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# The chi-squared test

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Introduction

The chi-squared test

Problem:

Mendel’s Law of  Independent Assortment makes a prediction about the ratios in which phenotypes occur. If we observe a population it is possible that we observe different results than those we can predict with the help of Mendel’s laws.

Example: Garden peas

We calculate/predict the phenotypic ratio of garden peas:

Y= yellow

y = green

R = round

r = wrinkled

First of all we create a test cross to predict the genotypes/phenotypes if the genotype of the F1 is YyRy:

 YR Yr yR yr YR YYRR YYRr YyRR YyRr Yr YYrR YYrr YyrR Yyrr yR yYRR yYRr yyRR YyRr yr YyRr Yyrr yyRr yyrr

Ratio: 9 (round yellow) : 3 (round green) : 3 (wrinkled yellow) : 1 (wrinkled green)

Genotypes: Not interesting for our topic

If we observe 556 peas

Middle

But as nature is not always obedient to mathematical predictions our observation has the following result:

 # round yellow 315 # round green 108 # wrinkled yellow 101 # wrinkled green 32

This result is the observed result, called O in the test.

Question: Is our prediction wrong or is the deviation from the expected result caused by chance?

In order to answer this question we have to use the so-called chi-squared test. It is used, in mathematical terms, to estimate the probability that differences between the observed result (O) and the expected result (E) are due to chance.

Before we go further into detail we have to take the following things –although seeming obvious – into consideration:

• the smaller the deviation the more sure can we be that our prediction is right
• the higher the deviation from the expected result the more sure we can be that our prediction was wrong
• Rule: If the probability that the deviation is due to chance is smaller than 5% we falsify our prediction; otherwise we accept it

Conclusion

for round yellow

0.101

for round yellow

0.218

The sum of those values, the calculation of x2, is 0.470. In order to find our the probability, we need a

statistical table

 probability 0.99 0.95 0.9 0.7 0.5 0.3 0.1 0.05 0.01 0.001 Result (sum) 0.115 0.35 0.58 0.71 1.39 3.66 6.25 7.82 11.34 16.27

Our value lies between 0.35 and 0.58. This means that with a probability of 90-95% the deviation is due to chance. Our prediction was right.

Questions: (to see if you have understood it!)

What happens if the expected result is identical with the observed result? (Use the formula?

Another observation of 556 garden peas had the following result:

 # round yellow 200 # round green 100 # wrinkled yellow 106 # wrinkled green 150

Would this observation falsify our predicted result? (Use the formula)

What are the requirements for approval/falsification of our predictions?

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