,6 15

1,3,7 11 2,4,6 12 4,5,7 16

1,4,5 10 2,4,7 13 4,6,7 17

1,4,6 11 2,5,6 13 5,6,7 18

1,4,7 12 2,5,7 14

Page 43: Two sample tests

1 1 2c.d.f . of W for n 3 and n 4

w F(w)

6 0.02856

7 0.05714

8 0.1143

9 0.2000

10 0.3142

11 0.4286

12 0.5714

13 0.6857

14 0.8000

15 0.8857

16 0.9429

17 0.9714

18 1.0000

Page 44: Two sample tests

0.1

1

0

Note: it is impossible to conduct a

2-sided =0.05 test. We will do a 2-sided

0.1 test. C {6,18}.

Observed W 1 2 4 7. do not

reject H . P value=2(0.05714)=0.1143.

Page 45: Two sample tests

1 2

11

1 21

N(N 1)Note : W W .

2

n (N 1)E(W ) .

2n n (N 1)

V(W ) .12

Page 46: Two sample tests

1 2

1 1

1

If n and n are large,

W E(W )z=

V(W )

will be approximately distributed

as (z).

Page 47: Two sample tests

1 2

1 21

q1 2

i i ii 1

This approximation is good for n ,n 12.

If there are ties:

n n (N 1)V(W )

12n n

{ t (t 1)(t 1)}12N(N 1)

Page 48: Two sample tests

Birthweights (lbs.)Drug Rank Placebo Rank

6.9 18 6.4 11

7.6 25.5 6.7 13

7.3 23.5 5.4 3

7.6 25.5 8.2 27.5

6.8 15 5.3 2

7.2 22 6.6 12

8.0 29 5.8 8.5

5.5 4 5.7 6.5

5.8 8.5 6.2 10

7.3 23.5 7.1 21

8.2 27.5 7.0 20

6.9 18 6.9 18

6.8 15 5.6 5

5.7 6.5 4.2 1

8.6 30 6.8 15

Page 49: Two sample tests

0 A

.05

d

2

d

H : 0; H : 0;

C {z 1.645}

15(31)E(W ) 232.5

2

15 (31)V(W no ties)= 581.25

12

Page 50: Two sample tests

1 2 3 4 5

6 7

q

i i ii 1

2

d

q 7; t 2; t 2; t 3; t 3; t 2;

t 2; t 2.

t (t 1)(t 1) 78

78(15)V(W adj. for ties) 581.25

12(31)(32)

581.25 1.47 579.78

Page 51: Two sample tests

d

0

W 291.5;

291.5 232.5z 2.45

579.78

reject H ; p=0.0071.

Page 52: Two sample tests

1 2

1i 2 j 1 2

1 1i 2 j

Mann-Whitney test:

Consider all n n possible pairs

(Y ,Y ); i=1,2,...,n ; j==1,2,...,n .

Let U # of pairs with Y Y .

Page 53: Two sample tests

1 21 1

It can be shown that:

n (N n 1)U W .

2Therefore the Mann-Whitney and

Wilcoxon tests are equivalent.

Page 54: Two sample tests

k

The Wilcoxon statistic is a special case

of a set of simple linear rank statistics.

Let R be the rank of the kth obs.

in the combined groups; k=1,2,...,N.

Page 55: Two sample tests

N

k k k k kk 1

k

Simple linear rank statistic:

S[a(R ),c ] c a(R ), where a(R ) is

a known function and c is a series

of constants.

Page 56: Two sample tests

N N

k k k kk 1 k 1

N

kk 1

N N2 2k k

k 1 k 1

It can be shown:

1E{S[a(R ),c ]} [ a(R )] c

N

a c

and

V(S)

1[ {a(R ) a} ] (c c)N 1

Page 57: Two sample tests

If N is large, the distribution of

S-E(S)z=

V(S)

is approximately (z).

Page 58: Two sample tests

k k

k

N

k k k k 1k 1

If a(R ) R , and

1 if kth obs in grp.1c

0 if kth obs. in grp.2

S[a(R ),c ] c R W .

Page 59: Two sample tests

k

1 kk

Normal Scores Test:

c as before,

Ra(R ) ( )

N 1

Page 60: Two sample tests

n kk

Savage Scores-logrank test

1 1 1a (R ) ...

N N 1 N R 1

1 1 1ln(c) 1 ...

2 3 c

Page 61: Two sample tests

n k k

k

Therefore:

a (R ) ln(N) ln(N R )

Nln( ).

N R

Page 62: Two sample tests

k

n kk

However, this is undefined if R N, so take

N 1a (R ) ln( )

N R 1

This form of the statistic is called the

logrank statistic and is used in survival

analysis.

Page 63: Two sample tests

1 0.1 0.095312 0.211111 0.2006713 0.336111 0.3184544 0.478968 0.4519855 0.645635 0.6061366 0.845635 0.7884577 1.095635 1.0116018 1.428968 1.2992839 1.928968 1.70474810 2.928968 2.397895

N=10Rank Savage Score Logrank score

Page 64: Two sample tests

Optimum LRS:

Distribution an(Rk)Normal Normal ScoresExponential Savage ScoresLogistic Wilcoxon Scores

Page 65: Two sample tests

1, 21

1 2 1

Permutation test:

N subjects randomly assigned to

N2 groups; n n . There are possible

n

assignments and each is equally likely.

Each of these assignments results in a

value of Y Y . Compute Y

2Y for

each possible outcome.

Page 66: Two sample tests

0

Compute the empirical distribution

under H of equal means. From the

edf, determine the critical region of

the test.

7Example: test scores; 35 possible

3

outcomes.

Page 67: Two sample tests

65 69 70 73 88 89 92 -17.50 65 69 73 70 88 89 92 -15.7565 69 88 70 73 89 92 -7.00 65 69 89 70 73 88 92 -6.4265 69 92 70 73 88 89 -4.67 65 70 73 69 88 89 92 -15.1765 70 88 69 73 89 92 -6.42 65 70 89 69 73 88 92 -5.8365 70 92 69 73 88 89 -4.08 65 73 88 69 70 89 92 -4.6765 73 89 69 70 88 92 -4.08 65 73 92 69 70 88 89 -2.3365 88 89 69 70 73 92 4.67 65 88 92 69 70 73 89 6.4265 89 92 69 70 73 92 6.00 69 70 73 65 88 89 92 -12.8369 70 88 65 73 89 92 -4.08 69 70 89 65 73 88 92 -3.5069 70 92 65 73 88 89 -1.75 69 73 88 65 70 89 92 -2.3369 73 89 65 70 88 92 -1.75 69 73 92 65 70 88 89 0.0069 88 89 65 70 73 92 7.00 69 88 92 65 70 73 89 8.7569 89 92 65 70 73 88 9.33 70 73 88 65 69 89 92 -1.7570 73 89 65 69 88 92 -1.17 70 73 92 65 69 88 89 0.5870 88 89 65 69 73 92 7.58 70 88 92 65 69 73 89 9.3370 89 92 65 69 73 88 9.92 73 88 89 65 69 70 92 9.3373 88 92 65 69 70 89 11.08 73 89 92 65 69 70 92 10.6788 89 92 65 69 70 73 20.42

Grp1 Grp2 diff. Grp1 Grp2 diff.

Page 68: Two sample tests

-17.50 0.029 6.41 0.714-15.75 0.057 7.00 0.743-15.17 0.086 7.58 0.771-12.83 0.114 8.75 0.800 -7.00 0.143 9.33 0.886 -6.42 0.200 9.92 0.914 -5.83 0.229 10.67 0.943 -4.67 0.286 11.08 0.971 -4.08 0.371 20.42 1.000 -3.50 0.400 -2.33 0.457 -1.75 0.543 -1.17 0.571 0.00 0.600 0.58 0.629 4.67 0.657 6.00 0.686

CDF of diff d P(Diff≤d) d (Diff≤d)

Page 69: Two sample tests

.1

1 2

0

Critical region:

C {d 17.5 or d=20.42},

where d=Y Y .

Observed d=-15.75; do not reject H .

Page 70: Two sample tests

Permutation Test

• No assumptions except random assignment

• Computations extensive if N is moderately large

• CLT type theorem shows that for large samples normal approximation is good.


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