AMS 572 Lecture Notes 7
Nov. 16th , 2009
Inference on several proportions—the Chi-square test
(large sample)
Def. Multinomial Experiment.
We have a total of n trials (sample size=n)
① For each trial, it will result in 1 of k possible outcomes.
② The probability of getting outcome i is pi , and
k
 p =1
i 1
i
③ These trials are independent.
e.g. Previous experience indicates that the probability of obtaining 1 healthy calf from a
mating is 0.83. Similarly, the probabilities of obtaining 0 and 2 healthy calves are 0.15 and
0.02 respectively. If the farmer breeds 3 dams from the herd, find the probability of getting
exact 3 health calves.
Def. Multinomial Distribution
Let X i be the number of trials resulted in i-th category out of a total of n trials and pi be the
probability of getting i-th category outcome, then
P( X 1  x1 , X 2  x2 ,..., X k  xk ) 
n!
p1x1 ... pk xk
x1 ! x2 !...xk !
Solution:
P(exact 3 health calves)= P( X1  1, X 2  1, X 3  1) + P( X 1  0, X 2  3, X 3  0)
=0.015+0.572=0.59
*Relations to the Binomial Distribution (k=2)
Category
1
2
Probability
p1 =p
p2 =1-p
# trials
X1 =x
X 2 =n-x
 P ( X 1  x, X 2  n  x ) 
n!
p1x1 p2 x2  ( nx ) p x (1  p)n  x
x !(n  x)!
Chi-square goodness of fit test
e.g1. Gregor Mendel (1822-1884) was an Austrian monk whose genetic theory is one of the
greatest scientific discovery of all time. In his famous experiment with garden peas, he
proposed a genetic model that would explain inheritance. In particular, he studied how the
shape (smooth or wrinkled) and color (yellow or green) of pea seeds are transmitted through
generations. His model shows that the second generation of peas from a certain ancestry
should have the following distribution.
wrinkledgreen
Theoretical probabilities
p1 
1
16
wrinkledgreen
p2 
n=556
General test:
Test whether the theoretical probability is correct
0
0
0
 H 0 : p1  p1 , p2  p2 ,..., pk  pk

 H a : H 0 is not true
3
16
smoothgreen
p3 
3
16
smoothyellow
p4 
9
16
( x  e )2
W0   i i
ei
i 1
k
T.S
H0
 k21
where xi is the observed number of observations in category i
ei is the expected count of the i-th category , ei  n  pi 0
At the significance level α, reject H 0 iff W0  k21,upper ,
Nov. 16th , 2009
Solution:
wrinkled-green
Theoretical
probabilities
p1 
Observed count out of
556
X1 =31
Expected counts
1
16
e1  556 
p2 
1
=34.75
16
1
3
3
9

 H 0 : p1  , p2  , p3  , p4 
16
16
16
16

 H a : H 0 is not true
( x  e )2
W0   i i
ei
i 1
k
T.S
wrinkledgreen
H0
 k21
2
=7.815
 0.604 < 3,0.05,upper
3
16
smoothgreen
p3 
3
16
smoothyellow
p4 
9
16
X 2 =102
X 3 =108
X 4 =315
e2 =104.25
e3 =104.25
e4 =312.75
 At significance level 0.05, we cannot reject H 0
e.g2. A classic tale involves four car-pooling students who missed a test and gave as an
excuse of a flat tire. On the make-up test, the professor asked the students to identify the
particular tire that went flat. If they really did not have a flat tire, would they be able to
identify the same tire?
To mimic this situation, 40 other students were asked to identify the tire they would select.
The data are:
Tire
Left front
Right front
Left rear
Right rear
11
15
8
6
Frequency
At α=0.05, please test whether each tire has the same chance to be selected?
Solution:
1

 H 0 : p1  p2  p3  p4 
4

 H a : H 0 is not true
n=40, ei =n pi =10
k
W0  
i 1
( xi  ei )2
2
 4.6  3,0.05,
upper  7.81
ei
 Fail to reject H 0 .
The chi-square goodness of fit test is an extension of the Z-test for
one population proportion.
Data: sample size n, x: successes with probability p
n-x: failures with probability 1-p
TS. Z 0 
pˆ  p0
H0
p0 (1  p0 )
n
~ N (0,1)
At α, reject H 0 iff | Z0 | Z /2
Success
p1  p0
Expected
Observed
W0 
Failure
p2  1  p0
e1  np0
X2  n  x
X 1 =x
( x  np0 ) 2 [n  x  n(1  p0 )]2 ( x  np0 ) 2 ( p0  1  p0 )]2
=

np0
n(1  p0 )
np0 (1  p0 )
x
(  p0 )2
( x  np0 ) 2
 n
 Z02
=
np0 (1  p0 ) p0 (1  p0 ) / n
iid
Recall: If Z1, Z2, ....Zn
k
N (0,1) , then W   Z i 2
1
When k=1, W  Z 2
12
Let Z~N(0,1), then W  Z 2 12
P(| Z | Z /2 )  P( Z 2  Z2 /2 )   = P(W  1,2 ,upper )
 The two tests are identical.
e2  n(1  p0 )
k 2 .
Nov. 23th , 2009
1. Fisher’s exact test:
Inference on 2 population proportions, 2 independent
samples
e.g1. The result of a randomized clinical trial for comparing Predsome and Predsome+VCR
drugs, is summarized below. Test if the success and failure probabilities are the same for the
two drugs.
Drug
Success
Failure
Row total
Pred
14
7
n1 =21
PVCR
38
4
n2 =42
m=58
n-m=11
n=63
“S”
“F”
Total
Sample1
x
n1 -x
n1
Sample2
y
n2 -y
n2
m=x+y
n-m
n
General setting:
 H 0 : p1  p2

 H a : p1  p2
(nk1 )(mn2k ) min( m,n1 ) (kn1 )(mn2k )
p  value  pU  P( X  x | X  Y  m)  
 
(nm )
(nm )
kx
k x
 H 0 : p1  p2

 H a : p1  p2
p  value  pL  P ( X  x | X  Y  m)  
x
( nk1 )(nm2 k )
(nk1 )(nm2 k )


( nm )
( nm )
kx
k  max(0,m  n2 )
 H 0 : p1  p2

 H a : p1  p2
p  value  2  min( PU , PL )
Solution:
 H 0 : p1  p2

 H a : p1  p2
p  value 
42
42
14
(21
(21
k )(52  k )
k )(52  k )


 (63 )  0.016 <0.05
63
(52
)
k  max(0,52  42)
k 10
52
14
 Reject H 0
SAS code:
Data trial;
input drug $ outcome$ count;
datalines;
pred S 14
pred F 7
PVCR S 38
PVCR F 4
;
run;
proc freq data=trial;
tables drug*outcome/chisq;
weight count;
run;
2. McNemar’s test
Inference on 2 population proportions- paired samples
e.g2. A preference poll of a panel of 75 voters was conducted before and after a TV debate
during the campaign for the 1980 presidential election between Jimmy Carter and Ronald
Reagan. Test whether there was a significant shift from Carter as a result of the TV debate.
Preference
Preference after
before
Carter
Reagan
Carter
28
13
Reagan
7
27
General setting:
Condition1
Condition2 response
response
Yes
No
Yes
A=a,
PA
B=b,
PB
No
C=c,
PC
D=d,
PD
PA + PB + PC + PD =1,
A+B+C+D=n, (A, B, C, D)~Multinomial
P1  PA  PB , P2  PA  PC
H 0 : p1  p2
 H 0 : pB  pC
P(B=k| B+C=m)~Bin(m,p=
pB
)
pB  pC
1
2
Under H 0 : p  , P(B=k| B+C=m)~Bin(m,p=1/2)
1

 H 0 : p  2
 H 0 : p1  p2

① 
H
:
p

p
2
 a 1
H : p  1
 a
2
1 m
p  value  pU  P ( B  b | B  C  m)  ( )m  (mk )
2 k b
1

 H 0 : p  2
 H 0 : p1  p2
② 

 H a : p1  p2
H : p  1
 a
2
1 b
p  value  pL  P( B  b | B  C  m)  ( )m  (mk )
2 k 0
1

H
:
p

0

H : p  p
2
③  0 1 2  
 H a : p1  p2
H : p  1
 a
2
p  value  2  min( PL , PU )
Solution:
1

 H 0 : p  2
 H 0 : p1  p2


H
:
p

p
2
 a 1
H : p  1
 a
2
1 20
p  value  ( ) 20  ( 20
k )  0.132
2 k 13
SAS code:
Data election;
input before $ after $ count;
datalines;
Carter Carter 28
Carter Reagan 13
Reagan Reagan 27
Reagan Carter 7
;
run;
proc freq data=election;
tables before*after/agree;
weight count;
run;