Psych 5500/6500
Introduction to Chi-Square
and Test for Goodness of Fit
Fall, 2008
1
Chi-square Test
Chi-square (χ²) can be used to test inferences about
the variance of the population from which a
sample was drawn. Let’s say you draw 30
scores from some population and you want to
test the following hypotheses.
H0: σ² = 6.25
Ha: σ²  6.25
2
The Chi-Square Distribution
Formula for computing the value of chi-square with N-1 degrees
of freedom.
χ
2
(N 1)

N - 1(est.σ

σ
2
)
2
Where: est. σ² comes from your data, and σ² is the value of the
population variance according to H0.
3
Expected value
2

N - 1(est.σ )
2
χ (N 1) 
2
σ
From this formula it is obvious that:
1. If H0 is true then the expected value of chi-square =
its degrees of freedom (i.e. N-1).
2. If the variance of the population from which the
sample was drawn is greater than the variance
proposed by H0 then chi-square will be greater than
N-1.
3. If the variance of the population is less than the value
proposed by H0 then the value of chi-square will be
4
less than N-1.
Sampling Distribution
The shape of the sampling distribution depends upon the df
5
Image from ‘Psychological Statistics Using SPSS for Windows’ by R. Gardner.
Testing H0
H0: σ² = 6.25
H1: σ²  6.25
The N of our sample is 30, so df= 29. If H0 is true then we
expect the value of χ² to be around 29 (but probably not
exactly equal to 29 due to chance)
How big or small χ² has to be to reject H0 can be found by
using the Chi-Square stat tool in the Oakley Stat Tools
link on the course web site:
χ²critical =16.05 (left tail, % above = 97.5)
χ²critical =45.72 (right tail, % above = 2.5)
6
Coming to a Conclusion
From our sample we estimate the variance of the
population from which the sample was drawn to
equal 4.55.
Is this significantly different than 6.25?
χ
2
(N 1)

N - 1est. 2 294.55


 21.11
σ
2
6.25
If H0 were true we would expect chi-square to equal 29, and there
would be a 95% chance it would be between 16.05 and 45.72.

2
( 29)
 21.11, p  .05
7
One-tail Tests
It is also possible to do one-tail tests, it just
involves having one rejection region (either
greater than d.f. or less than d.f.).
8
Assumption of Normality
The χ² test assumes that the data are normally
distributed. Unlike tests concerning the mean, the
central limit theorem does not come into play for
χ² (i.e. having a large N does not allow you to
ignore this assumption). For χ² to be valid it is
important that the population be normally
distributed.
9
Confidence Interval of σ²
χ² gives us a way to create a confidence interval of the
true value of the variance of the population.
N 1est.σ2
χ
2
critical,upper_boun d
σ 
2
N 1est.σ2
χ
2
critical,lower_bound
95% confidence interval
29(4.25)   2  29(4.25)
45.72
16.05
2.70  σ 2  7.70
10
Chi-square Tests of Frequencies
We know turn to another use of chi-square, as a
means of testing hypotheses about frequencies.
These tests are among the relative few that can
be used on nominal data. The two tests we will
look at are:
1. Goodness of fit
2. Test for association
11
Comparing Sample and Population Distributions
using ‘Goodness of Fit’
This is used to determine whether or not the
frequency of scores in a sample differs from
what would be predicted by H0. As rejecting H0
indicates there is a difference, it really should be
called the ‘Badness of Fit’ test.
12
Example 1: Fair Die?
We have a six-sided die and we want to know if it is
‘fair’ (i.e. each side will come up an equal number of
times).
H0 is that it is a fair die, if H0 is true then the
probability of any one side coming up = 1/6 = 0.1667
(approx)
Ha is that it is not a fair die and the probability of each
side coming up is not 0.1667.
Our approach will be to sample from the die, and
compare the frequencies of how many times each
side appears to the expected frequencies if H0 were
true.
13
Our Hypotheses
H0: Every side has an equal probability of occurring.
HA: Every side does not have an equal probability of
occurring.
I suppose I could have written these in a more
mathematical manner that would fit every use of this
test, but I think conceptually this way is clearer.
14
Symbols
We plan on rolling the die 100 times, and counting how frequently
each side occurs.
N = the size of the sample = 100
j = the outcome category
Oj = the observed frequency of each outcome category.
Ej = the expected frequency of each outcome category if H0 were
true.
15
Determining Ej
If H0 is true then each outcome has a 0.1667
probability of occurring, if we roll the die 100
times each side is expected to occur 100*0.1667
= 16.67 times.
Thus for each outcome category the expected
frequency if H0 is true (i.e. Ej) would be:
Ej = 16.67
16
Observed vs Expected
Frequencies
Side
Observed
Frequency
Expected
Frequency
1
2
3
4
5
6
10
30
5
14
25
16
16.67
16.67
16.67
16.67
16.67
16.67
N=
100
100
17
Testing H0
There were some differences between the observed
frequencies and what we would have expected if
H0 were true.
How to test whether those differences are
statistically signficant?
 O  E 
j
j
j
won’t work because it will always
equal zero.
18
Pearson Chi-square statistic
 O
j  Ej
2
j

O
j
 Ej
Ej
2
This could work, but the following
one is preferred.
This has two properties, one is that
it weights more heavily changes in
cells that have a small expected
frequency, the second is that under
many circumstances it approximately
fits the theoretical Chi-square
distribution, with j-1 degrees of freedom.
19
Chi-square Distribution for
Goodness of Fit
• Chi-square is a distribution that has a table
where we can look up critical values assuming
H0 is true.
• If H0 true, the expected value of Chi-square is:
2
χ  degrees of freedom j -1
• If H0 is false, the expected value of Chi-square
is:
2
χ  degrees of freedom
20
Computing Chi-square
and the resultant p value
2
2
2






10

16
.
67
30

16
.
67
5

16
.
67
2 



16.67
16.67
16.67
14  16.672  25  16.672  16  16.672  26.12
16.67
16.67
16.67
If H0 were true we would expect chi-square to equal j-1= 6-1=5. If
H0 is false then chi-square should be greater than 5. Chi square is
greater than 5, is it enough greater that it would only occur 5% of
the time or less if H0 were true?
21
Chi-square Critical Values
and the resultant p value
The test for goodness of fit is always one-tailed (if
H0 is false then chi-square will be greater than its
df). The critical value for the test may be obtained
from the table I provide in class or through the Chisquare statistical tool I wrote. The tool has the
advantage of giving us the exact p value of our
obtained value of chi-square.
With df=5, χ²critical =11.07
χ²(5)=26.12, p=.0001
22
Example 2: Smoking
In a previous year the following proportion of men in
some population smoked the following number of packs
of cigarettes per day.
Category (# of packs)
Proportion
none
.43
one
.17
two
.24
three
.10
four or more
.06
23
This year we sample 863 men and we are interested in
knowing whether the new data fit (have the same
distribution as) the previous data.
N=863
j=5
H0: distribution hasn’t changed
Ha: distribution has changed
Expected value of χ² if H0 is true = 4, critical value of
χ²=9.49
24
Determining Ej
If H0 is true then the same proportion of men should fall
into each category this year as well. To use the Chisquare test, we will need to turn those proportions into
what frequencies we would expect in each category if
H0 were true.
Category (# of
packs)
Expected
Proportion
Expected frequency out
of 863 men
none
.43
.43 x 863 = 371.09
one
.17
.17 x 863 = 146.71
two
.24
.24 x 863 = 207.12
three
.10
.10 x 863 = 86.3
four or more
.06
.06 x 863 = 51.78
25
Category (# of
packs)
Observed
frequency
Expected
frequency
none
406
371.09
one
164
146.71
two
189
207.12
three
78
86.3
four or more
26
51.78
N=863
N=863
2
2
2






406

371
.
09
164

146
.
71
189

207
.
12
2 



371.09
146.71
78  86.32  26  51.782  20.54
86.3
51.78
207.12
 2 (4)  20.54, p  .0004
26
Assumptions
1) Each and every sample observation falls into
one and only one category.
2) The outcomes for the N respective observations
in the sample are independent.
3) Deviations (Oj – Ej) are normally distributed.
The next slide takes a closer look at this.
27
Assumption of Normality
So we are assuming that the deviations of the
expected from observed frequencies (Oj – Ej) are
normally distributed. This assumption becomes
untenable when the expected frequency in a
group is small. For example, if the expected
frequency is ‘1’, then there is obviously more
room for the observed frequency to be greater
than ‘1’ then there is for it to be less than ‘1’, so
the distribution of (Oj – Ej) will be neither
normal nor symmetrical.
28
Solution
A variety of approaches are available for handling
problems that occur when the expected frequency in
any one cell is small (e.g. less than ‘5’). The various
approaches have their supporters and critics. I’ll simply
refer you to a review article and let you explore this
area if you plan on doing Chi-square:
Delucchi, K. L. (1993). On the use and misuse of Chisquare. In G. Keren & C. Lewis (eds.). A handbook for
data analysis in the behavioral sciences: Statistical
Issues. Hillsdale, NJ: Lawrence Erlbaum.
29
Fitting the Sample Distribution to the Normal
Distribtuion
An interesting use of chi-square goodness of fit is to
determine if the population from which you sampled is
normally distributed.
Compare the obtained frequencies in your data to what
you would expect if the scores came from a normal
population.
H0: frequencies fit normal distribution
HA: frequencies do not fit normal distribution.
30
One (flawed) approach
Change the data to standard scores and compare to the following
proportions from the normal table:
Category
Expected proportion if
from normal population
z = 3.00 or more
z = 2.00 to 3:00
z = 1.00 to 2.00
z = 0 to 1.00
z = 0 to –1.00
z = -1.00 to –2.00
z = -2.00 to – 3.00
z = -3.00 or more
.0013
.0215
.1359
.3413
.3413
.1359
.0215
.0013
31
If N = 100 then the expected frequencies would be
Category
Expected frequencies
z = 3.00 or more
z = 2.00 to 3:00
z = 1.00 to 2.00
z = 0 to 1.00
z = 0 to –1.00
z = -1.00 to –2.00
z = -2.00 to – 3.00
z = -3.00 or more
.13
2.15
13.59
34.13
34.13
13.59
2.15
.13
32
A better approach
Categories of z values
z = 1.15 or more
Expected proportions if from
normal distribution
.125
z = 0.68 to 1.15
.125
z = 0.32 to 0.68
.125
z = 0 to 0.32
.125
z = 0 to -0.32
.125
z = -0.32 to -0.68
.125
z = -0.68 to -1.15
.125
z = -1.15 or less
.125
33
Example
Categories of z
values
Observed
Frequencies of
z in the sample
Expected
Proportions
Expected
Frequencies
(N x 0.125)
z = 1.15 or more
10
.125
17.5
z = 0.68 to 1.15
25
.125
17.5
z = 0.32 to 0.68
15
.125
17.5
z = 0 to 0.32
30
.125
17.5
z = 0 to -0.32
15
.125
17.5
z = -0.32 to -0.68
20
.125
17.5
z = -0.68 to -1.15
5
.125
17.5
z = -1.15 or less
20
.125
17.5
N=140
34
2
2
2
2








10

17
.
5
25

17
.
5
15

17
.
5
30

17
.
5
2 



17.5
17.5
17.5
17.5

15  17.52 20  17.52 5  17.52 20  17.52




17.5
d.f.=7
17.5
17.5
17.5
 25.71
χ²(7) = 25.71, p=.0006
35