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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
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)
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.
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.
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
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
2 21 2
1 2
If we randomly assign subjects to
interventions, it is reasonable to assume
. Then the optimum allocation
is n n .
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
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)
1 2
2 21 2
1 21 2
Therefore if a=1 and b=-1:
Y Y is approximately distributed as
N( , ).n n
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
.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
2nn
21p
212121
t~
n1
n1
s
)()YY(
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.
0 1 2
A 1 2
28,
.05
H : (1=drug; 2=placebo)
H :
C {t t }
C {t 1.7}
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
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.
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.
k
1 2
2k
2 1,
1 2
In general:
If X ~ ; k=1,2,
XY= ~ F
X
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 /
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 .
1 22 1
, ,, ,1
Note :
1F .
F
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 .
2 21 2What do we do if ?
Fisher Behrens problem.
1. Satterwaite approximation.
2. Transform the data.
3. Nonparametric methods.
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
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.
1 2
Notes:
(1) We transform both Y and Y .
(2) I rarely use transformations.
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 ).
0 1 2
1 22 21 2
1 2
To test H : , we use
(Y Y )z= .
s sn n
0
1 2
Under H , z is approximately distributed
as a N(0,1) variate. The approximation
gets better as n ,n .
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).
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).
0 T A A T A
T A
.05
H : ; H :
n 63; n 35.
C {z 1.96}.
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
If the sample sizes are small and the Y's
are not normally distributed:
1. Transform the Y's.
2. Nonparametric method
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.
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.
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.
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.
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
Group1 Group2
65 89
73 70
69 92
88
7There are 35 possible outcomes
3
of the study.
Grades on second exam
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
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
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.
1 2
11
1 21
N(N 1)Note : W W .
2
n (N 1)E(W ) .
2n n (N 1)
V(W ) .12
1 2
1 1
1
If n and n are large,
W E(W )z=
V(W )
will be approximately distributed
as (z).
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)
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
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
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
d
0
W 291.5;
291.5 232.5z 2.45
579.78
reject H ; p=0.0071.
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 .
1 21 1
It can be shown that:
n (N n 1)U W .
2Therefore the Mann-Whitney and
Wilcoxon tests are equivalent.
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.
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.
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
If N is large, the distribution of
S-E(S)z=
V(S)
is approximately (z).
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 .
k
1 kk
Normal Scores Test:
c as before,
Ra(R ) ( )
N 1
n kk
Savage Scores-logrank test
1 1 1a (R ) ...
N N 1 N R 1
1 1 1ln(c) 1 ...
2 3 c
n k k
k
Therefore:
a (R ) ln(N) ln(N R )
Nln( ).
N R
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.
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
Optimum LRS:
Distribution an(Rk)Normal Normal ScoresExponential Savage ScoresLogistic Wilcoxon Scores
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.
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.
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.
-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)
.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 .
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.