Question

When counting seeds, Gregor noticed that some of them were wrinkled and some round. He wanted to know if the number of wrinkled seeds he found was unusual. After doing a little research he found that 26% of pea seeds are expected to be wrinkled normally. Out of the 60 seeds he grew, he found 12 wrinkled ones.

Perform a chi square test and give your results.

a) X2 (chi square)

b) Degrees of Freedom

c) p-value

Answer 1

Observed values:

Wrinkled = 12

Round = 60-12 = 48

Expected values:

Wrinkled = 26/100 *60 = 15.6

Round = 60-15.6 = 44.4

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
Round	48	44.4	3.6	12.9600	0.2919
female	12	15.6	-3.6	12.9600	0.8308
Total	60	60			1.1227

a) X2 (chi square) = 1.1227

b) Degrees of Freedom (DF)

DF = No. of phenotypes - 1

DF = 2-1=1

c) p-value = 3.84

The Chi-square value of 1.1227 is less than the p-value of 3.84, so the null hypothesis is accepted.