Chi Square Analysis Basics
A Chi square analysis is one that is used so that we could determine if counts are true to is to compare observed values with expected values and
do a type of statistical test known as a “goodness of fit” test. This type of statistical test allows us to determine if any differences between our
observed measurements and our expected are simply due to chance or some other reason.
Steps:
1.
2.
State the null hypothesis.
a. A null hypothesis is the prediction that:
i. something is not present,
ii. that a treatment will have no effect, or
iii. that there is no difference between treatment and control.
b. Another way of saying this is the hypothesis that an observed pattern of data and an expected pattern are effectively the same,
differing only by chance, not because they are truly different.
i. The null hypothesis state there is no difference between what we expect and what we observe.
Calculate the X2 statistic, which is calculated in the following way:
X2 = Σ(sum of) (O-E)2
E
1.
2.
Where O is the observed (actual count) and
E is the expected number (if no effect)

Can find this by multiplying the total observed by expected frequencies
The main thing to note about this formula is that, when all else is equal, the value of X2 increases as the difference between the observed and
expected values increase.
Use this data table to help organize your data. You can add or take away rows based on the categories
(treatments/experimental and control groups) you have
Category
Observed (O)
Expected (E)
O-E
(O-E)2
(O-E)2
E
1 (positive control group)
2 (negative control group)
3 (experimental group 1)
4 (experimental group 2)
Totals
3.
Find your degrees of freedom
a. For this statistical test the degrees of freedom equal the number of categories minus one:
degrees of freedom = number of categories –1
4.
Determine the probability that the difference between the observed and expected values occurred simply by chance.
a. Scan down the column to your degrees of freedom the row corresponding to the p-value requested (almost always
0.05 for standard biology)
i. Note that the chi-square increases as the probability decreases.
ii. If your X2 = 3.36 with 4 degrees of freedom, there is a 50% chance that your results were due to chance
BUT if X2 = 11.35 with the same degrees of freedom, you have a 1% chance of results coming from chance
b. If you were asked to provide the p-value:
i. Estimate the probability based on degrees of freedom and closest degree of freedom in chart
Example problem:
In peas, smooth seeds (R) are dominant over wrinkled (r) seeds. In the P generation, a plant homozygous for smooth seeds is crossed with a plant with wrinkled seeds.
The resulting F1 plants are crossed. The seeds of the observed F2 generation were:
Smooth =
5474
Wrinkled = 1850
Does the data fit the predicted phenotypic ratio?
1.
First fill in the observed and find your total sample size
Phenotype
Observed (O)
Expected (E)
Smooth
Wrinkled
Totals
2.
(O-E)2
(O-E)2
E
O-E
(O-E)2
(O-E)2
E
5474
1850
7324
The totals for O & E should be the same
Phenotype
Observed (O)
Smooth
Wrinkled
Totals
O-E
5474
1850
7324
Expected (E)
7324
3.
Now, multiply the Total Expected by the expected frequency of each type.
a.
For genetics problems, we get the expected from the Punnett Squares. So, you would need to draw a Punnett Square or have their ratios
memorized!
P gen= RRxrr = 100% Rr (these are the F1)
F1 gen = RrxRr = 25%RR:50%Rr:25%rr (these are the F2) or 75% smooth and 25% wrinkled
Phenotype
Observed (O)
Expected (E)
(O-E)2
(O-E)2
E
O-E
(O-E)2
(O-E)2
E
O-E
(O-E)2
(O-E)2
E
O-E
Smooth
5474
5493 (0.75x7324)
Wrinkled
1850
1831 (0.25x7324)
7324
7324
Totals
4. Now, subtract observed from expected. You can ignore signs since you will square anyway 😊
Phenotype
Smooth
Wrinkled
Totals
5. Now square
Phenotype
Observed (O)
5474
1850
7324
Expected (E)
5493
1831
7324
Observed (O)
19
19
Expected (E)
Smooth
5474
5493
19
361
Wrinkled
1850
1831
19
361
7324
7324
Totals
6. Now divide by expected for that row. You will end with different values here because your denominator is different for each.
Phenotype
Observed (O)
Expected (E)
O-E
(O-E)2
Smooth
Wrinkled
Totals
7. Now add
Phenotype
5474
1850
7324
Smooth
Wrinkled
Totals
5474
1850
7324
5.
5493
1831
7324
Observed (O)
19
19
Expected (E)
5493
1831
7324
361
361
0.0657
0.1971
(O-E)2
O-E
19
19
(O-E)2
E
361
361
(O-E)2
E
0.0657
0.1971
0.2628
Find your degrees of freedom and read in the p=0.05 column.
Degrees freedom = 2 – 1 = 1 (you had two categories- smooth and wrinkled)
We fail to reject our null hypothesis that there is no difference between the expected number of each phenotype and the observed number of each phenotype since our
X2 = 0.2628 is less than the critical value of 3.85. If we were to estimate our p-value, we would say that there is somewhere around a 60% chance that observations
came from random chance and not the experiment itself.