Question

For questions 2 and 3, write your null hypothesis, perform a chi-square analysis, and answer the questions.

2) Null Hypothesis:

Class Data Observed Expected O - E (O – E)^2 (O – E)2/E   

Green 315

Polka-dot 85

Degrees of freedom =

X2 =

p =

Fail to Reject or Reject null hypothesis?

Answer 1

Null hypothesis states that there is no significant difference between observed and expected frequencies.

Chi- square analysis:-

1. First we predict the expected frequencies. Here the expected ratio is 3:1 and the expected frequencies will be 300 and 100 as the total no. Of offspring given is 400.

2. Subtract expected frequencies from the observed frequency. O-E

3. (O-E)^2 and then (O-E)^2/E.

4. We compare the chi square value from the chi square table by taking p<0.05 and degree of freedom = 1

Degree of freedom = no. Of phenotype - 1X^2 = 3.00

From the chi square table we can see the critical value = 3.841

When the chi square value exceeds the critical value then we can reject the null hypothesis but here we can see that the chi square value is less than the critical value so we fail to reject the null hypothesis.

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