Download Statistical Analysis

BS 105
Statistical Analysis
Chi-Square Test
Statistics can be used to determine if differences among groups are significant, or simply
the result of chance. A common statistical method used to determine if observed experimental
data are a good approximation to the expected or theoretical data is the Chi-square test. In
short, this test can determine if deviations from the expected values are due to chance alone,
or to the effect of the independent variable.
When conducting an experiment, a researcher first states the hypothesis about how the
experiment will turn out. The hypothesis is typically stated as two possible outcomes, the null
hypothesis (H0) of no effect and the alternate hypothesis (H1) of an effect. Based upon the
data, the researcher will then accept one hypothesis and reject the other. An example is given
below.
A study was conducted of freshmen students to see if there would be an improvement in individual
test scores when the students were placed in supervised study groups at the start of the semester.
Half of the students were assigned to study groups and half were not.

Null hypothesis (H0) =
Placement in study groups will result in no improvement in student test
scores.

Alternate hypothesis (H1) =
Placement in study groups will result in improvement in student test
scores.
To determine if the observed data fall within acceptable limits; a Chi-square analysis is
performed to test the validity of the null hypothesis; that there is no statistically significant
difference between the observed and expected data. If the Chi-square analysis indicates that
the data vary too much from the expected results, then the alternate hypothesis is accepted.
The formula for Chi-square is:
Χ2 = ∑ (o-e)2
e
where o = observed number of individuals
e = expected number of individuals
∑ = sum of the values (in this case, the differences, squared, divided by the number
expected)
An Example Experiment and Sample Calculations:
In mammals, sex determination (male vs. female) of the offspring occurs at fertilization.
This is not the case for many species of reptiles, which show temperature-dependent sex
determination (TSD). In TSD reptiles, the temperature at which the eggs or embryos are
incubated during development determines the sex or gender of the offspring. The relationship
between incubation temperatures and gender is not consistent across TSD reptile species. In
some species, cool incubation temperatures result in more males and in other species more
females. Similarly, in some species warm incubation temperatures result in more males and in
other species more females.
The paragraph above is based on experimental evidence and is therefore scientifically
supported. The paragraph below is based on fiction – for the purpose of illustrating the use of
Chi-square in evaluation of data.
A researcher is working with a turtle species for which there is limited information available on
reproduction. The researcher knows that eggs incubated at 29oC will produce equal numbers of male and female
offspring. The researcher wants to determine if this is a TSD species and if so, what the effect of warm incubation
temperatures might be on sex of the offspring. The researcher proposes two hypotheses:

Null hypothesis (H0) =
Incubation of eggs at a warm temperature will not affect the sex ratio of
offspring.

Alternate hypothesis (H1) =
Incubation of eggs at a warm temperature will result in an altered sex
ratio of offspring.
The researcher obtains 200 eggs of the turtle species and incubates them at 31.5 oC. At hatching there
are 80 male and 120 female offspring. The researcher performs a Chi-square analysis of the data, as indicated
below:
(o-e)2
e
Observed (o)
Expected (e)
(o-e)
(o-e)2
Male
80
100
-20
400
4.00
Female
120
100
20
400
4.00
Sex/Gender
Χ2 =
8.00
The Chi-square value is simply the sum of the final column. This Χ2 value is then
compared to the following table.
Critical Values of the Chi-square Distribution
DEGREES OF FREEDOM (df)
Probability (p)
1
2
3
4
5
0.05
3.84
5.99
7.82
9.49
11.1
0.01
6.64
9.21
11.3
13.2
15.1
0.001
10.8
13.8
16.3
18.5
20.5
How to use the Critical Values Table:
1. Determine the degrees of freedom (df) for the experiment. This is simply the number
of categories minus 1. Since there are two possible categories for this example (male
and female), the degrees of freedom is 1 (2 – 1).
2. Find the p value. Under the 1 df column find the critical value in the probability (p) = 0.01
row: it is 6.64. What does this mean? If the calculated Chi-square value is greater
than or equal to the critical value from the table, then the null hypothesis is
rejected. Since our calculated Χ2 value is 8.00 and 8.00 > 6.64, we reject the null
hypothesis that there is no statistically significant difference between the observed and
expected data. In other words, chance alone cannot explain the deviations the
researcher observed. In the sciences the minimum probability for rejecting a null
hypothesis is generally 0.05.
3. The results of the turtle experiment are said to be significant at a probability of p = 0.01.
This means that only 1% of the time would you expect to see similar data if the null
hypothesis was correct. Therefore, you are 99% certain that the data do not fit the
expected 1:1 ratio.
4. If the calculated value was 4.5 then the null hypothesis would still be rejected, but this
time at a probability of p = 0.05 (4.5 > 3.84, but < 6.64). This means that less than 5% of
the time would you expect to collect the observed data if the null hypothesis was correct.
Stated differently, you would be 95% sure that the data do not fit the expected 1:1 ratio.
5. Since these data do not fit the expected 1:1 ratio, you must consider reasons for this
variation. The logical conclusion here is that the warmer temperature has caused a
significant difference in the number of male vs. female offspring. In other words, the
data support the alternate hypothesis.
Applying this information to the “Pill Bug” Experiment:
1. The number of categories in each part of the experiment is always two (acid vs. base;
food vs. no-food, wet vs. dry, light vs. shade).
2. Determine the expected values! Since we have no reason to expect anything otherwise,
we assume that half of the pill bugs will be in one chamber and half in the other (in other
words, a 1:1 ratio), regardless of the particular treatment. So…total the number of pill
bugs for a given treatment (this is actually done for you on the data sheet) and divide
that number in half to calculate the expected number of pill bugs in each side of the
choice chamber.