Chi-Square and the Null Hypothesis
The chi-square method of statistical analysis is used when the variable has counted data.
This means that specimens examined can be classified into one of two (or more)
categories. Even though it is possible to calculate percentages for such data, actual
numbers rather than percentages are used for chi-square analysis.
Null Hypothesis
There is no statistically significant difference between the expected data (control
group) and the observed data (experimental group).

If the null hypothesis is either accepted or rejected through calculation of a Pvalue (probability value) using Chi-Square analysis.

If the null hypothesis is rejected, the alternate hypothesis is accepted and there is
a statistically significant difference between the expected and observed data.
Example 1 - Testing the Effect of a Variable
In drug testing experiments, the goal of the Chi-square test is to reject the null
hypothesis.
The test is used to show that a drug does make a difference in the results that are
obtained. For example, if 88% of people taking a drug achieved a lower cholesterol level
and 52% of people taking a placebo achieved a lower cholesterol level during a blind
study, did the drug make a significant difference? If the null hypothesis is rejected, it
means that the drug was effective!
Probability
.99
.05
If the P Value determined by the Chisquare test falls between 0.99 and 0.05,
then the null hypothesis is supported,
and the drug did not make a difference.
If the P value determined by the Chi-square
falls between 0.05 and 0.01, then the null
hypothesis is rejected, and the drug did
make a difference in lowering cholesterol
The drug worked!
.01
Example 2 – Validation of Genetics Experiments
In genetics experiments, the goal of the Chi-square test is to support the null hypothesis.
The test is used to show that results of a genetic cross are close enough to perfect results
to be considered valid. For example, if the offspring of a cross consists of 48% males and
52% females, is this close enough to a 1:1 ratio to consider the results valid and
unaffected by some environmental factor? Probably – but the Chi-square test will
quantitatively support it!
P Value
.99
.05
If the P Value determined by the Chisquare test falls between 0.99 and 0.05,
then the null hypothesis is supported,
and there is not difference between the
actual results and ideal results.
The genetic cross is valid – data
obtained was close enough!
.01
In statistics, instead of saying the null
hypothesis is supported, it is said that….
“the chi square analysis failed to reject the
null hypothesis”.
Example Problem
Many species of slugs feed on plant material and can destroy crops such as lettuce. It is
known that slugs have negative chemotaxis to certain chemicals and will not eat lettuce
plants sprayed with these chemicals. A currently used spray made of marigold flower
extract repels slugs with approximately 75% efficiency. A new spray made of stinging
nettle extract has been developed and its effectiveness needs to be determined before the
company that developed the new spray can claim that is it a better alternative can be used
in marketing.
Experimental
Data
Marigold
Spray
(Control)
Stinging Nettle
Spray
(Experimental)
Lettuce Seedlings
Protected by Spray
Lettuce Seedlings
NOT Protected by
Spray
Total Sample Size
300
100
400
304
96
400
Step 1
Use the Chi-Square Formula to determine the Chi-Square Value (X2).
Chi-Square Formula (AP Formula Sheet)
Step 2
Determine the Degrees of Freedom for the Analysis
DF = number of classes -1
DF = 2-1
DF = 1
Step 3
Look the Chi-Square Value up in a Chi-Square Table using the correct number of
Degrees of Freedom (DF) to determine the Probability (P).
Chi-Square Table (AP Formula Sheet)
X2 = 0.214
DF = 1
So, P is greater than 0.05.
Accept the Null Hypothesis
There is no statistically significant difference between the effectiveness of the new
stinging nettle spray and the currently used marigold spray in repelling slugs from
lettuce.
Practice Problems
1. A vaccine was developed which was intended to protect mice
against a particularly virulent (deadly) bacterial disease. A
group of 55 mice was given the vaccine and then challenged
with a heavy dose of the virulent bacterium. Another group of
55 mice, which had not been vaccinated, was challenged with
the same dose of the virulent bacterium. In the group that was
vaccinated, 29 mice contracted the disease and died, while 35
mice in the control group died. Use the Chi-Square analysis
method to determine if the vaccine is affective. As well, be sure
to state a null-hypothesis and indicate if it is supported or
refuted by the Chi-Square analysis.
Experimental Data
Control Group
(no vaccine)
Experimental Group
(vaccine)
# of Mice Surviving
or
# of Mice Killed
2. An experiment was performed to determine the
degree to which the isopod named Porcellio laevis
(common pillbug) can acclimate to different
temperatures. Of 50 specimens held at 15oC, 20
died. Meanwhile, of 50 specimens held at normal
environmental conditions of 25oC, 2 died. Analyze
the data to see if temperature has an affect the
survival of pillbugs using Chi-Square analysis. As
well, be sure to state a null hypothesis and indicate
if it is supported or refuted by the Chi-Square
analysis.
3. A drug believed to have
teratogenic properties was given to
a group of 65 pregnant female rats,
and it was later noted that 23
females in this experimental group
produced litters containing at least
one malformed offspring. From a
group of 85 females used as a
control, 12 litters contained at least
one malformed offspring. On the
basis of these data, what conclusion
can be drawn relative to the teratogenic properties of the drug? Use the Chi-Square
analysis method to analyze the data and indicate the null hypothesis in question.
Note: The numbers of individuals in the experimental and control groups must be the
same in order to carry out Chi-Square analysis. To adjust the data, calculate the
percentage of normal and malformed offspring in the control group. Then, multiply these
percentages by the total number of offspring in the experimental group. This will provide
a comparable number of control individuals for use in the Chi-Square analysis.
4. In a genetics experiment, two
Drosophila melanogaster (fruit flies)
are crossed. The offspring are expected
to have a 50:50 male to female ratio.
However, if an environmental factor
selects against (kills) one of the sexes
of flies preferentially, the observed
number of offspring will not be in a
50:50 male to female ratio.
Analyze the data below to determine if
an environmental factor is selecting
against males. In doing so,
include the following components
to your answer.



Null Hypothesis
Chi Square Analysis
Conclusion