By Jon Baker, National
Marine Fisheries Service and BioLab
When designing and
conducting an experiment, a biologist will have a predicted or expected outcome
based on his/her hypothesis. However the results do not always match the
expected results perfectly and the problem of how to interpret the results
arises. That is why biologists turn to
statistics to help them make decisions about the results of their experiments. For example, suppose you are performing a
guinea pig heredity experiment. You
crossed a pure-breeding black with a pure-breeding white guinea pig and all the
progeny were black. You then crossed
the offspring of the first mating and knowing of Mendel’s principles you
predict that a 3:1 ratio, black to white, should result. After the guinea pigs are born you find 164
guinea pigs, 117 black and 47 white.
This is not exactly a 3:1 ratio.
Given that there were 164 guinea pigs born you would expect 123 black
and 41 white. The question then becomes
are these data sufficiently close to the predicted result to say that they
agree? This is when statistics must be
used.
The
first thing to do is form a statement of the hypothesis being tested. In this experiment you believe that the
guinea pig cross will result in a 3:1 ratio and you write a hypothesis that
predicts just that. It is called the null
hypothesis (Ho)
because it predicts no difference between the experimental results and your
prediction. Write a very precise and
clear statement such as, “The sample data come from a population having a 3:1
ratio of black to white guinea pigs”.
You also write an alternate hypothesis (HA) that states there is a difference. For example, “the sample data come from a
population not having a 3:1 ratio of black to white guinea pigs”. If your statistical analysis shows that the Ho is false then the HA is assumed to be true. You must state a null hypothesis and an alternate hypothesis for
every statistical test you perform.
This assures that all possible outcomes are accounted for by the two
hypotheses.
The
statistic to use when analyzing genetic data of this type, when we are
comparing two distributions of data, is the Chi-Square (X2) Goodness of Fit.
The two distributions of data are the observed experimental results (117
black and 47 white) and the expected hypothetical distribution (3:1 or 123
black and 41 white). This statistic
will measure how far a sample distribution deviates from the hypothetical distribution.
Let's examine the equation and see how to it is used. Here is the equation.
c2 = S ( O – E )2 /E
O = observed value of each class of data
E = expected value of each class of data
Our data has two classes of
data, black guinea pigs and white guinea pigs and we know what the numbers are
for the observed and expected data. All
there is to do is plug in the numbers!
c2 = (117 – 123)2 /123 + (47 – 41)2 /41
c2 = (117 – 123)2 /123 + (47 – 41)2 /41
c2 = (-6)2 /123 + (6)2 /41
c2 = (36) /123 + (36) /41
c2 = .29 + .88
c2 = 1.17
Notice what happened; we
found the arithmetic difference between the observed and expected values for
each class of data and since that was squared, the negative number was
eliminated. The number 1.17 is the test statistic and is compared to values in
a table to determine which hypothesis to accept. The example below demonstrates the proper way to present and
perform a statistical test on paper.
Ho : The sample data come from a population having a
3:1 ratio of black to white guinea pigs.
HA : The sample data come from a population not having
a 3:1 ratio of black to white guinea pigs.
|
|
Black
|
White
|
n |
|
Expected |
123 |
41 |
164 |
|
Observed |
117 |
47 |
|
n = total of number of
samples
c2 = (-6)2 /123 + (6)2 /41
c2 = .29 + .88
c2 = 1.17
n = k – 1
n = 2 – 1
n = 1
0.25 < P < 0.50
Whoa! It was going well until all this nonsense at
the end. Well it is not that bad, let
me explain. Well first of all n is something known as degrees of freedom. It is calculated by subtracting 1 from the
number of classes in the data set (k).
In the example above there are two classes of data, black and white
guinea pigs. (We will not take the time
to explain degrees of freedom here because it is a complicated concept best
saved for a college statistics class.
Just know that you must calculate and use it to find your place in the
Chi-square critical values table, which will be demonstrated next.)
Once
the test statistic (c2) and
degrees of freedom (n) have been calculated it is
time to compare the value to the Chi-square critical values table. This table will tell us the probability
that chance caused the amount of difference we saw between our data and expected
results. Generally, if that probability
turns out to be less than 0.05, then it is agreed that chance did not cause the
difference. Now there is nothing
special about 0.05, it is just the level that most scientists have agreed
on. This is what is meant by statistical
significance. If your results reveal
that the probability that chance caused the difference between your data and
the expected results is less than 0.05 then scientists will agree that chance
probably is not the cause of the difference.
So some other mechanism is working to produce the results. It is up to you to determine what that
mechanism is, the statistics will not tell you. In the example above we have calculated a test statistic of 1.17
and a n of 1.
Look
at the Chi-square critical values table your teacher has provided. You will first notice that there are many
columns. The first column is headed
with a n and the others with the numbers 0.999, 0.995 and so
on until ending with .001. The n column is the degrees of freedom and all other columns
are probabilities. To find the
probability of obtaining a X2
statistic of 1.17 read down the degrees of freedom column until you reach the
degrees of freedom calculated for your test.
Then read across that row until you reach the number you calculated. It is unlikely that you will find your
number, rather it will be bracketed between two values. In this example, our value of 1.17 is
between 0.455 and 1.323. Reading the
top of the columns we find the probabilities 0.50 and 0.25 respectively. This means that our statistic has a
probability some where between .50 and .25.
Stated symbolically it is 0.25 < P < 0.50. What this means is that the chance that the
difference we observed was due to random sampling is located between 0.25 and
0.50. A more detailed table will be
needed to acquire a more precise probability.
Now,
how to interpret the results? Well with
a P value of 0.25<P<0.50 the Chi-square test reveals that there is a
pretty good probability that chance could be responsible for the experimental
results. This means that our
experimental results agree with the predicted outcome. Therefore we fail to reject the null
hypothesis (Ho)
and conclude that the sample data come from a population having a 3:1 ratio of
black to white guinea pigs. This is
consistent with a Mendelian explanation for the data.
Questions:
Now do a thought experiment
and change the observed frequency values:
What will happen to the
Chi-Square (X2) statistic
as the disagreement between the observed and the expected becomes larger and
smaller? What do you find? Try it when changing the total number of
samples.
In
hypothesis testing there are four outcomes.
See the table below.
|
|
Null Hypothesis |
||
|
Accepted |
Rejected |
||
|
Null Hypothesis |
True |
Correct Decision |
Type I error |
|
False |
Type II error |
Correct Decision |
|
Before you carry out a test
you must first determine the magnitude of Type I error you will accept. A Type I error is one in which you reject a
true hypothesis (see the table). By
doing this you are setting the significance level. This is the point at which you will accept or reject the null
hypothesis. It is this level that you
will have to defend from peers reviewing your work. A typical level in the sciences is .05 or 5%. What this says is that "the difference
observed in my data from what is expected has only a .05 (or less) probability
of being due to random error (chance)".
Stated another way - "there is only a 5% chance that chance caused
the difference seen in the data."
So if you get a Chi-Square (X2)
statistic of .05 or less we reject the null hypothesis and say that the
difference we see is likely due to some factor other than chance.
Now your teacher will show
you how to calculate a Chi-Square (X2)
statistic.