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 5.5 5.2 6.8 4.6 6.5 5.3 5.1 6.1 4.4 5.5 4.1 7.9 4.1 5.1 5.9 5.4 6.1 6.9 5.4 5.1 6.6 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. Ho: ?1 = ?2 = ?3; H1: Exactly two means are equal.Ho: ?1 = ?2 = ?3; H1: Not all the means are equal. Ho: ?1 = ?2 = ?3; H1: All three means are different.Ho: ?1 = ?2 = ?3; H1: At least two means are equal. (b) Find SSTOT, SSBET, and SSW and check that SSTOT = SSBET + SSW. (Use 3 decimal places.) SSTOT = SSBET = SSW = Find d.f.BET, d.f.W, MSBET, and MSW. (Use 4 decimal places for MSBET, and MSW.) 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. P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.001 < P-value < 0.010P-value < 0.001

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Description: 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 tak

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
5.5	5.2	6.8
4.6	6.5	5.3
5.1	6.1	4.4
5.5	4.1	7.9
4.1	5.1	5.9
5.4	6.1	6.9
5.4	5.1
	6.6

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.

H_o: ?₁ = ?₂ = ?₃; H₁: Exactly two means are equal.H_o: ?₁ = ?₂ = ?₃; H₁: Not all the means are equal. H_o: ?₁ = ?₂ = ?₃; H₁: All three means are different.H_o: ?₁ = ?₂ = ?₃; 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.)

SS_TOT	=
SS_BET	=
SS_W	=

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

df_BET	=
df_W	=
MS_BET	=
MS_W	=

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.

P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.001 < P-value < 0.010P-value < 0.001

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value is greater than the level of significance at ? = 0.05, we do not reject H₀.Since the P-value is less than or equal to the level of significance at ? = 0.05, we reject H₀. Since the P-value is greater than the level of significance at ? = 0.05, we reject H₀.Since the P-value is less than or equal to the level of significance at ? = 0.05, we do not reject H₀.

(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	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