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In: Statistics and Probability

Geneticists examined the distribution of seed coat color in cultivated amaranth grains, Amaranthus caudatus. Crossing black-seeded...

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 328 75 29

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 5% significance level.
(a) Find the value of the test statistic.
(b) Fiind the critical value.
(c)

What is the conclusion?

options for c)

(A) 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). (B) We conclude that the data is consistent with the theory since
the answer in (a) is greater than the answer in (b). (C) We conclude that the theory is true since the answer in (a)
is less than or equal to the answer in (b). (D) We conclude that the observed frequencies contradict the theory
since the answer in (a) is less than or equal to the answer in (b). (E) We cannot conclude that the observed frequencies contradict the theory
since the answer is (a) is greater than the answer in (b). (F) We conclude that the theory is true 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 bigger than the answer in (b).

Solutions

Expert Solution

Computational table:

Colour Oi % Ei (Oi-Ei) (Oi-Ei)^2 (Oi-Ei)^2/Ei
black 328 0.75 =0.75*432 = 324 4 16 0.049
brown 75 0.1875 = 0.1875*432 = 81 -6 36 0.444
pale 29 0.0625 =0.0625*432 = 27 2 4 0.148
Total 432 1 432 0.642

Test statistic:

Degrees of Freedom: k-1 = 3-1 = 2

Where, K = Number of Categories = 3

b)

Critical value: For = 0.05

C)

Test statistic < critical value, i.e 0.642 < 5.99, That is Fail toReject Ho at 5% level of significance.

ANSWER: A

(A) 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)

orchestra answered 2 years ago
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