Question

Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by R. A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179 -188). Measurements of sepal length in centimeters from random samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows below.

I	II	III
5.7	5.1	6.9
4.2	6.5	5.5
4.6	6.6	4.7
5.2	4.6	7.8
4.4	5.3	5.2
5.4	6.3	6.6
5.5	5.9
	6.3

Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a 5% level of significance.

(a) What is the level of significance?
State the null and alternate hypotheses.(from the following)
H_o: μ₁ = μ₂ = μ₃; H₁: Not all the means are equal.

H_o: μ₁ = μ₂ = μ₃; H₁: All three means are different.    

H_o: μ₁ = μ₂ = μ₃; H₁: Exactly two means are equal.

H_o: μ₁ = μ₂ = μ₃; H₁: At least two means are equal.

(b) Find SS_TOT, SS_BET, and SS_W and check that SS_TOT = SS_BET + SS_W. (Use 3 decimal places.)

SSTOT	=
SSBET	=
SSW	=

Find d.f._BET, d.f._W, MS_BET, and MS_W. (Use 4 decimal places for MS_BET, and MS_W.)

dfBET	=
dfW	=
MSBET	=
MSW	=

Find the value of the sample F statistic. (Use 2 decimal places.)


What are the degrees of freedom?
(numerator)
(denominator)

(c) Find the P-value of the sample test statistic.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
(e) Interpret your conclusion in the context of the application.

(f) Make a summary table for your ANOVA test.

Source of Variation	Sum of Squares	Degrees of Freedom	MS	F Ratio	P Value	Test Decision
Between groups					---Select from the following--- p-value > 0.100 -- 0.050 < p-value < 0.100 -- 0.025 < p-value < 0.050 -- 0.010 < p-value < 0.025 -- 0.001 < p-value < 0.010 -- p-value < 0.001	---Select from the following--- Do not reject H0. -- Reject H0.
Within groups
Total

Answer 1

a)

level of significance =0.05

H_o: μ₁ = μ₂ = μ₃; H₁: Not all the means are equal.

b)

SSTOT	17.327
SSBET	4.503
SSW	12.823

dfBET	2
dfW	18
MSBET	2.2517
MSW	0.7124

value of test statistic for factor A =

3.16

df(numerator) =		2
df(Denominator) =		18

c)

p value =0.0666

d)

fail to reject the null hypothesis

e) can not conclude that population means differs

f)

Source	SS	df	MS	F	P value
Between	4.503	2	2.2517	3.16	0.0666
Within	12.823	18	0.7124
Total	17.327	20

do not reject ho