The claim is that the observed genotype offspring frequencieshave the distributi
ID: 2914126 • Letter: T
Question
The claim is that the observed genotype offspring frequencieshave the distribution of pAA = 0.25, pAa =0.50, pzz = 0.25.
If the claim is false, then one of the following is truepAA 0.25, pAa 0.50, paa 0.25.
H0: pAA = 0.25, pAa =0.50, pzz = 0.25 This is really important because we'lluse these proportions to calculate our expected frequencies. SopAA = 0.25 will be 36.25, and pAa= 0.50,and this will be the E for the second row (50% or double the 25%used in row 1), and then pzz = 0.25, which is the 36.25once again. H1: At least one of theproportions is not equal to the given claimed value.
= 0.05
Genotype
Observed Frequency O
Expected Frequency E
O - E
(O – E)2
(O – E)2/E
AA
20
36.25
-16.25
264.0625
7.28448
Aa
90
72.5
17.5
306.25
4.22414
aa
35
36.25
-1.25
-1.5625
-0.04310
Sum of columns
145
145
0
568.75
11.46552
Genotype
Observed Frequency O
Expected Frequency E
O - E
(O – E)2
(O – E)2/E
AA
20
36.25
-16.25
264.0625
7.28448
Aa
90
72.5
17.5
306.25
4.22414
aa
35
36.25
-1.25
-1.5625
-0.04310
Sum of columns
145
145
0
568.75
11.46552
Explanation / Answer
You already mentioned the calculated value ofchi-square = 11.46552 The critical value of chi-square at 0.025(sincethis is a two tailed therefore we have consider/2=0.05/2=0.025), the level of significance and thedf=(3-1=2) is 7.38 Since the calcualted value of chi-squaer isgreater than the critical value of chi-squaer so we reject the nullhypothesis and conclude that the observed genotype offspringfrequencies does not fit the expected distribution of 25% forAA, 50% for Aa, and 25% for aa.Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.