Chi Square Tests
Chapter 17
Nonparametric Statistics
• A special class of hypothesis tests
• Used when assumptions for parametric tests
are not met
– Review: What are the assumptions for
parametric tests?
Assumptions for Parametric Tests
• Dependent variable is a scale variable  interval or
ratio
– If the dependent variable is ordinal or nominal, it is a nonparametric test
• Participants are randomly selected
– If there is no randomization, it is a non-parametric test
• The underlying population distribution is normal
– If the shape is not normal, it is a non-parametric test
When to Use
Nonparametric Tests
• When the dependent variable is nominal
– What are ordinal, nominal, interval, and ratio
scales of measurement?
• Used when either the dependent or
independent variable is ordinal
• Used when the sample size is small
• Used when underlying population is not
normal
Limitations of Nonparametric
Tests
• Cannot easily use confidence intervals or
effect sizes
• Have less statistical power than parametric
tests
• Nominal and ordinal data provide less
information
• More likely to commit type II error
– Review: What is type I error? Type II error?
Chi-Square Test for
Goodness-of-Fit
• Nonparametric test when we have one
nominal variable
– These variables, also called "attribute variables" or
"categorical variables," classify observations into a
small number of categories. A good rule of thumb is
that an individual observation of a nominal variable is
usually a word, not a number
– Examples of nominal variables include sex (the
possible values are male or female), genotype (values
are AA, Aa, or aa), or ankle condition (values are
normal, sprained, torn ligament, or broken)
Chi-Square Test for
Goodness-of-Fit
• Nonparametric test when we have one
nominal variable
– Measurement v. Nominal: Imagine recording each
observation in a lab notebook. If you record a number
(width, height, speed, errors) it’s a measurement, if you
record a label it’s nominal (sex, popularity, beauty)
Examples of When to Use Chi-Square
• The observed counts of numbers of observations in
each category are compared with the expected
counts, which are calculated using some kind of
theoretical expectation, such as a 1:1 sex ratio, or
4:2:1 population density in following example.
• Example: looking at an area of shore that had 59% of the area
covered in sand, 28% mud and 13% rocks (4:2:1); if seagulls
were standing in random places, your null hypothesis would
be that 59% of the seagulls were standing on sand, 28% on
mud and 13% on rocks (4:2:1).
Examples of Chi-Square
• Does the count of the Observed match the count of
the Expected?
• Mendel crossed peas that were heterozygotes for Smooth/wrinkled, where
Smooth is dominant. The expected ratio in the offspring is 3 Smooth: 1
wrinkled. He observed 423 Smooth and 133 wrinkled.
• The expected frequency of Smooth is calculated by multiplying the sample
size (556) by the expected proportion (0.75) to yield 417. The same is done
for green to yield 139. The number of degrees of freedom when an extrinsic
hypothesis is used is the number of values of the nominal variable minus
one. In this case, there are two values (Smooth and wrinkled), so there is
one degree of freedom.
• The result is chi-square=0.35, 1 d.f., P=0.557, indicating that the null
hypothesis cannot be rejected; there is no significant difference between the
observed and expected frequencies.
Examples of Chi-Square
• Does the count of the Observed match the count of
the Expected?
• Mannan and Meslow (1984) studied bird foraging behavior in a forest in
Oregon. In a managed forest, 54% of the canopy volume was Douglas fir,
40% was ponderosa pine, 5% was grand fir, and 1% was western larch.
They made 156 observations of foraging by red-breasted nuthatches; 70
observations (45% of the total) in Douglas fir, 79 (51%) in ponderosa pine,
3 (2%) in grand fir, and 4 (3%) in western larch. The biological null
hypothesis is that the birds forage randomly, without regard to what species
of tree they're in; the statistical null hypothesis is that the proportions of
foraging events are equal to the proportions of canopy volume. The
difference in proportions is significant (chi-square=13.593, 3 d.f.,
P=0.0035).
How the test works
• The test statistic is calculated by taking an observed
number (O), subtracting the expected number (E), then
squaring this difference. The larger the deviation from
the null hypothesis, the larger the difference between
observed and expected is.
2


(
O

E
)
2
Χ  

 E 
• Squaring the differences makes them all positive. Each
difference is divided by the expected number, and these
standardized ratios are summed: the more differences
between what you would expect and what you get the
bigger the number.
Chi-Square Test for
Goodness-of-Fit
• The six steps of hypothesis testing
Question: Are the best soccer players born early
rather than later in the year ?
1. Identify
2. State the hypotheses
3. Characteristics of the comparison distribution
4. Critical values
5. Calculate
6. Decide
Chi-Square Test for
Goodness-of-Fit
• The six steps of hypothesis testing
1. Identify Pop. Distribution & Assumptions
a)
Two populations, one distribution that matches expected
outcomes and another where distribution matches observed
outcomes. E.g., great soccer players are born evenly throughout
year, great soccer players born in first half of year.
b) Comparison distribution is chi-square
c) First assumption, variable of interest is nominal, birth month.
Second, independence of observation, that is each observation
fits in only one category, no soccer player has two birth months.
Third, random selection of pop ( in this case, they are only
Germans, and only elite). Fourth, large enough sample size,
ideally 5 times the number of cells (in this case N= 56 > 10 (2 x
5).
Chi-Square Test for
Goodness-of-Fit
• State the hypotheses: does the Observed
count of elite soccer player Birth Months
match the Expected count of elite soccer
player Birth Months
• Null: Match
• Alternative: No match
Chi-Square Test for
Goodness-of-Fit
• Characteristics of the comparison
distribution
df  2  k  1
df  2  2  1  1
• Only two categories of soccer players
Chi-Square Test for
Goodness-of-Fit
• Critical values
Chi-Square Test for
Goodness-of-Fit
• Calculate
Chi-Square Test for
Goodness-of-Fit
• Calculate
Making a Decision
A more typical Chi-Square
• Evenly divided expected frequencies
– Can you think of examples where you would
expect evenly divided expected frequencies in
the population?
• Chi-square test for independence
– Analyzes 2 nominal variables
– The six steps of hypothesis testing
1. Identify
2. State the hypotheses
3. Characteristics of the comparison distribution
4. Critical values
5. Calculate
6. Decide
The Cutoff for a Chi-Square Test for Independence
The Decision
Cramer’s V (phi)
• The effect size for chi-square test for
independence
X2

( N )( df row/ column )
Graphing Chi-Squared
Percentages
Relative Risk
> We can quantify the size of an effect
with chi square through relative risk,
also called relative likelihood.
> By making a ratio of two conditional
proportions, we can say, for example,
that one group is three times as likely to
show some outcome or, conversely, that
the other group is one-third as likely to
show that outcome.
Adjusted Standardized
Residuals
> The difference between the observed
frequency and the expected frequency for a
cell in a chi-square research design, divided
by the standard error; also called adjusted
residual.
Formulae
 (O  E ) 
Χ  

 E 
2
2
df row  krow  1
df column  kcolumn  1
df X 2  (df row )( df column )
Determining the Cutoff for a Chi-Square Statistic