Chi-Squared tests (?2):

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Chi-Squared tests (?2)
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Chi-Squared tests (?2) Use with nominal
(categorical) data when all you have is the
frequency with which certain events have occurred.
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The ?2 Goodness of Fit test Compares an
observed frequency distribution with an expected
frequency distribution. Useful when you have the
observed frequencies for a number of mutually-
exclusive categories, and you want to decide if
they have occurred equally frequently.
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Which soap-powder name do shoppers like
best? Each of 100 shoppers picks the powder name
they like most. Number of shoppers picking
each name (observed frequencies) Washo
Scruba Musty Stainzoff Beeo total 40
35 5 10 10
100 Expected frequency for each category
is total no.observations / number of
categories 100 / 5 20.
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The formula for Chi-Square
Washo Scruba Musty Stainzoff
Beeo total O 40 35 5
10 10 100 E 20 20
20 20 20
100 (O-E) 20 15 -15
-10 -10 (O-E) 2 400
225 225 100 100 20
11.25 11.25 5
5 ?2 52.5
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Chi-squared is the sum of the squared differences
between each observed frequency and its
associated expected frequency. The bigger the
value of ?2, the greater the difference between
observed and expected frequencies. But how big
does ?2 have to be, to be regarded as big? Is
52.5 big?
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We compare our obtained ?2 value to ?2 values
which would be obtained by chance. To do this,
we need the degrees of freedom this is the
number of categories (or cells) minus one. We
have a ?2 value of 52.5, with 5-1 4 d.f.
Tables show how likely various values of ?2 are
to occur by chance. e.g. probability
level d.f. .05 .01 .001 1 3.84 6.63 10.83
2 5.99 9.21 13.82 3 7.81 11.34 16.27 4 9.49 13.
28 18.46 5 11.07 etc. etc. 52.5 is bigger than
18.46, a value of ?2 which will occur by chance
less than 1 times in a 1000 (plt.001).
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The sampling distribution of chi-square Frequency
with which ?2 values occur purely by chance
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Our obtained ?2 52.5, with 4 d.f., p lt .001.
A ?2 value this large is highly unlikely to
have arisen by chance. It appears that the
distribution of shoppers choices across
soap-powder names is not random. Some names get
picked more than we would expect by chance and
some get picked less.
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The ?2 test of association between two
independent variables Another common use of ?2
is to determine whether there is an association
between two independent variables. Is there an
association between gender (male or female IV A)
and soap powder (Washo, Musty, etc. IV B)?
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This gives a 2 x 5 contingency table. Data for a
random sample of 100 shoppers, 70 men and 30
women Washoe Scrubbup Musty
Stainoff Nogunge total male 10
12 5 3 40 70 female
6 2 1 20 1
30 totals 16 14 6 23
41 100
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To calculate expected frequencies E row
total column total grand total Work out
the expected frequency for each cell
e.g. 11.2 (16 70)/100 6.9 (23 30)/100,
etc.
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Using exactly the same formula as before, we get
?2 52.94. d.f. (number of rows - 1)
(number of columns - 1). We have two rows and
five columns, so d.f. (2-1) (5-1) 4
d.f. Use the same table to assess the chances of
obtaining a Chi-Squared value as large as this by
chance again plt .001. Conclusion our observed
frequencies are significantly different from the
frequencies we would expect to obtain if there
were no association between the two variables
i.e. the pattern of name preferences is different
for men and women.
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Chi-Square test merely tells you that there is
some relationship (an association) between the
two variables in question it does not tell you
anything about the causal relationship between
the two variables. Here, it is reasonable to
assume that gender causes people to pick
different soap powder names it's unlikely that
soap powder names cause people to be male or
female. However, in principle the direction of
causality could equally well go in either
direction.
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Assumptions of the Chi-Square test 1.
Observations must be independent each subject
must contribute to one and only one category.
Otherwise the test results are completely
invalid. 2. Problems arise when expected
frequencies are very small. Chi-Square should not
be used if more than 20 of the expected
frequencies have a value of less than 5. (It does
not matter what the observed frequencies are).
Two solutions combine some categories (if this
is meaningful in your experiment), OR obtain more
data (make the sample size bigger).
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?2 test of association - the one- d.f.
case Preferred TV programme Stenders
Corrie Row total Origin North 13 10 23
South 5 24 29 Column
total 18 34 52 With 1 d.f. (as with a 2 x 2
table), the obtained ?2 value is inflated some
statisticians advocate using "Yates' Correction
for Continuity" to make the ?2 test more
conservative (i.e. make ?2 value smaller and
hence less likely to be significant).
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Same procedure as before, except (a) take the
absolute value of O - E (i.e., ignore any
negative signs). (b) Subtract 0.5 from each O-E,
before squaring it.
Without Yates Correction ?2 8.74. With
Yates Correction ?2 7.09.
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Why you should avoid using Chi-Square if you
can Design studies so that you can avoid using
Chi-Square! Frequency data give little
information about participants' performance all
you have is knowledge about which category
someone is in, a very crude measure. It's much
more informative to obtain one or more scores per
participant scores give you more information
about performance than categorical data (and can
be used with better statistical tests). e.g. IQ
which is better - to know participants are
bright or dim, or have their actual IQ scores?
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?2 Goodness of Fit test on "fast food" data,
using SPSS Are all brands mentioned equally
frequently? Analyze gt Nonparametric Testsgt Legacy
Dialogs gt Chi-Square
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?2 test of association on "fast food" data, using
SPSS Is there an association between gender and
brand first mentioned? Analyze gt Descriptive
Statistics gt Crosstabs...
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?2 test of association on "fast food" data
(continued) Is there an association between
gender and brand first mentioned?
11 response categories - gives too many expected
frequencies lt 5. Therefore confined analysis to
Burger King, KFC and McDonalds. (Use "Select
Cases" on "Data" menu to filter out unwanted
response categories).
Conclusion no significant association between
gender and brand first mentioned. (?2 (2)
0.28, p .87)