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 inAnnals of Eugenics(Vol. 7, part 2, pp. 179 -188). Measurements of sepal length in centimeters from random samples ofIris setosa(I),Iris versicolor(II), andIris virginica(III) are as follows below.

I	II	III
5.2	5.3	6.6
4.4	6.7	5.6
5.2	6.5	4.1
5.3	4.3	7.2
4.9	5.8	5.1
5.2	6.2	6.4
5.8	5.3
	6.7

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. State the null and alternate hypotheses. H_o:₁=₂=₃;H₁: All three means are different.

H_o:₁=₂=₃;H₁: Not all the means are equal.

H_o:₁=₂=₃;H₁: At least two means are equal.

H_o:₁=₂=₃;H₁: Exactly two means are equal. (b) FindSS_TOT, SS_BET, andSS_Wand check thatSS_TOT= SS_BET+ SS_W. (Use 3 decimal places.)

SSTOT	=
SSBET	=
SSW	=

Findd.f._BET, d.f._W, MS_BET, andMS_W. (Use 4 decimal places forMS_BET, andMS_W.)

dfBET	=
dfW	=
MSBET	=
MSW	=

Find the value of the sampleFstatistic. (Use 2 decimal places.) What are the degrees of freedom? (numerator) (denominator) (c) Find theP-value of the sample test statistic. 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 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since theP-value is greater than the level of significance at= 0.05, we do not rejectH₀.

Since theP-value is less than or equal to the level of significance at= 0.05, we rejectH₀.

Since theP-value is greater than the level of significance at= 0.05, we rejectH₀.

Since theP-value is less than or equal to the level of significance at= 0.05, we do not rejectH₀.

(e) Interpret your conclusion in the context of the application. At the 5% level of significance there is insufficient evidence to conclude that the means are not all equal.

At the 5% level of significance there is sufficient evidence to conclude that the means are all equal.

At the 5% level of significance there is insufficient evidence to conclude that the means are all equal.

At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal. (f) Make a summary table for your ANOVA test.

Source of Variation	Sum of Squares	Degrees of Freedom	MS	F Ratio	PValue	Test Decision
Between groups					-	.
Within groups
Total