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
5.1	5.5	6.6
4.7	6.8	5.4
5.3	6.2	4.4
5.9	4.9	7.2
4.1	5.3	5.1
5.4	6.9	6.5
5.7	5.6
	6.4

(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)=

(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--- 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--- Do not reject H0. Reject H0.
Within groups
Total

Answer 1

Applying ANOVA"

Source of Variation	SS	df	MS	F	P-value
Between Groups	1.5098	2	0.7549	1.0651	0.3654
Within Groups	12.7569	18	0.7087
Total	14.2667	20

b)

SSTOT	14.2667
SSBET	1.5098
SSW	12.7569

dfBET	2
dfW	18
MSBET	0.7549
MSW	0.7087

  value of the sample F statistic =1.065

(numerator)=2
(denominator)=18

Source of Variation	SS	df	MS	F	P-value	Decision
Between Groups	1.5098	2	0.7549	1.0651	p value >0.10	fail to reject Ho
Within Groups	12.7569	18	0.7087
Total	14.2667	20