Question

18) 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	5.6	6.9
4.6	6.7	5.3
4.5	6.6	4.3
5.4	4.5	7.8
4.1	5.9	5.3
5.6	6.2	6.1
5.1	5.1
	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?

(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

Within groups
Total

Answer 1

Firstly, let's write down the null & the alternate hypothesis.

• Null Hypothesis - There is no difference among the population means of the sepal length
• Alternate Hypothesis - There is a difference among the population mean of the sepal length

(a) Given that the significance level is 5%, therefore the level of significance is 5%. Hence, the cut off value will be 0.05

(b)

• SSTOT = 18.02
• SSBET = 3.806
• SSW = 14.214

• dfBET = 2
• dfW = 18
• MSBET = 1.9031
• MSW = 0.7896

Degree of freedom numerator = dfBET = 2

Degree of freedom denominator = dfW = 18

F statistic value = 2.41

(f)

Source of Variation	Sum of Squares	DOF	Mean Squares	F Statistic value sample	F Statistics value from F Table	Test Decision
Between	3.80625	2	1.903125	2.410078269	3.369	Fail to reject the null hypothesis, therefore, there is no difference among the population means of the sepal length
Within	14.21375	18	0.789652778
Total	18.02	20	0.901