(15 points) Based on the genotypes of parents, offsprings are expected to have g
ID: 3318503 • Letter: #
Question
(15 points) Based on the genotypes of parents, offsprings are expected to have genotypes distributed in such a way that 25% have genotypes denoted by AA, 50% have genotypes denoted by Aa, and 25% have genotypes denoted by aa, when 145 offsprings are obtained, it is found that 20 of them have AA genotypes, 90 have Aa genotypes, and 35 have aa genotypes. Test the claim that the observed genotype offspring frequencies fit the expected distribution of 25% for AA, 50% for Aa, and 25% for aa. Use a significance level of 0.05.Explanation / Answer
Solution:
Here, we have to use Chi square test for goodness of fit. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The observed genotype offspring frequencies fit the expected distribution of 25% for AA, 50% for Aa, and 25% for aa.
Alternative hypothesis: Ha: The observed genotype offspring frequencies do not fit the expected distribution of 25% for AA, 50% for Aa, and 25% for aa.
We are given level of significance or alpha value as 0.05. ( = 0.05).
The test statistic formula is given as below:
Chi square = [(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
Table for calculations is given as below:
No.
O
Prob.
E
(O - E)
(O - E)^2/E
1
20
0.25
36.25
-16.25
7.284482759
2
90
0.50
72.5
17.5
4.224137931
3
35
0.25
36.25
-1.25
0.043103448
Total
145
1.00
145
0
11.55172414
P-value =
0.003101523
Chi square = [(O – E)^2/E] = 11.55172414
P-value = 0.003101523 (by using Chi square table or excel)
= 0.05
P-value < = 0.05
So, we reject the null hypothesis that observed genotype offspring frequencies fit the expected distribution of 25% for AA, 50% for Aa, and 25% for aa.
There is insufficient evidence to conclude that observed genotype offspring frequencies fit the expected distribution of 25% for AA, 50% for Aa, and 25% for aa.
No.
O
Prob.
E
(O - E)
(O - E)^2/E
1
20
0.25
36.25
-16.25
7.284482759
2
90
0.50
72.5
17.5
4.224137931
3
35
0.25
36.25
-1.25
0.043103448
Total
145
1.00
145
0
11.55172414
P-value =
0.003101523
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