1
1
Lecture 4
ChiSquare Test
2
This lecture covers
?
C
rosstabulation
?
C
hiSquare Test
2
3
Crosstabulation
?
C
rosstabulations
are also called
contingency tables or twoway frequency tables. In their simplest form they are the count of the categories of one variable for each category of another variable.
?
F
or example, we might like to examine a
crosstabulation
o
f age of woman with
whether they have ever had a child.
4
3
56
?
W
e call this a table because it has
7 rows and 2 columns. A table with r rows and c columns is an
table
. The
table shows the relationship between two catego
rical va
riables
. The explanatory
variable is the treatment (the drugs). The respon
se variabl
e
is su
ccess (no
relapse
)
or failure (relapse). The twoway table gives the counts for all 6 combinations of values of these variables. Each of the count
s o
ccu
pies a
cell
of the table.
rc
×
72
×
4
7
?
W
e can also calculate row or column
percentages. The following table shows column percentages. It presents the percentage age distribution for each category of WCEB. We can see straight away that the NO category is younger than the YES catego
ry.
8
5year age group * whether have any child
Crosstabulation
% withi
n
whether have any chi
l
d
51.
9%
.0
%
10.
3%
36.
0%
7.
3%
13.
0%
7.
9%
22.
1%
19.
3%
1.
5%
22.
4%
18.
3%
1.
3%
14.
4%
11.
8%
.6
%
18.
2%
14.
7%
.9
%
15.
5%
12.
6%
100.
0%
100.
0%
100.
0%
1519202425293034353940444549
5yearagegroupTotal
no
yes
whether have any
child
Total
5
9
?
W
hen creating crosstabulations
it is
standard practice to use the dependent variable as the rows and the independent variable as the columns.
?
W
e can create a crosstabulation
w
ith three
variables. For example, we may want to see the age distribution for WCEB for urban and rural area separately. This is shown in the following table.
10
Rural and urban age distribution for
WCEB
5

ye
a
r
a
g
e
gr
oup *
whe
t
he
r
ha
ve
a
n
y child *
pla
c
e
of
r
e
side
nce
C
r
o
sst
abulat
io
n
% wit
h
in
wh
et
h
e
r
h
a
v
e
an
y
ch
ild
57.
3%
.1
%
10.
7%
33.
9%
8.
4%
13.
2%
5.
0%
22.
9%
19.
6%
1.
4%
22.
8%
18.
8%
1.
4%
12.
9%
10.
8%
.3
%
17.
9%
14.
6%
.8
%
15.
0%
12.
3%
100.
0%
100.
0%
100.
0%
37.
4%
8.
9%
41.
5%
3.
3%
12.
4%
15.
6%
19.
3%
18.
4%
1.
9%
21.
1%
16.
5%
1.
1%
19.
9%
15.
4%
1.
5%
19.
1%
14.
9%
1.
1%
17.
3%
13.
5%
100.
0%
100.
0%
100.
0%
15
19
20
24
25
29
30
34
35
39
40
44
45
49
5
ye
ar
agegrou
p
To
tal
15
19
20
24
25
29
30
34
35
39
40
44
45
49
5
ye
ar
agegrou
p
To
tal
place of res
iden
c
e
ruralurb
a
n
no
yes
w
h
e
t
he
r have
any
chi
ld
To
tal
5
year ag
e g
r
o
u
p
* whet
her hav
e
any child
* p
l
ace o
f
resid
e
nce
C
r
o
sst
abulat
io
n
% wi
thi
n
5year age gro
u
p
99.
5%
.5
%
100.
0%
48.
0%
52.
0%
100.
0%
4.
8%
95.
2%
100.
0%
1.
4%
98.
6%
100.
0%
2.
4%
97.
6%
100.
0%
.4
%
99.
6%
100.
0%
1.
3%
98.
7%
100.
0%
18.
7%
81.
3%
100.
0%
100.
0%
100.
0%
79.
4%
20.
6%
100.
0%
20.
1%
79.
9%
100.
0%
2.
7%
97.
3%
100.
0%
1.
7%
98.
3%
100.
0%
2.
4%
97.
6%
100.
0%
2.
0%
98.
0%
100.
0%
23.
8%
76.
2%
100.
0%
15
19
20
24
25
29
30
34
35
39
40
44
45
49
5
ye
ar
agegrou
p
To
tal
15
19
20
24
25
29
30
34
35
39
40
44
45
49
5
ye
ar
agegrou
p
To
tal
place of res
iden
c
e
ruralurb
a
n
no
yes
wh
et
h
e
r
h
a
v
e
an
y
chi
ld
To
tal
6
11
?
T
he question is: Is there a significant
relationship between woman’s age and having ever had a child?5

y
e
ar
ag
e
g
r
o
u
p
45
49
40
4
4
35

3
9
30
34
25
2
9
20

2
4
15
19
M e a n w h e t he r ha v e a n y c h i l d
1.
2
1.
0.8.6.4.2
0.
0
ag
e
49
47
45
43
41
39
37
35
33
31
29
27
25
23
21
19
17
15
M e an w h e t h e r h a v e an y c h i l d
1.
2
1.
0.8.6.4.2
0.
0
Please examine this question by your own after class. I will discuss some other examples.
12
Example 1: Treating cocaine
addiction
?
T
his is a threeyear study on medication to
help cocaine addicts stay off cocaine: D, L, and P. Each treatment was randomly assigned with 24 subjects. The counts and proportions who avoided relapse into cocaine use during the study:
7
13
0.167
4
24
P
3
0.250
6
24
L
2
0.583
14
24
D
1
Proportion
No relaps
e
Subjects
Treatment
Group
14
?
T
he sample proportions of subjects who
stayed off cocaine are quite different. Are these data good evidence that the proportions of successes for the three treatments differ in the population of all cocaine addicts?
Does su
cce
ss di
ffer signi
ficantly
between the treatments?Is there a significant relationship between treatment and success?
8
15
Here is the twoway table of the
cocaine addiction data:
16
?
W
e want to test the null hypothesis that
there are no differences among the proportions of success
es for addicts given
the three treatments:
?
T
he alternative hypothesis is that there is
some difference, that not all three proportions are equal:
01
2
3
:
Hp
p
p
==
11
2
3
:
not a
l
l
of
,
,
a
n
d
a
r
e
e
qua
l
Hp
p
p
9
17
?
T
o test , we compare the observed
counts in a twoway table with the expected counts, the counts we would expect if were true. If the observed counts are far from the expected counts, that is evidence against .
0
H
0
H
0
H
18
Expected counts
?
T
he expected count in any cell of a two
way table when is true is
0
H
r
o
w
tota
l
c
ol
um
n
tota
l
e
xpe
c
t
e
d
c
ount
ta
bl
e
tota
l
×
=
10
19
?
I
n more formal language, if we have n
independent tries and the probability of a success on each try is p, we expect npsuccesses. If we draw an SRS of n individuals from a population in which the proportion of successes
is p, we expect np
succe
sses in the
sample.
That’s t
h
e fact
behind the formula for expected counts in a twoway table.
20
?
L
et’s ap
ply this f
a
ct to th
e co
caine stud
y.
The twoway table with row and column totals is
11
21
?
W
e will find the expected count for the cell
in row 1 and column 1. The proportion of all 72 subjects who succeed in avoiding a relapse is
count
of suc
c
e
sses
c
o
l
u
m
n
1 total
2
4
1
ta
bl
e
tota
l
t
a
b
l
e
tota
l
7
2
3
==
=
Think of this as p, the overall proportion of successes. If is
true, we expect this
same proportion of successes in all three groups.
0
H
22
?
S
o the expected count of successes
among the 24 subjects who took D is
1
24
8
3
np
=×
=
This expected count has the form:
row 1 total
c
ol
um
n 1 total
2
4
2
4
ta
bl
e
tota
l
7
2
××
=
12
23
Observed versus expected counts
24
?
B
ecause 1/3 of all subjects succeed, we
expect 1/3 of the 24 subjects in each group to avoid a relapse if there are no differences among the treatments. In fact, D has more successes (14) and fewer failures (10) than expected. The P has fewer su
cce
sse
s
(4) and more rel
a
pse
s
(20). D does much better than P, with L in between.
13
25
The ChiSquare Test
?
T
he statistical test that tells us whether
those differences are statistically significant compares the observed and expe
cte
d
count
s
. The te
st st
atist
i
c that
makes the comparison is the chisquare statistic.
26
Chisquare statistic
?
T
he chisquare statistic is a measure of
how far the observed counts in a twoway table are from the expected counts. The formula for the statistic is
2
2
(obser
v
e
d count
e
xpected count)
expe
cted count
χ
?
=∑
14
27
?
T
he chisquare statistic is a sum of terms,
one for each cell in the table. In the cocaine example, 14 of the D group succeeded in avoiding a relapse. The expected count for this cell is 8. So the component of the chisquare statistic from this cell is
2
2
(obse
r
v
e
d c
ount
expe
c
t
e
d
c
ount)
e
xpec
t
e
d
count
(1
4
8
)
3
6
4.5
88
?
?
==
=
28
?
T
hink of the chisquare statistic as a
measure of the distance of the observed counts from the expected counts. Like any distance
, it is always
zero or positive, and
it is zero only when
the observed counts
are exactly equal to the expected counts. Large values of
are evidence against
because they say that the observed counts are far from what we would expect if were true.
2
χ
2
χ
0
H
0
H
15
29
The chisquare distribution
?
T
he chisquare distributions are a family of
distributions that take onl
y po
sitive values
and are skewed to the right. A specific chisquare distribution is specified by giving its degrees of freedom.
?
T
he chisquare test for a twoway table
with r ro
ws and c colum
n
s u
s
es critical
values from the chisquare distribution with (r1
)(c1
)
degree
s of fre
e
d
om. Th
e Pvalu
e
is the area to the right of under the chisquare density curve.
2
χ
30
Figure 1 shows the density curves for three members of the chisquare family of distributions.
16
31
There are three major properties of a chisquare distribution?
C
hisquare is either 0 or positive, never
negative.
?
A
chisquare distribution in not symmetrical.
Its skewness is positive.
As the number of
degree
s of freed
o
m incre
a
se
s, ch
isquar
e
approaches a symmetric distribution.
?
T
here is a particular distribution for each
degree of freedom.
32
Table E gives cri
t
ical
values for chi
squa
re
distributions. Use Table E to find Pvalue for a chi
squa
re
test.
17
33
?
W
e use the formula to calculate chisquare
statistic:
2
2
22
2
22
2
(o
bs
erv
e
d
cou
n
t
e
x
p
ect
ed
co
un
t
)
ex
p
ect
ed
co
u
n
t
(
1
4
8
)
(
10
16)
(
6
8
)
81
6
8
(
1
8
16)
(
4
8
)
(
2
0
16)
16
8
1
6
4.50
2.25
0.500
0
.25
2
.00
1
.00
10.50
χ
?
=∑
??
?
=+
+
??
?
++
+
=+
+
+
+
+
=
34
?
T
he twoway table has 3 rows and 2
columns. That is, r=3, c=2. The chisquare statistic therefore has
degrees of freedom
(r1)(c1)=(3
1
)(21)=(2)(1
)=2.
?
L
ook in the df=2 row of Table E. The chi
square statistic =10.5 falls between the 0.01 and 0.05 critical values. Remember that the chisquare test is always onesided. So the Pvalue of =10.5 is between 0.01 and 0.05. The Pvalue is equal to 0.005 when rounded to three decimal places.
2
χ
2
χ
18
35
Using SPSS we can ea
sily find the Pvalu
e.
36
?
If we want our significance level to be < 0.05, the critical value is 5.99. To reject the null hypothesis at the 0.05 level the value of chisquare needs to be greater than 5.99. If it were less than 5.99 the null hypothesis would be accepted.
?
I
n this example, the value of chisquare is
10.5, which is greater than 5.99, so we reject the null hypothesis in this case. We can conclude that there is a statistically significant difference in effects of the treatments (drugs).
19
37
Using Crosstabs
in SPSS
?
C
alculating the expected counts and then the
chisquare statistic by hand is a bit timeconsuming. We can avoid this trouble by using SPSS’s crosstabs. But you need to arrange the data in the following format:
38
20
3940
21
41
Example 2: Abortion and
educational level
?
T
he 1997 survey data:
educational level * abortion C
r
osstabulation
Count
229
765
994
374
942
1316
465
715
1180
223
191
414
61
48
109
1352
2661
4013
illiterateprimaryj
unior high
senior highc
o
llege+
e
d
uc
ational
levelTota
l
yes
no
abortion
Tota
l
42
22
43
?
O
ne of the most useful properties of chi
square is that it tests the hypothesis “the row and column variables are not related to each other”
i
n a twoway table.
?
I
n this example, the null hypothesis:
: there is no relationship
between education and ever having had an abortion.
0
H
44
Using SPSS
23
4546
ChiSquare Tests
183.743
a
4
.000
180.879
4
.000
175.137
1
.000
4013
Pearson ChiSquareLikelihood RatioLinearbyLinearAssociationN of Valid Cases
Value
df
Asymp. Sig.
(2sided)
0 cells (.0%) have expected count less than 5. Theminimum expected count is 36.72.
a.
24
47
Uses of the chisquare test
Use the chisquare test to test the null hypothesis
: there is no relationship between
two categorical variables
when you have a twoway table from one of these situations:
0
H
48
?
A
single SRS, with each individual
classified according to both of two categorical variables.
?
I
ndependent SRSs
from each of several
populations, with each individual classified according to one categorical variable. (The other va
riable sa
ys which
sample the
individual come
s from.
)
25
49
Cell counts required for the chi
square test
?
Y
ou can safely use the chisquare test
with critical values from the chisquare distribution when no more than 20% of the expected counts are less than 5 and all individual expected counts are 1 or greater. In particular, all four expected counts in a
table should be 5 or greater.
22
×
50
Example 3: SRB
?
E
ffect of place of residence on SRB.
dummy for urban * PAR_OC Crosstabulation
3474
3055
6529
53.2%
46.8%
100.0%
613
526
1139
53.8%
46.2%
100.0%
4087
3581
7668
53.3%
46.7%
100.0%
Count% wit
h
in dummy
for
u
r
ba
n
Count% wit
h
in dummy
for
u
r
ba
n
Count% wit
h
in dummy
for
u
r
ba
n
.001.00
d
u
mmy
f
o
r
urbanTotal
mal
e
fe
m
a
le
PAR_OC
Total
26
51
ChiSquare Tests
.145
b
1
.703
.122
1
.727
.145
1
.703
.723
.364
.145
1
.703
7668
Pearson ChiSquareContinuity Correction
a
Likelihood RatioFisher's Exa
c
t
T
e
st
L
i
ne
arbyL
i
ne
ar
Associa
t
ion
N of Valid Cases
Value
df
Asymp. Sig.
(2sided)
Exact Sig.
(2sided)
Exact Sig.
(1sided)
Computed only for a 2x2 table
a.
0 cells (.0%) ha
ve ex
pect
ed count
less t
h
a
n
5. T
h
e m
i
nim
u
m
ex
pect
ed count
is
531.92.
b.
52
?
E
ffect of parity on SRB:
PAR_NUM * PAR_OC Cr
osstabulation
1958
1815
3773
51.9%
48.1%
100.0%
1127
903
2030
55.5%
44.5%
100.0%
3085
2718
5803
53.2%
46.8%
100.0%
Co
unt
% within P
A
R_NUM
Co
unt
% within P
A
R_NUM
Co
unt
% within P
A
R_NUM
12
PA
R_N
U
M
To
t
a
l
male
fema
le
PA
R_O
C
To
t
a
l
ChiSquare Tests
6.955
b
1
.008
6.810
1
.009
6.963
1
.008
.009
.005
6.953
1
.008
5803
P
e
a
r
so
n Chi

Squa
re
Co
nti
nui
ty
Co
rrec
t
i
o
n
a
Likelih
ood Rat
i
o
Fisher's Exact TestLin
e
ar
byLin
e
ar
AssociationN of Valid Cases
Valu
e
df
Asymp. Sig.
(2sided)
Ex
act
Sig.
(2sided)
Ex
act
Sig.
(1sided)
Comput
ed only for a 2x2 t
a
ble
a.
0 cells (.0%) h
ave ex
pect
ed cou
n
t
less t
h
an
5. T
h
e min
i
mu
m ex
pect
ed cou
n
t
is
950.81.
b.