## 5. Chi Squared Test

The Chi Squared test is to test the null hypothesis.  The chi squared test will evaluate if there is a significant difference between your observed results and the expected results you predicted.

The formula is:

X2∑   (O – E)2

E

• O = observed number of individuals with (red-eyed/white-eyed) phenotype
• E = expected number of individuals with (red-eyed/white-eyed) phenotype
• Σ= sum of deviations for all phenotypes

EXAMPLE CHI SQUARED TEST

Looking at sex-linked genes in fruit fly experiment.  You crossed a heterozygous dominant red-eyed female with a white-eyed male and they produced 182 offspring.  From this offspring you got the following phenotypes:

You expect to get:

• 78 flies that are female with red eyes
• 52 flies that are male and have red eyes
• 52 flies that are male and have white eyes

What you actually get:

• 80 flies that are female and homozygous dominant for red eyes
• 56 flies that are male and have red eyes
• 46 flies that are male and have white eyes

Use this table for calculations:   (keep to two decimal places) Use the Chi Squared Table given below to look at the probability from your value.  The table ranges in p-values from 0.25 to 0.005. Degrees of freedom (df) are also needed to look up your result.  This is the number of categories you have minus 1.  We have four categories: female red eyes, male red eyes and male white eyes.  Therefore 3 – 1 = 2.

In the Chi – Squared test, the critical value is when p = 0.05 (5%)

If you value leads to a p-value MORE than p=0.05 (i.e. more than 5%) then you can accept your null hypothesis that the data is not statistically different.

However, such as in our example using the degrees of freedom and are value we can say using our table that the closest value is 1.05.  This value at 2 degrees of freedom is not on the table but by looking at the trend, you can see that the p-value will be above 0.25 (25%) and therefore we can accept the null hypothesis that the data is not statistically different.

However, if the results indicated a p-value that was LOWER than 0.05 then you would say that you  can reject the null hypothesis and there is a statistically significant difference between the observed and expected results.   This means that there must be another influencing factor on the results because the experiment has not conformed to the usual laws of sex-linked inheritance.