Question

A teacher is explaining to her class the concepts of genetics regarding eye color. She assumes that each category will have the same number of students in it. Use a 0.05 level of significance. Here are the results for the class. Observed Eye Colors: Blue=3; Green=2; Brown= 12, Hazel= 3, Expected Eye Colors: All the categories of eye colors are the same. What are the expected values for the blue eye color? What are the expected values for the green eye color? What are the expected values for the brown eye color? What are the expected values for the hazel eye color

Answer 1

There are a total of 3+2+12+3 =20 students.

The expected value for each eye color(Blue,Green,Brown,Hazel) is simply 5 (Teacher assumes that each category will have the same number of students in it)

The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
Blue	3	20*0.25=5	(3-5)2/5 = 0.8
Green	2	20*0.25=5	(2-5)2/5 = 1.8
Brown	12	20*0.25=5	(12-5)2/5 = 9.8
Hazel	3	20*0.25=5	(3-5)2/5 = 0.8
Sum =	20	20	13.2

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ha​: Some of the population proportions differ from the values stated in the null hypothesis

This corresponds to a Chi-Square test for Goodness of Fit.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=4−1=3, so then the rejection region for this test is

(3) Test Statistics

The Chi-Squared statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that , it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level.

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