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Social Science Statistics Module I
Gwilym Pryce
Lecture 7
Two-Way Tables
Slides available from Statistics & SPSS page of
www.gpryce.com
1
Notices:
• Register
Aims and Objectives:
• Aim:
– This session introduces methods of examining
relationships between categorical variables
• Objectives:
– By the end of this session you should be able to:
• Understand how to examine relationships between
categorical variables using:
» 2 way tables
» Chi square test for independence.
Plan:
•
•
•
•
1. Independent events
2. Contingent events
3. Chi square test for independence
4. Further Study
1. Probability of two Independent events occurring
• If knowing that one event occurs does not
affect the outcome of another event, we say
those two outcomes are independent.
• And if A and B are independent, and we
know the probability of each of them
occurring, we can calculate the probability of
them both occurring
Example: You have a two sided die and a coin, find Pr(1 and H).
• Answer: ½ x ½ = ¼
• Rule: P(A  B) = P(A) x P(B)
e.g. You have one fair coin which you toss twice:
what’s the probability of getting two heads?
• Suppose:
• A = 1st toss is a head
• B = 2nd toss is a head
– what is the probability of A  B?
• Answer: A and B are independent and are
not disjoint (i.e. not mutually exclusive). P(A)
= 0.5 and P(B) = 0.5.
P (A  B) = 0.5 x 0.5 = 0.25.
2. Probability of two contingent events occurring
• If knowing that one event occurs does change the
probability that the other occurs, then two events are
not independent and are said to be contingent upon
each other
• If events are contingent then we can say that there is
some kind of relationship between them
• So testing for contingency is one way of testing for a
relationship
Example of contingent events:
• There is a 70% chance that a child will go to
university if his/her parents are middle class, but only
a 10% chance if his/her parents are working class.
Given that there is a 60% chance of a child’s parents
being working class:
• What are the chances that a child will be born working class
and go to University?
• What proportion of people at university will be from working
working class backgrounds?
A tricky one...
10% go to University
60%
Working Class
90% Not goto
University
Child
40%
70% go to University
Middle Class
30% Not goto
University
This diagram illustrates graphically how the probability of going to
university is contingent upon the social class of your parents.
6% of all children are both working class and end up going to
University
10% go to University
Working Class
90% do not go to
University
60%
6% of all children
are working class
and go to University
54% of all children
are working class
and do not go to University
Child
70% go to University
40%
Middle Class
30% do not go to
University
28% of all children
are Middle class
and go to University
12% of all children
are Middle class
and do not go to University
% = as percent of all children
Working class
Middle class
Go to University
6%
28%
Do not go to
University
54%
12%
% at Uni from WC parents?
10% go to University
Working Class
90% do not go to
University
60%
6% of all children
are working class
and go to University
54% of all children
are working class
and do not go to University
Child
70% go to University
40%
Middle Class
30% do not go to
University
28% of all children
are Middle class
and go to University
12% of all children
are Middle class
and do not go to University
• Of all children, only 34% end up at university (6% WC;
28% MC)
– i.e. 6 out of every 34 University students are from WC
parents:
•
6/
34
= 17.6% of University students are WC
• Probability theory states that:
– if x and y are independent, then the probability of
events x and y simultaneously occurring is simply equal
to the product of the two events occurring:
Pr ob( x  y )  Pr ob( x)  Pr ob( y )
• But, if x and y are not independent, then:
Prob(x  y) = Prob(x)  Prob(y given that x has occurred)
Test for independence
• We can use these two rules to test whether events
are independent
– Does the distribution of observations across possible
outcomes resemble
• the random distribution we would get if events were
independent?
– I.e. if we assume independence and calculate the expected
number of of cases in each category, do these figures
correspond fairly closely to the actual distribution of outcomes
found in our data?
• Or a distribution of outcomes more akin to contingency
– i.e. one event contingent on the other
Example 1: Is there a relationship between social class and
education? We might test this by looking at categories in our data
of WC, MC, University, no University.
Suppose we have 300 observations distributed as follows:
Working class
Go to University
Middle class
18
84
162
36
Do not go to University
Given this distribution, would you say these two variables are independent?
• To do the test for independence we need to compare
expected with observed.
• But how do we calculate ei, the expected number of
observations in category i?
– ei = number of cases expected in cell i assuming that the
two categorical variables are independent
– ei is calculated simply as: the probability of an observation
falling into category i under the independence assumption,
multiplied by the total number of observations.
• I.e. No contingency
So, if UNIY or UNIN and WC or MC are
independent (i.e. assuming H0) then:
Prob(UNIY  WC) = Prob(UNIY)Prob(WC)
so the expected number of cases for each of the four
mutually exclusive categories are as follows:
Working class
Go to University
Do not go to
University
Middle class
P(UNIY) x P(WC) x n
P(UNIY) x P(MC) x n
P(UNIN) x P(WC) x n
P(UNIN) x P(MC) x n
• But how do we work out:
Prob(UNIY)
and
Prob(WC)
which are needed to calcluate Prob(UNIY  WC):
Prob(UNIY  WC) = Prob(UNIY)Prob(WC)
•
Answer: we assume independence and so estimate them from the data by
simply dividing the total observations by the total number in the given
category:
E.g. Prob(UNIY)
= Total no. cases UNIY
= (18 + 84) / 300 = 0.34
 All observations
• Prob(WC) is calculated the same way:
E.g. Prob(WC)
•
Prob(UNIY  WC)
= Total no. cases WC
= (18 + 162)/300 = 0.6
= .34 x .6 x 300 = 61.2
 All observations
Working class
Go to
University
Do not go to
University
Middle class
P(UNIY) x P(WC) x n
= (no. at Uni / n)
x (no. WC/n)
xn
P(UNIY) x P(MC) x n
= (no. at Uni / n)
x (no. MC/n)
xn
P(UNIN) x P(WC) x n
= (no.not Uni / n)
x (no. WC/n)
xn
P(UNIN) x P(MC) x n
= (no. not Uni / n)
x (no. MC/n)
xn
Go to University
Do not go to
University
Working class
Middle class
18
84
102
162
36
198
180
120
300
Go to University
Do not go to
University
Working class
Middle class
P(UNIY) x P(WC) x n
= (102 / 300)
x (180 /300)
x 300
P(UNIY) x P(MC) x n
= (102 / 300)
x (120 /300)
x 300
P(UNIN) x P(WC) x n
= (198 / 300)
x (180 /300)
x 300
P(UNIN) x P(MC) x n
= (198 / 300)
x (120 /300)
x 300
Expected count in each category:
Working class
Go to
University
Do not go
to
University
(102 / 300) x (180 /300) x 300
Middle class
= .34 x .6 x 300 = 61.2
(102 / 300) x (120 /300) x 300
= .34 x .4 x 300 = 40.8
(198 / 300) x (180 /300) x 300
= .66 x .6 x 300 = 118.8
(198 / 300) x (120 /300) x 300
= .66 x .4 x 300 = 79.2
We have the actual count (I.e. from our data set):
Go to University
Do not go to
University
Working class
Middle class
18
84
162
36
And the expected count:
(I.e. the numbers we’d expect if we assume class & education to be independent of each other):
Working class
Middle class
Go to University
61.2
40.8
Do not go to
University
118.8
79.2
What does this table tell you?
Go to University
Working class
Middle class
18
84
61.2
40.8
162
36
118.8
79.2
Actual count
Expected count
Do not go to University
Actual count
Expected count
• It tells you that if class and education were indeed
independent of each other
• I.e. the outcome of one does not affect the chances of
outcome of the other
– Then you’d expect a lot more working class people in
the data to have gone to university than actually
recorded (61 people, rather than 18)
– Conversely, you’d expect far fewer middle class people
to have gone to university (half the number actually
recorded – 41 people rather than 80).
But remember, all this is based on a sample, not the entire
population…
• Q/ Is this discrepancy due to sampling
variation alone or does it indicate that we
must reject the assumption of
independence?
– To answer this within the standardised hypothesis
testing framework we need to know the chances of
false rejection
3. Chi-square test for independence
(non-parametric -- I.e. no presuppositions re distribution of variables;
sample size not relevant)
(1) H0: expected = actual  x & y are independent
» I.e. Prob(x) is not affected by whether or not y occurs;
H1: expected  actual  there is some relationship
»I.e. Prob(x) is affected by y occurring.
(2) a = 0.05
2


(
o

e
)
2
i
i

   
ei
i 1 

k = no. of categories
df  (r  1)(c  1)
oi = actual no. of sample observations
in the ith category
r = no. of rows in table
d = no. of parameters that have to be
estimated from the sample data.
k
c = no. of colums “ “
ei = expected (given H0) no. of sample
observations in the ith category
Chi-square distribution changes shape for different df:
(3) Reject H0 iff P < a
(4) Calculate P:
• P = Prob(2 > 2c)
– N.B. Chi-square tests are always an upper tail test
 2 Tables: are usually set up like a t-table with df down the
side, and the probabilities listed along the top row, with
values of 2c actually in the body of the table. So look up
2c in the body of the table for the relevant df and then find
the upper tail probability that heads that column.
– SPSS: - CDF.CHISQ(2c,df) calculates Prob(2 < 2c), so
use the following syntax:
» COMPUTE chi_prob = 1 - CDF.CHISQ(2c,df).
» EXECUTE.
Do a chi-square test on the following table:
Working class
Go to University
Middle class
18
84
61.2
40.8
162
36
118.8
79.2
Actual count
Expected count
Do not go to University
Actual count
Expected count
(1) H0: expected = actual
 class and Higher Education are independent
H1: expected  actual
 there is some relationship between class and
Higher Education
(2)
State the formula & calc 2 :
2


(
o

e
)
2
i
i

   
ei
i 1 

k
2 =
(
(18 - 61.2)2 / 61.2
+
(84 - 40.8)2/ 40.8
+
(162-118.8)2 / 118.8
+
(36 - 79.2)2/ 79.2
)
2
= ((18 - 61.2)2 / 61.2
+ (84 - 40.8)2/ 40.8 +
(162-118.8)2 /118.8 +
(36 - 79.2)2/ 79.2 )
= 30.49 + 45.74 + 15.71 + 23.56
= 115.51
df = (r-1)(c-1) = 1
Sig = P(2 > 115.51) = 0
(3) Reject H0 iff P < a
(4) Calculate P:
COMPUTE chi_prob = 1 - CDF.CHISQ(115.51,1).
EXECUTE.
Sig = P(2 > 115.51) = 0
 Reject H0
Caveat:
• As with the 2 proportions tests, the chi-square test is,
•
“an approximate method that becomes more accurate as
the counts in the cells of the table get larger” (Moore,
Basic Practice of Statistics, 2000, p. 485)
• Cell counts required for the Chi-square test:
• “You can safely use the chi-square test with critical values
from the chi-square 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 2x2 table should be 5 or greater”
(Moore, Basic Practice of Statistics, 2000, p. 485)
Example 2: Is there a relationship between whether a
borrower is a first time buyer and whether they live in
Durham or Cumberland?
• Only real problem is how do we calculate ei the expected number
of observations in category i?
– (I.e. number of cases expected in i assuming that the variables
are independent)
• the formula for ei is the probability of an observation falling into
category i multiplied by the total number of observations.
As noted earlier:
• Probability theory states that:
– if x and y are independent, then the probability of
events x and y simultaneously occurring is simply equal
to the product of the two events occurring:
Pr ob( x  y )  Pr ob( x)  Pr ob( y )
• But, if x and y are not independent, then:
Prob(x  y) = Prob(x)  Prob(y given that x has occurred)
• So, if FTBY or N and CountyD or C are independent (i.e.
assuming H0) then:
Prob(FTBY  CountyD) = Prob(FTBY)Prob(CountyD)
• so the expected number of cases for each of the four mutually
exclusive categories are as follows:
CountyC
FTBN
FTBY
CountyD
Pr(FTB N )  Pr(County C )
Pr( FTBN )  Pr(CountyD )
n
n
Pr( FTBY )  Pr(County C )
Pr( FTB Y )  Pr( County D )
n
n
Prob(FTBN) = Total no. cases FTBN  All observations
CountyC
FTBN
 Total FTB N   Total County C   Total FTB N   Total County D 


 


n
n
n
n

 
 

 
n
FTBY
CountyD
 Total FTBY

n

n
  Total CountyC   Total FTB Y    Total County D 


n
n
 

n
 
 
n
n
This gives us the expected count:
o
b
E
o
t
c
o
m
r
h
f
N
1
9
0
y
Y
9
1
0
T
0
0
0
To obtain this table in SPSS, go to Analyse, Descriptive
Statistics, Crosstabs, Cells, and choose expected count rather
than observed
u
n
o
u
t
c
o
m
r
h
t
f
N
C
i
3
4
7
y
E
1
9
0
Y
C
4
5
9
E
9
1
0
T
C
7
9
6
E
0
0
0
• What does this table tell you?
– Does it suggest that the probability of being an
FTB independent of location?
– Or does it suggest that the two factors are
contingent on each other in some way?
– Can it tell you anything about the direction of
causation?
– What about sampling variation?
2


(
o

e
)
2
i
i

   
ei
i 1 

k
u
n
o
u
t
c
o
m
r
h
t
f
N
C
i
3
4
7
y
E
1
9
0
Y
C
4
5
9
E
9
1
0
T
C
7
9
6
E
0
0
0
 (oi  ei ) 2 

   
ei
i 1 

 (203  197.1) 2   (154  159.9) 2   (104  109.9) 2   (95  89.1) 2 
  
  
  

 
197.1
159.9
109.9
89.1

 
 
 

k
2
 1.094071
Summary of Hypothesis test:
– (1)
– (2)
H0: FTB and County are independent
H1: there is some relationship
a = 0.05
2

(
o

e
)
i
 2    i
ei
i 1 
 1.094071
k
– (3) Reject H0 iff P < a
– (4) Calculate P:
• P = Prob(2 > 2c) = 0.29557
 df  (r  1)(c  1)

 (2  1)( 2  1)

1
Do not reject H0
I.e. if we were to reject H0, there would be a 1 in 3 chance of us rejecting it
incorrectly, and so we cannot do so. In other words, FTB and County are
independent.
Contingency Tables in SPSS:
• Click Cells button to select counts & %s
• If you select all three (row, column and total), you will end up
with:
u
n
o
u
t
c
u m
o
r
h
t
f
N
C
i r
3
4
7
y
=
%
%
%
%
b
%
%
%
%
C
%
%
%
%
Y
C
4
5
9
%
%
%
%
b
%
%
%
%
C
%
%
%
%
T
C
7
9
6
%
%
%
%
b
%
%
%
%
C
%
%
%
%
• Click the Statistics button to choose which stats you
want.
• If you click Chi-square, the results of a range of tests
will be listed…
We have been calculating the Pearson Chi-square:
Chi-Square Tests
Pearson Chi-Square
Continuity Correction a
Likelihood Ratio
Fis her's Exact Test
Linear-by-Linear
As sociation
N of Valid Cases
Value
1.094b
.916
1.093
1.092
df
1
1
1
1
As ymp. Sig.
(2-sided)
.296
.339
.296
Exact Sig.
(2-sided)
Exact Sig.
(1-sided)
.328
.169
.296
556
a. Computed only for a 2x2 table
b. 0 cells (.0%) have expected count less than 5. The minimum expected count is
89.12.
4. For further study:
• The Pearson Chi square test only tests for the
existence of a relationship
• It tells you little about the strength of the relationship
• SPSS includes a raft of measures that try to measure
the level of association between categorical variables.
• Click on the name of one of the statistics and SPSS will
give you a brief definition (see below)
• In the lab exercises, take a look at these statistics and
copy and paste the definitions along side your answers
– Right click on the definition and select Copy. Then open up a
Word document and paste along with your output.
Nominal variables:
•
Contingency coefficient:
• “A measure of association based on chi-square. The value ranges between zero and 1, with
zero indicating no association between the row and column variables and values close to 1
indicating a high degree of association between the variables. The maximum value possible
depends on the number of rows and columns in a table.”
•
Phi and Cramer’s V:
• “Phi is a chi-square based measure of association that involves dividing the chi-square
statistic by the sample size and taking the square root of the result. Cramer's V is a measure
of association based on chi-square.”
•
Lambda:
• “A measure of association which reflects the proportional reduction in error when values of
the independent variable are used to predict values of the dependent variable. A value of 1
means that the independent variable perfectly predicts the dependent variable. A value of 0
means that the independent variable is no help in predicting the dependent variable.”
•
Uncertainty coefficient:
• “A measure of association that indicates the proportional reduction in error when values of
one variable are used to predict values of the other variable. For example, a value of 0.83
indicates that knowledge of one variable reduces error in predicting values of the other
variable by 83%. The program calculates both symmetric and asymmetric versions of the
uncertainty coefficient.”
Ordinal Variables:
•
Gamma:
• A symmetric measure of association between two ordinal variables that ranges between
negative 1 and 1. Values close to an absolute value of 1 indicate a strong relationship
between the two variables. Values close to zero indicate little or no relationship. For 2-way
tables, zero-order gammas are displayed. For 3-way to n-way tables, conditional gammas
are displayed.
•
Somers’ d:
• “A measure of association between two ordinal variables that ranges from -1 to 1. Values
close to an absolute value of 1 indicate a strong relationship between the two variables,
and values close to 0 indicate little or no relationship between the variables. Somers' d is
an asymmetric extension of gamma that differs only in the inclusion of the number of pairs
not tied on the independent variable. A symmetric version of this statistic is also calculated.”
•
Kendall’s tau-b:
• “A nonparametric measure of association for ordinal or ranked variables that take ties into
account. The sign of the coefficient indicates the direction of the relationship, and its
absolute value indicates the strength, with larger absolute values indicating stronger
relationships. Possible values range from -1 to 1, but a value of -1 or +1 can only be
obtained from square tables.”
•
Kendall’s tau-c:
• “A nonparametric measure of association for ordinal variables that ignores ties. The sign of
the coefficient indicates the direction of the relationship, and its absolute value indicates the
strength, with larger absolute values indicating stronger relationships. Possible values
range from -1 to 1, but a value of -1 or +1 can only be obtained from square tables.”
Correlations:
• Pearson correlation coefficient r
• “a measure of linear association between two variables”
• Spearman correlation coefficient
• “a measure of association between rank orders. Values of
both range between -1 (a perfect negative relationship)
and +1 (a perfect positive relationship). A value of 0
indicates no linear relationship.”
When you have a dependent variable measured on an
interval scale & an independent variable with a limited
number of categories:
• Eta:
– “A measure of association that ranges from 0 to 1, with 0 indicating no
association between the row and column variables and values close to
1 indicating a high degree of association. Eta is appropriate for a
dependent variable measured on an interval scale (e.g., income) and
an independent variable with a limited number of categories (e.g.,
gender). Two eta values are computed: one treats the row variable as
the interval variable; the other treats the column variable as the interval
variable.”