Question

A geneticist hypothesizes that half of a given population will have brown eyes and the remaining half will split evenly between blue- and green-eyed people. use the following sample of sixty individuals to test the geneticist's claim at a 5% of significance.

Brown eyes	Green eyes	Blue Eyes
34	15	11

A. Hypothesis:

B. Degrees of freedom:

C. Expected number:

D. Test statistic:

E. Approximate p-value.

F. Decision:

G. what is the appropriate distribution? your choices are : Chi-squared, normal or t-distribution.

Answer 1

A) null hypothesis: Ho: p_brown =0.50 , p_green =0.25 , p_{blue eyes} =0.25

Alternate hpothesis: Ha: at least one of above proportion is not correct

b)

degree of freedom =categories-1=

3

c)

Expected number for brown eyes =np=60*0.5 =30

Expected number for green eyes =np=60*0.25 =15

Expected number for blue eyes =np=60*0.25 =15

d)

Applying goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
brown	0.50	34.00	30.00	0.73	0.533
green	0.25	15.00	15.00	0.00	0.000
blue	0.25	11.00	15.00	-1.03	1.067
total	1.000	60	60		1.600

  Test statistic: =1.600
e) p value =0.4493

F) Decision: fail to reject Ho

G)  Chi-squared