IM911
Quantitative Research Methods
Bivariate Analysis:
Cross-tabulations and Chi-square
Thursday 18th February 2016
A reminder...
• Hypothesis testing involves testing the NULL HYPOTHESIS that there
is no difference/effect/relationship, e.g. variables are unrelated.
• In testing such a hypothesis, we test whether the relationship or
difference that we have observed in our sample is likely to have
occurred assuming the null hypothesis is true for the population.
• The sampling distribution (a distribution reflecting the full range of
possible samples) plays an important role in inferential statistics, by
allowing us to work out how much variation is likely to have been
produced by sampling error (e.g. if levels of happiness vary a lot,
then small gender differences are quite likely to reflect sampling error).
• When we find a very low probability (p<0.05, or less than 5%) that we
would have found what we found in our sample if the null hypothesis
were true, we can infer that it is unlikely to be true.
• And therefore, we can infer that the ALTERNATIVE HYPOTHESIS
(saying that there is a difference/effect) is likely to be true.
• This is a sort of backwards logic. So… if you find it easier to think
forwards, the simpler version (although less correct, technically) is that
if we find that p<0.05 then we have identified a relationship/effect.
So far...
So far we have examined at tests that have had interval-level
variables (such as income, or years spent at an address) as
dependent variables.
Specifically we looked at tests that investigated:
• Whether a population has a mean that is different from a
suggested mean (z-tests, or ‘one-sample t-tests’ in SPSS).
e.g. Do people on average stay at an address for 10 years?
• Whether two groups have population means that differ from one
another (t-tests, or ‘independent samples t-test’ in SPSS). e.g. Is
there a gender difference in the mean time spent at one’s current
address?
• Whether the different categories of a variable (e.g. social class)
have population means that differ in some way (ANOVA, or ‘one
way ANOVA’ in SPSS). e.g. Do people in different social classes
have different average lengths of time at their addresses?
Categorical data analysis
• Today we are going to look at relationships between categorical
variables (e.g. gender, ‘race’, religious denomination).
• When both variables are categorical we cannot produce means.
Instead we construct contingency tables that show the frequency
with which cases fall into each combination of categories – e.g.
‘man’ and ‘Christian’ (we cannot do this directly with continuous
variables, e.g. age, as people often fall into numerous categories,
leading to tables that are enormous and unmanageable).
• When we conduct statistical analyses of cross-tabulated data, we
are trying to work out whether there is any systematic
relationship between the variables being analyzed or whether
any ‘patterns’ are ‘random’, i.e. only reflect sampling error.
• Therefore the tests that we do compare what we find (observe)
with what would be expected if there were no relationship – i.e.
given the null hypothesis of no relationship in the population.
From: Phoenix, A. 1991. Young Mothers? Cambridge: Polity Press.
Table 3.2 in Chapter 3: ‘How the Women Came to be Mothers’ (p61)
MARITAL STATUS (AT CONCEPTION) by ORIENTATION TO PREGNANCY
Wanted to Did not
conceive
mind
Had not
thought
about it
Important TOTAL
not to
Single
4 (8%)
Cohabiting
4 (44%)
2 (22%)
1 (11%)
2 (22%)
9
Married
9 (53%)
6 (35%)
0 (0%)
2 (12%)
17
TOTAL
17
12 (23%) 13 (25%) 24 (45%)
20
14
28
53
79
From: Jupp, P. 1993. ‘Cremation or burial?
Contemporary choice in city and village’. In Clark, D.
(ed.) The Sociology of Death. Oxford: Blackwell.
(Derived from Tables 5 and 6 on pages 177 and 178).
OCCUPATIONAL CLASS by DISPOSAL CHOICE
Cremation
Burial
TOTAL
Working class
20 (59%)
14 (41%)
34
Middle class
21 (88%)
3 (13%)
24
TOTAL
41
17
58
What can be learned from these
cross-tabulations?
Do you think that the patterns in the
MARITAL STATUS by ORIENTATION TO PREGNANCY
and
OCCUPATIONAL CLASS by DISPOSAL CHOICE
cross-tabulations provide sufficient evidence
to conclude that relationships exist?
Where can the relationship, if any, be found in the
MARITAL STATUS (AT CONCEPTION)
by ORIENTATION TO PREGNANCY
cross-tabulation?
How do you work out whether a gender
difference is likely to be due to chance?
We use chi-square (2) to look at the
difference between what we observe
and what would be likely if there were
no difference except that generated by
chance (i.e. sampling error):
2 =

(Observedij – Expectedij)2
Expectedij
•
•
•
•
•
It may or may not be clear from
this slide that the above formula
simply (or not so simply?!)
represents the process
described to the right, so
working through an example
may help...
For each cell in the table: we work out
the frequency that we would expect and
see how much the observed frequency
differs from this.
We then square this difference and
divide by the expected frequency.
We then sum these values.
The observed frequency for each cell is
what we see.
The expected frequency is the
frequency that you would get in each cell
if men and women were exactly as likely
as each other to fall into each of the
categories of the other variable.
DEGREE SUBJECT by GENDER:
‘BLACK ‘ GRADUATES.
Subject
Arts
Sciences
Social Sciences
Education
Male
16 (47%)
29 (58%)
42 (53%)
3 (19%)
Female
18 (53%)
21 (42%)
38 (47%)
13 (81%)
Total
34
50
80
16
TOTAL
90 (50%)
90 (50%)
180
The above table is based on a random sample of 180 ‘Black’ graduates
born in the UK and aged 25-34 in 1991. (Data adapted from the 1991
Census SARs; ‘Black’ includes Black-African, Black-Caribbean and
Black-Other).
‘Expected’ frequencies
Subject
Male
Female
Arts
Sciences
Social Sciences
Education
17
25
40
8
17
25
40
8
TOTAL
90 (50%)
(50%)
(50%)
(50%)
(50%)
Total
(50%)
(50%)
(50%)
(50%)
34
50
80
16
90 (50%)
180
Differences (‘Observed’ minus ‘Expected’)
Subject
Male
Female
Total
Arts
Sciences
Social Sciences
Education
-1
4
2
-5
1
-4
-2
5
0
0
0
0
TOTAL
0
0
0
Squared differences
Subject
Arts
Sciences
Social Sciences
Education
Male
Female
1
16
4
25
1
16
4
25
... divided by ‘Expected’ values and summed
Subject
Arts
Sciences
Social Sciences
Education
Male
Female
1/17 = 0.06
16/25 = 0.64
4/40 = 0.10
25/8 = 3.13
1/17 = 0.06
16/25 = 0.64
4/40 = 0.10
25/8 = 3.13
0.06 + 0.06 + 0.64 + 0.64 + 0.10 + 0.10 + 3.13 + 3.13 = 7.86
7.86 is the value of the chi-square statistic (2) for the original table
(cross-tabulation).
Degrees of freedom
If a cross-tabulation has R rows and C columns, the corresponding chisquare statistic has (R-1) x (C-1) ‘degrees of freedom’
i.e. in this case it has (4 - 1) x (2 - 1) = 3 degrees of freedom.
Degrees of freedom can be thought of as sources of variation. In an
independent samples t-test, the number of sources of variation depends
on the numbers of cases in the samples being compared. In a chisquare test, however, it depends on the number of cells in the crosstabulation being examined.
Chi-square distributions for
1 to 5 degrees of freedom
Here, chi-square
values of more than 6
are relatively rare.
However higher
values of chi-square
are more common
when the degrees of
freedom (k in this
chart) is larger.
Chi-square tables
reflect this: it can be
seen in these that the
values for p=0.05 (the
point at which only
5% of cases lie to the
right) increase as the
degrees of freedom
increase.
Checking the chi-square statistic
• To check whether a result is statistically significant (i.e. unlikely to simply
reflect sampling error) we look up critical values of chi-square in a table.
• For 3 d.f. the critical values are 7.81 (p=0.05) and 11.34 (p=0.01).
• Because our chi-square value (7.86) is bigger than 7.81 we can say that it
is significant at p<0.05. (Since it is not bigger than 11.34 we cannot say
that it is significant at p<0.01).
• What this means is that we would find a difference in our sample of at least
the magnitude of the difference that we observed in the distributions of
proportions for men and for women between 1% and 5% of the time if
there were no real difference between the distributions of proportions in the
population (which is the null hypothesis).
• This is rare enough that we consider the null hypothesis to be unlikely. We
can therefore reject the null hypothesis and accept the alternative
hypothesis of a relationship between gender and subject.
• When you use SPSS, it produces an estimate of the precise probability of
obtaining a chi-square statistic at least as big as (in this case) 7.86 by
chance. A p-value of less than 0.05 is said to be statistically significant
(and to imply a significant relationship between the two variables).
A note on using chi-square
• Chi-square tests can be safely used if each cell in the
table has an expected value of at least 5. Opinions
differ as to what is appropriate where this is not the
case: one ‘rule of thumb’ is that no expected values
should be less than 1 and no more than 20% of
expected values should be less than 5.
• The categories must be discrete, i.e. mutually
exclusive. No case should fall into more than one
category (which might pose some difficulties when
carrying out analyses focusing on degree subjects!)
Chi-square statistics
MARITAL STATUS (AT CONCEPTION)
26 = 24.86
by ORIENTATION TO PREGNANCY
(p < 0.001)
(but the sparseness of cases means that the chi-square statistic is invalid!)
OCCUPATIONAL CLASS
21 = 5.58
by DISPOSAL CHOICE
(p < 0.05)
(but the chi-square statistic for a 2x2 cross-tabulation needs – arguably –
to be adjusted using ‘Yates’ correction for continuity’, giving an adjusted
value of 4.29 (p <0.05))
GENDER by SUBJECT:
OTHER GRADUATES
23 = 1542
(p < 0.0001)
Marital Status (at Conception) by Orientation to Pregnancy:
Collapsed/reduced versions of cross-tabulation
Single
Cohab./Married
Wanted to Did not Had not Important
conceive
mind
thought + not to
about it
4 (8%) 12 (23%)
37 (70%)
13 (50%) 8 (31%)
5 (19%)
22 = 23.46 (p < 0.0001)
Cohabiting
Married
Wanted to Did not Had not Important
conceive
mind
thought not to
about it
4 (44%) 2 (22%) 1 (11%) 2 (22%)
9 (53%) 6 (35%) 0 (0%)
2 (12%)
23 = 2.72 (p > 0.05) [N.B. ‘Sparse’ table: chi-square invalid]
Strength of association
• Chi-square tells us whether there is a ‘significant’
relationship between two variables (or whether a
relationship exists that would have been unlikely to have
been found by chance).
• However it does not tell us in a clear-cut way how strong
this association is, since the size of the chi-square statistic
depends in part on the sample size (as well as the crosstabulation shape)
• We will therefore look at a different measure (which we
‘met’ in Week 3) that does tell us about the strength of
association:
• Cramér’s V. This is a chi-square-based measure that tells
us the strength of association in a cross-tabulation. No
association is represented by 0; ‘perfect’ association by 1.
Example: Sex, Age and Sport
Data from Young People’s Social Attitudes Study 2003 (available from Nesstar!!)
a
YP played sport as part of sports club YP * Age band Crosstabulation
Boys:
YP played sport as part
of s ports club YP
Yes
No
Total
Count
% within Age band
Count
% within Age band
Count
% within Age band
12-13
62
65.3%
33
34.7%
95
100.0%
Age band
14-16
68
54.4%
57
45.6%
125
100.0%
17-19
34
41.0%
49
59.0%
83
100.0%
Total
164
54.1%
139
45.9%
303
100.0%
a. YP sex hous ehold grid [BSA2003] YP22 = Male
a
YP played sport as part of sports club YP * Age band Crosstabulation
Girls:
YP played sport as part
of s ports club YP
Yes
No
Total
Count
% within Age band
Count
% within Age band
Count
% within Age band
a. YP sex hous ehold grid [BSA2003] YP22 = Female
12-13
55
44.4%
69
55.6%
124
100.0%
Age band
14-16
48
32.0%
102
68.0%
150
100.0%
17-19
11
12.8%
75
87.2%
86
100.0%
Total
114
31.7%
246
68.3%
360
100.0%
What can we say about these tables?
•
•
•
It looks like boys play sport as part of a club
more than girls do.
And it looks like both boys and girls become
less likely to play sport as part of a club as
they get older.
But is the association between age and
sports club membership stronger/weaker for
boys than it is for girls?
Does age significantly affect sports club
membership for boys?
a
YP played sport as part of sports club YP * Age band Crosstabulation
YP played sport as part
of s ports club YP
Yes
No
Total
Count
% within Age band
Count
% within Age band
Count
% within Age band
12-13
62
65.3%
33
34.7%
95
100.0%
Age band
14-16
68
54.4%
57
45.6%
125
100.0%
17-19
34
41.0%
49
59.0%
83
100.0%
Total
164
54.1%
139
45.9%
303
100.0%
a. YP sex hous ehol d gri d [BSA2003] YP22 = Male
To answer this question we work out chi-square, by calculating:
χ2=
=

(Observedij – Expectedij)2
Expectedij
(62 – (95*164)/303)2 + (33 – (95*139)/303)2 + (68 – (125*164)/303)2 +
(95*164)/303
(95*139)/303
(125*164)/303
(57 – (125*139)/303)2 + (34 – (83*164)/303)2 + (49 – (83*139)/303)2 +
(125*139)/303
(83*164)/303
(83*139)/303
=
(62-51.4)2/51.4 + (33-43.6)2/43.6 + (68-67.7)2/67.7 + (57-57.3)2/57.3 +
(34-44.9)2/44.9 + (49-38.1)2/38.1
=
2.2 +2.6 + 0 + 0 +2.6 + 3.1
= 10.5
Does age significantly affect sports club
membership for boys?
• Chi-square = 10.5.
• d.f. = (r-1) x (c-1) = 1 x 2 = 2
• If we look up the .05 value for 2 degrees of freedom it is 5.99, and the
.01 value is 9.21. Since 10.5 is bigger than both of these, it is
significant at (p < 0.01).
• The following SPSS output confirms that, in fact, the p-value is .005,
which is less than 0.01 (N.B. the chi-square value
without rounding
Chi-Square Testsb
is 10.541).
Pears on Chi-Square
Likelihood Ratio
Linear-by-Linear
Ass ociation
N of Valid Cases
Value
10.541a
10.626
10.457
2
2
Asymp. Sig.
(2-s ided)
.005
.005
1
.001
df
303
a. 0 cells (.0%) have expected count les s than 5. The
minimum expected count is 38.08.
b. YP s ex hous ehold grid [BSA2003] YP22 = Male
And for girls…?
• Work out the chi-square statistic and test
whether it is significant for girls…
a
YP played sport as part of sports club YP * Age band Crosstabulation
YP played sport as part
of s ports club YP
Yes
No
Total
Count
% within Age band
Count
% within Age band
Count
% within Age band
a. YP sex hous ehold grid [BSA2003] YP22 = Female
12-13
55
44.4%
69
55.6%
124
100.0%
Age band
14-16
48
32.0%
102
68.0%
150
100.0%
17-19
11
12.8%
75
87.2%
86
100.0%
Total
114
31.7%
246
68.3%
360
100.0%
And for girls…?
• The SPSS output for the chi-square test for girls shows a chisquare value of 23.394.
• The p-value for this chi-square statistic ,with 2 degrees of freedom
is rounded to 0.000 (SPSS only shows you results to 3 decimal
places). This means that it is less than 0.001 (or, more precisely,
that it is less than 0.0005).
• Therefore “Age has a significant effect on whether or not girls play
sport in clubs (p < 0.001)”:
Chi-Square Testsb
Pears on Chi-Square
Likelihood Ratio
Linear-by-Linear
Ass ociation
N of Valid Cases
Value
23.394a
25.370
22.862
2
2
Asymp. Sig.
(2-s ided)
.000
.000
1
.000
df
360
a. 0 cells (.0%) have expected count les s than 5. The
minimum expected count is 27.23.
b. YP s ex hous ehold grid [BSA2003] YP22 = Female
Is the effect of age stronger/weaker
for boys than it is for girls?
• The chi-square statistic is bigger for girls than for boys, however
the sample of girls is also bigger (360 as compared to 303) so this
will have affected the relative size of the two values.
• To work out the strength of association, we need to correct for
both sample size and for the table shape (since this also affects
the magnitude of chi-square statistics). A frequently-used
measure of association is Cramér’s V:
2
N ( L  1)
where 2 is chi-square,
N is the sample size,
and L is the lesser (smaller) of the number of rows and number of columns.
Note:
In any table where either the number of rows or the number of columns is equal to 2,
Cramér’s V is equal to another measure of association, referred to as phi (or Φ).
Comparing strength of association between age
and involvement in sport for boys and for girls
a
YP played sport as part of sports club YP * Age band Crosstabulation
Boys:
YP played sport as part
of s ports club YP
χ2 = 10.541
Yes
No
Total
Count
% within Age band
Count
% within Age band
Count
% within Age band
12-13
62
65.3%
33
34.7%
95
100.0%
Age band
14-16
68
54.4%
57
45.6%
125
100.0%
17-19
34
41.0%
49
59.0%
83
100.0%
Total
164
54.1%
139
45.9%
303
100.0%
a. YP sex hous ehold grid [BSA2003] YP22 = Male
a
YP played sport as part of sports club YP * Age band Crosstabulation
Girls :
YP played sport as part
of s ports club YP
χ2 = 23.394
Yes
No
Total
Count
% within Age band
Count
% within Age band
Count
% within Age band
12-13
55
44.4%
69
55.6%
124
100.0%
Age band
14-16
48
32.0%
102
68.0%
150
100.0%
17-19
11
12.8%
75
87.2%
86
100.0%
Total
114
31.7%
246
68.3%
360
100.0%
a. YP sex hous ehold grid [BSA2003] YP22 = Female
Cramér’s V values for the two tables are therefore:
2
N ( L  1)
Boys
10.5
 0.186
303(2  1)
Girls
23.9
 0.258
360(2  1)
Cramér’s V in SPSS
Symmetric Measuresc
• We can also see
Cramér’s V in SPSS
output
• The value for boys is
above and that for girls
is below – note that the
values of Cramér’s V
are the same as those
we worked out (0.186
and 0.258), with small
differences due to
rounding error.
Nominal by
Nominal
Phi
Cramer's V
N of Valid Cas es
Value
.187
.187
303
Approx. Sig.
.005
.005
a. Not as s uming the null hypothes is.
b. Using the as ymptotic standard error as suming the null
hypothesis .
c. YP sex hous ehold grid [BSA2003] YP22 = Male
Symmetric Measuresc
Nominal by
Nominal
Phi
Cramer's V
N of Valid Cas es
Value
.255
.255
360
Approx. Sig.
.000
.000
a. Not as s uming the null hypothes is.
b. Using the as ymptotic standard error as suming the null
hypothesis .
c. YP sex hous ehold grid [BSA2003] YP22 = Female
What do the results mean
substantively?
• We can say that age has a significant effect on boys’
participation in sport.
• And that age has a significant and somewhat stronger
effect on girls’ participation in sport.
• Hence both girls and boys are likely to decrease their
participation in sports clubs as they get older but this
effect is more pronounced among girls than among boys.
• There is thus a (small-ish) gender difference in the
relationship between age and participation in sport.
Two more Cramér’s V values…
Subject and Gender: ‘Black’ graduates =
Subject and Gender: Other graduates
=
√
√
7.86
180(2-1)
= 0.209
1542
=
17094(2-1)
0.300
But could this difference just reflect sampling error?
Log-linear model: Test for difference between form of
Subject/Gender relationship for ‘Black’ graduates and for Other
graduates:
23 = 3.67 (p > 0.05)
i.e. Not enough evidence to conclude subject ‘gendering’ varies
between ‘Black’ graduates and Other graduates