Question

Problem #1: Geneticists examined the distribution of seed coat color in cultivated amaranth grains, Amaranthus caudatus. Crossing black-seeded and pale-seeded A. caudatus populations gave the following counts of black, brown, and pale seeds in the second generation. Seed Coat Color black brown pale Seed Count 311 85 30 According to genetics laws, dominant epistasis should lead to 12/16  of all such seeds being black, 3/16 brown, and 1/16 pale. We want to test this theory at the 10% significance level.
Problem #1(a):	Find the value of the test statistic (correct to 3 decimals)
Problem #1(b):	Fiind the critical value (correct to 3 decimals)
Problem #1(c):	What is the conclusion? (A) We conclude that the theory is true since the answer in (a) is less than or equal to the answer in (b). (B) We cannot conclude that the observed frequencies contradict the theory since the answer is (a) is greater than the answer in (b). (C) We cannot conclude that the observed frequencies contradict the theory since the answer in (a) is less than or equal to the answer in (b). (D) We conclude that the theory is true since the answer in (a) is greater than the answer in (b). (E) We conclude that the observed frequencies contradict the theory since the answer in (a) is bigger than the answer in (b). (F) We conclude that the data is consistent with the theory since the answer in (a) is greater than the answer in (b). (G) We conclude that the observed frequencies contradict the theory since the answer in (a) is less than or equal to the answer in (b). Problem #1(c):   Select A B C D E F G conclusion

Answer 1

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: Data follows genetics laws.

Alternative hypothesis: H_a: Data do not follow genetics laws.

We are given level of significance = α = 0.10

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

We are given

N = 3

Degrees of freedom = df = N – 1 = 2

α = 0.10

Critical value = 4.60517019

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Color	Exp. Prob.	O	E	(O - E)^2	(O - E)^2/E
Black	0.75	311	319.5	72.25	0.226134585
Brown	0.1875	85	79.875	26.265625	0.328834116
Pale	0.0625	30	26.625	11.390625	0.427816901
Total	1	426	426		0.982785603

Chi square = ∑[(O – E)^2/E] = 0.982785603

P-value = 0.611773724

(By using Chi square table or excel)

P-value > α = 0.10

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the given data follows genetics laws.

Problem 1(a)

Test statistic = 0.983

Problem 1(b)

Critical value = 4.605

Problem 1(c)

Conclusion:

(A) We conclude that the theory is true since the answer in (a)
is less than or equal to the answer in (b).