The claim is that the observed genotype offspring frequencieshave the distributi

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.

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