Question

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 307 75 32

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 1% significance level.
(a) [2 marks] Find the value of the test statistic.
(b) [1 mark] Fiind the critical value.
(c)

[1 mark] What is the conclusion?

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

Solutions

Expert Solution

a)

applying chi square goodness of fit test:
           relative observed Expected residual Chi square
category frequency(p) Oi Ei=total*p R2i=(Oi-Ei)/√Ei R2i=(Oi-Ei)2/Ei
1    3/4 307 310.50 -0.20 0.039
2    3/16 75 77.63 -0.30 0.089
3    1/16 32 25.88 1.20 1.450
total 1.000 414 414 1.5781
test statistic X2 = 1.5781

b)

degree of freedom =categories-1= 2
for 0.01 level and 2 df :crtiical value X2 = 9.2103

c)

  (F) 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).


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