Question

Statistics: Chi-squared Introduction:

Answer 1

We perform a Chi-square analysis to find out if the difference between observed and expected is due to chance or not. In this example, 1) FD=1 / 2) X² = 2.22 / 3) Fail to reject the null / 4) 75.

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Since a complete introduction to Chi-square was provided, we will proceed with the problem analysis.

1st Cross: Yellow   x   green

Parentals)  YY    x      yy

F1) 100% Yy, yellow

2nd Cross:  Yellow   x   yellow (From F1)

Parentals)    Yy    x     Yy

F2) 3/4 = 75% of the progeny is expected to be yellow

     1/4 = 25% of the progeny is expected to be green

We know from the data, that there are 1500 individuals in F2. So we can get the expected number of individuals from this data.

100 % of the progeny ----------------------- 1500 individuals

75 % yellow ------------------------------------X = 1125 individuals

25% green -------------------------------------X = 375 individuals

We assume the population is in H-W equilibrium, but we observe differences in what we expect to see and what we actually see. We want to know why.

• H₀ = there is no significant difference between the observed results and the expected results. The difference in observed and expected results is by chance.

• H₁ = The difference in observed and expected results is not just a matter of chance.

Now we will make a table resuming information

                               Yellow                       Green

Observed                 1150                            350

Expected                  1125                            375

(Obs-Exp)²/Exp        0.555                         1.666

• X² = Σ(Obs-Exp)²/Exp = 0.555 + 1.666 = 2.221

• Y = 2

• df = Y-1 = 2 - 1 = 1

• Significance level / probability value = 0.05

• Table value / Critical value = 3.841

• P₀.₀₅ > X²

• 3.841 > 2.221

We can see that the table value is greater than the X² value, so there is not enough evidence to reject the null hypothesis.

The genotypes might be in equilibrium, and there might be independent assortment.

This results suggest that the difference between the observed individuals and the expected individuals is by random chances.

So now let us answer the questions,

1) Determine the degrees of Freedom of the phenotypic ratio for this genetic cross ⇒ DF = 2 - 1 = 1

   

2) What is the result of your Chi-squared analysis? ⇒ X² = 2.22

3) Do we fail to reject or reject the null hypothesis? ⇒

We Fail to reject the null hypothesis. There is not enough evidence to reject it.

4) If we flipped a coin 100 times. Which of the following is the minimum number necessary to achieve significance? 75 is the minimum number necessary to achieve significance

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