Question

The following data are from a completely randomized design.

	Treatment
	A	B	C
	163	142	125
	143	156	121
	165	124	138
	144	143	139
	147	136	149
	168	151	138
Sample mean	155	142	135
Sample variance	132.4	127.6	105.2

(a)

Compute the sum of squares between treatments.

(b)

Compute the mean square between treatments.

(c)

Compute the sum of squares due to error.

(d)

Compute the mean square due to error. (Round your answer to two decimal places.)

(e)

Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments
Error
Total

(f)

At the α = 0.05 level of significance, test whether the means for the three treatments are equal.

State the null and alternative hypotheses.

H₀: μ_A ≠ μ_B ≠ μ_C
H_a: μ_A = μ_B = μ_CH₀: μ_A = μ_B = μ_C
H_a: μ_A ≠ μ_B ≠ μ_C    H₀: Not all the population means are equal.
H_a: μ_A = μ_B = μ_CH₀: At least two of the population means are equal.
H_a: At least two of the population means are different.H₀: μ_A = μ_B = μ_C
H_a: Not all the population means are equal.

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

Reject H₀. There is sufficient evidence to conclude that the means for the three treatments are not equal.Reject H₀. There is not sufficient evidence to conclude that the means for the three treatments are not equal.    Do not reject H₀. There is not sufficient evidence to conclude that the means for the three treatments are not equal.Do not reject H₀. There is sufficient evidence to conclude that the means for the three treatments are not equal.

Answer 1

The statistical software output for this problem is :

(a)

The sum of squares between treatments = 1236

(b)

The mean square between treatments = 618

(c)

The sum of squares due to error = 1826

(d)

The mean square due to error = 121.73

(e)

ANOVA table

Source	DF	SS	MS	F-Stat	P-value
Columns	2	1236	618	5.08	0.0207
Error	15	1826	121.733
Total	17	3062

(f)

H_a: μ_A = μ_B = μ_C

H_a: At least two of the population means are different.

Test statistics = 5.08

P-value = 0.0207

Reject H₀. There is sufficient evidence to conclude that the means for the three

treatments are not equal.