A genetics engineer was attempting to cross a tiger and a cheetah. She crossed 1

Question

A genetics engineer was attempting to cross a tiger and a cheetah. She crossed 150 tigers and cheetahs. Based on prior research she expected that 25% of the animals would have a phenotypic outcome of stripes only: 10% would have 3 spots only: and 50% would have both stripes and spots and 15% would have no stripes and no spots. When the cross was performed and she counted the individual cubs she found 45 with stripes only, 41 with spots only, 34 with both, and 30 without stripes or spots.         

a.         State the null and research hypothesis.         

b.         Set the level of risk

c.         Select the appropriate test

d.         Compute the obtained value.

e.         Find the critical value

f.          Compare the obtained value and the critical value.

g.         State your decision.

Explanation / Answer

a)null hypothessis: distribution follows prior probability distribution.

alternate hypothesis:distribution does not follows prior probability distribution.

b) level of risk =0.05

c)chi sqaure goodness of fit

d)

from above test stat obtained value =71.48

e) critical value for 3 degree of freedom and 0.05 level =7.8147

f) obtained value is higher then critical value.

g) we reject null hypothesis and conclude that distribution does not follows prior probability distribution.

observed Expected Chi square Probability O E=total*p =(O-E)^2/E stripes only 0.250 45.000 37.50 1.50 spots only 0.100 41.000 15.00 45.07 strips and spots 0.500 34.000 75.00 22.41 no strips and no spots 0.150 30.000 22.50 2.50 1 150 150 71.4800

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