Cross Tabs and Chi-Squared
Testing for a Relationship
Between Nominal/Ordinal
Variables
Cross Tabs and Chi-Squared
The test you choose depends on level of measurement:
Independent
Dependent
Statistical Test
Dichotomous
Interval-ratio
Dichotomous
Independent Samples t-test
Nominal
Dichotomous
Nominal
Dichotomous
Cross Tabs
Nominal
Dichotomous
Interval-ratio
Dichotomous
ANOVA
Interval-ratio
Dichotomous
Interval-ratio
Correlation and OLS Regression
Cross Tabs and Chi-Squared
We are asking whether there is a relationship
between two nominal (or ordinal) variables—this
includes dichotomous variables.
(Even though one may use cross tabs for ordinal
variables, it is generally better to treat them as
interval variables and use more powerful
statistical techniques whenever you can.)
Cross Tabs and Chi-Squared
Cross tabs and Chi-Squared will tell you whether classification on one
nominal or ordinal variable is related to classification on a second
nominal or ordinal variable.
For Example:
Are rural Americans more likely to vote Republican in presidential races
than urban Americans?
Classification of Region
Party Vote
Are white people more likely to drive SUV’s than blacks or Hispanics?
Race
Type of Vehicle
Cross Tabs and Chi-Squared
The statistical focus will be on the number
of people in a sample who are classified
in patterned ways on two variables.
Why?
Means and standard deviations are
meaningless for nominal variables.
Cross Tabs and Chi-Squared
The procedure starts with a “cross classification” of
the cases in categories of each variable.
Example:
Data on male and female support for SJSU football
from 650 students put into a matrix
Female:
Male:
Total:
Yes
185
80
265
No
200
65
265
Maybe
65
55
120
Total
450
200
650
Cross Tabs and Chi-Squared
In the example, I can see that the campus is divided on the
issue.
But are there associations between sex and attitudes?
Example:
Data on male and female support for SJSU football from
650 students put into a matrix
Female:
Male:
Total:
Yes
185
80
265
No
200
65
265
Maybe
65
55
120
Total
450
200
650
Cross Tabs and Chi-Squared
But are there associations between sex and attitudes?
An easy way to get more information is to convert the
frequencies to percentages.
Example:
Data on male and female support for SJSU football from 650
students put into a matrix
Yes
No
Female: 185
(41%)
200
(44%)
Male:
80
(40%)
65
(33%)
Total:
265
(41%)
265
(41%)
*percentages do not add to 100 due to rounding
Maybe
65
55
120
(14%)
(28%)
(18%)
Total
450
200
650
(99%)*
(101%)
(100%)
Cross Tabs and Chi-Squared
We can see that in the sample men are less likely to
oppose football, but no more likely to say “yes”
than women—men are more likely to say “maybe”
Example:
Data on male and female support for SJSU football
from 650 students put into a matrix
Yes
No
Maybe
Female: 185
(41%) 200
(44%) 65
Male:
80
(40%) 65
(33%) 55
Total:
265
(41%) 265
(41%) 120
*percentages do not add to 100 due to rounding
(14%)
(28%)
(18%)
Total
450
200
650
(99%)*
(101%)
(100%)
Cross Tabs and Chi-Squared
Data on male and female support for SJSU football from 650 students put
into a matrix
Yes
No
Female: 185
(41%)
200
(44%)
Male:
80
(40%)
65
(33%)
Total:
265
(41%)
265
(41%)
*percentages do not add to 100 due to rounding
Maybe
65
55
120
(14%)
(28%)
(18%)
Total
450
200
650
(99%)*
(101%)
(100%)
Using percentages to describe relationships is valid statistical analysis: These are
descriptive statistics! However, they are not inferential statistics.
What can we say about the population?
Could we have gotten sample statistics like these from a population where there is no
association between sex and attitudes about starting football?
This is where the Chi-Squared Test of Independence comes in handy.
Cross Tabs and Chi-Squared
The whole idea behind the Chi-Squared test of
independence is to determine whether the
patterns of frequencies in your cross
classification table could have occurred by
chance, or whether they represent systematic
assignment to particular cells.
For example, were women more likely to answer
“no” than men or could the deviation in
responses by sex have occurred because of
random sampling or chance alone?
Cross Tabs and Chi-Squared
A number called Chi-Squared, 2, tells us whether the
numbers in our sample deviate from what would be
expected by chance.
It’s formula:
2 =  ((fo - fe)2 / fe)
fo= observed frequency in each cell
fe= expected frequency in each cell
A bigger 2 will result as our sample data deviates more
and more from what would be expected by chance.
A big 2 will imply that there is a relationship between our
two nominal variables.
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
Calculating 2 begins with the concept of a deviation of
observed data from what is expected by chance alone.
Deviation in 2 = Observed frequency – Expected frequency
Observed frequency is just the number of cases in each cell of the
cross classification table. For example, 185 women said “yes,” they
support football at SJSU. 185 is the observed frequency.
Expected frequency is the number of cases that would be in a cell of
the cross classification table if people in each group of one variable
had a propensity to answer the same as each other on the second
variable.
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
Data on male and female support for SJSU football from 650 students
Yes
No
Maybe
Female:
185
200
65
Male:
80
65
55
Total:
265
265
120
Total
450
200
650
Expected frequency (if our variables were unrelated):
Since females comprise 69.2% of the sample, we’d expect 69.2% of
the “Yes” answers to come from females, 69.2% of the “No” answers
to come from females, and 69.2% of the “Maybe” answers to come
from females. On the other hand, 30.8% of the “Yes,” “No,” and
“Maybe” answers should come from Men.
Therefore, to calculate expected frequency for each cell you do this:
fe = cell’s row total / table total * cell’s column total or
fe = cell’s column total / table total * cell’s row total
The idea is that you find the percent of persons in one category on the
first variable, and “expect” to find that percent of those people in the
other variable’s categories.
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
Data on male and female support for SJSU football from 650 students
Yes
No
Maybe
Female:
185
200
65
Male:
80
65
55
Total:
265
265
120
Now you know how to calculate the expected frequency (and the
observed frequency is obvious).
fe1 = (450/650) * 265 = 183.5
fe4 = (200/650) * 265 = 81.5
fe2 = (450/650) * 265 = 183.5
fe5 = (200/650) * 265 = 81.5
fe3 = (450/650) * 120 = 83.1
fe6 = (200/650) * 120 = 36.9
You already saw how to calculate the deviations too.
Dc = fo – fe
D1 = 185 – 183.5 = 1.5
D4 = 80 – 81.5 = -1.5
D2 = 200 – 183.5 = 16.5
D5 = 65 – 81.5 = -16.5
D3 = 65 – 83.1 = -18.1
D4 = 55 – 36.9 = 18.1
Total
450
200
650
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
Data on male and female support for SJSU football from 650 students
Yes
No
Maybe
Total
Female:
185
200
65
450
Male:
80
65
55
200
Total:
265
265
120
650
Deviations:
Dc = fo – fe
D1 = 185 – 183.5 = 1.5
D2 = 200 – 183.5 = 16.5
D3 = 65 – 83.1 = -18.1
D4 = 80 – 81.5 = -1.5
D5 = 65 – 81.5 = -16.5
D4 = 55 – 36.9 = 18.1
Now, we want to add up the deviations…
What would happen if we added these deviations together?
To get rid of negative deviations, we square each one (like in computing
standard deviations).
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
Data on male and female support for SJSU football from 650 students
Yes
No
Maybe
Female:
185
200
65
Male:
80
65
55
Total:
265
265
120
Deviations:
Dc = fo – fe
D1 = 185 – 183.5 = 1.5
D2 = 200 – 183.5 = 16.5
D3 = 65 – 83.1 = -18.1
Total
450
200
650
D4 = 80 – 81.5 = -1.5
D5 = 65 – 81.5 = -16.5
D4 = 55 – 36.9 = 18.1
To get rid of negative deviations, we square each one (like in standard
deviations).
(D1)2 = (1.5)2 =
2.25
(D4)2 = (-1.5)2 =
2.25
(D2)2 = (16.5)2 = 272.25
(D5)2 = (-16.5)2 = 272.25
(D3)2 = (-18.1)2 = 327.61
(D6)2 = (18.1)2 = 327.61
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
Squared Deviations:
(D1)2 = (1.5)2 =
2.25
(D2)2 = (16.5)2 = 272.25
(D3)2 = (-18.1)2 = 327.61
(D4)2 = (-1.5)2 =
2.25
(D5)2 = (-16.5)2 = 272.25
(D6)2 = (18.1)2 = 327.61
Just how large is each of these squared deviations?
The next step is to give the deviations a “metric.” The deviations are compared relative
to the what was expected. In other words, we divide by what was expected.
You’ve already calculated what was expected in each cell:
fe1 = (450/650) * 265 = 183.5
fe4 = (200/650) * 265 = 81.5
fe2 = (450/650) * 265 = 183.5
fe5 = (200/650) * 265 = 81.5
fe3 = (450/650) * 120 = 83.1
fe6 = (200/650) * 120 = 36.9
Relative Deviations-squared—Small values indicate little deviation from what was
expected, while larger values indicate much deviation from what was expected:
(D1)2 / fe1 = 2.25 / 183.5 = 0.012
(D4)2 / fe4 = 2.25 / 81.5 = 0.028
(D2)2 / fe2 = 272.25 / 183.5 = 1.484
(D5)2 / fe5 = 272.25 / 81.5 = 3.340
(D3)2 / fe3 = 327.61 / 83.1 = 3.942
(D6)2 / fe6 = 327.61 / 36.9 = 8.878
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
Relative Deviations-squared—Small values indicate little deviation from what was
expected, while larger values indicate much deviation from what was expected:
(D1)2 / fe1 = 2.25 / 183.5 = 0.012
(D4)2 / fe4 = 2.25 / 81.5 = 0.028
(D2)2 / fe2 = 272.25 / 183.5 = 1.484
(D5)2 / fe5 = 272.25 / 81.5 = 3.340
(D3)2 / fe3 = 327.61 / 83.1 = 3.942
(D6)2 / fe6 = 327.61 / 36.9 = 8.878
The next step will be to see what the total relative deviations-squared are:
Sum of
Relative Deviations-squared = 0.012 + 1.484 + 3.942 + 0.028 + 3.340 + 8.878 = 17.684
This number is also what we call Chi-Squared or 2.
So…
Of what good is knowing this number?
2 =  ((fo - fe)2 / fe)
Cross Tabs and Chi-Squared
This value, 2, would form an identifiable
shape in repeated sampling if the two
variables were unrelated to each other.
That shape depends only on the number of
rows and columns. We technically refer to
this as the “degrees of freedom.”
For 2, df =(#rows – 1)*(#columns – 1)
Cross Tabs and Chi-Squared
For 2, df =(#rows – 1)*(#columns – 1)
2 distributions:
df = 5
FYI:
df = 10
df = 20
df = 1
1
5
10
20
This should remind you of
the normal distribution,
except that, it changes
shape depending on the
nature of your variables.
Cross Tabs and Chi-Squared
Think of the Power!!!!
We can use the known
properties of the 2
distribution to identify the
probability that we would
get our sample’s 2 if our
variables were unrelated!
This is exciting!
Cross Tabs and Chi-Squared
5% of 2 values
If our 2 in a particular analysis were under the
shaded area or beyond, what could we say
about the population given our sample?
Cross Tabs and Chi-Squared
5% of 2 values
Answer: We’d reject the null, saying that it is highly unlikely
that we could get such a large chi-squared value from a
population where the two variables are unrelated.
Cross Tabs and Chi-Squared
5% of 2 values
So, what is the critical 2 value?
Cross Tabs and Chi-Squared
That depends on the particular problem because the
distribution changes depending on the number of rows
and columns.
df = 5
df = 10
df = 20
df = 1
1
5
10
20
Critical 2 ‘s
Cross Tabs and Chi-Squared
df = 1, critical 2 = 3.84
df = 5, critical 2 = 11.07
df = 10, critical 2 = 18.31
df = 20, critical 2 = 31.41
According to Table C,
with -level = .05, if:
df = 5
df = 10
df = 20
df = 1
1
5
10
20
Cross Tabs and Chi-Squared
In our football problem above, we had a chi-squared of 17.68 in a cross
classification table with 2 rows and 3 columns.
Our chi-squared distribution for that table would have
df = (2 – 1) * (3 – 1) = 2.
According to Table C, with -level = .05, Critical Chi-Squared is: 5.99.
Since 17.68 > 5.99, we reject the null.
We reject that our sample could have come from a population where
sex was not related to attitudes toward football.
Cross Tabs and Chi-Squared
Now let’s get formal…
7 steps to Chi-squared test of independence:
1. Set -level (e.g., .05)
2. Find Critical 2 (depends on df and -level)
3. The null and alternative hypotheses:
Ho: The two nominal variables are independent
Ha: The two variables are dependent on each other
4. Collect Data
5. Calculate 2 : 2 =  ((fo - fe)2 / fe)
6. Make decision about the null hypothesis
7. Report the P-value
Cross Tabs and Chi-Squared
Afterwards, what have you found?
If Chi-Squared is not significant, your variables are unrelated.
If Chi-Squared is significant, your variables are related.
That’s All!
Chi-Squared cannot tell you anything like the strength or direction of
association. For purely nominal variables, there is no “direction” of
association.
Chi-Squared is a large-sample test. If dealing with small samples, look
up appropriate tests. (A condition of the test: no expected frequency
lower than 5 in each cell)
The larger the sample size, the easier it is for Chi-Squared to be
significant.
2 x 2 table Chi-Square gives same result as Independent Samples ttest for proportion and ANOVA.
Cross Tabs and Chi-Squared
If you want to know how you depart from independence,
you may:
1.
2.
Check percentages (conditional distributions) in your
cross classification table.
Do a residual analysis:
The difference between observed and expected counts
in a cell behaves like a significance test when divided
by a standard error for the difference.
That s.e. = fe*(1-cell’s row )*(1 – cell’s column )
fo – fe
Z=
s.e.
Cross Tabs and Chi-Squared
Residual Analysis:
Let’s do cell 5!
s.e. = fe*(1-cell’s row )*(1 – cell’s column )
fo – fe
#5 row  = 200/650 = .308, column  = 265/650 = .408
Z=
s.e.
s.e. = 81.5 * (.692) * (.592) = 5.78
Z = 65 – 81.5 / 5.78 = -2.85; 2.85 > 1.96, there is a significant difference in cell 5
Data on male and female support for SJSU football from 650 students
Yes
No
Maybe
Female:
185
200
65
Male:
80
65
55
Total:
265
265
120
fe1 = (450/650) * 265 = 183.5
fe2 = (450/650) * 265 = 183.5
fe3 = (450/650) * 120 = 83.1
Deviations:
Dc = fo – fe
D1 = 185 – 183.5 = 1.5
D2 = 200 – 183.5 = 16.5
D3 = 65 – 83.1 = -18.1
fe4 = (200/650) * 265 = 81.5
fe5 = (200/650) * 265 = 81.5
fe6 = (200/650) * 120 = 36.9
D4 = 80 – 81.5 = -1.5
D5 = 65 – 81.5 = -16.5
D4 = 55 – 36.9 = 18.1
Total
450
200
650
Cross Tabs and Chi-Squared
Further topics you could explore:
Strength of Association
• Discussing outcomes in terms of difference of proportions
• Reporting Odds Ratios (likelihood of a group giving one answer versus
other answers or the group giving an answer relative to other groups giving
that answer)
Strength and Direction of Association for Ordinal Variables
• Gamma (an inferential statistic, so check for significance)
Ranges from -1 to +1
Valence indicates direction of relationship
Magnitude indicates strength of relationship
Chi-squared and Gamma can disagree when there is a nonrandom pattern
that has no direction. Chi-squared will catch it, gamma won’t.
• Kendall’s tau-b
• Somer’s d