Question

You may need to use the appropriate technology to answer this question.

Consider the experimental results for the following randomized block design. Make the calculations necessary to set up the analysis of variance table.

		Treatments
		A	B	C
Blocks	1	10	9	8
	2	12	6	5
	3	18	16	14
	4	20	18	18
	5	8	7	8

Use α = 0.05 to test for any significant differences.

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: μ_A = μ_B = μ_C

H_a: Not all the population means are equal.H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.

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

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

p-value =

State your conclusion.

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

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

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

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

Answer 1

(a)

Correct option

H₀: μ_A = μ_B = μ_C

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

(b)

From the given data, the following Table is calculated:

	Treatment A	Treatment B	Treatment C
n	5	5	5
Sum	68	56	53
Mean	13.6	11.2	10.6
	1032	746	673
Std. Dev.	5.177	5.45	5.273
SS	107.2	118.8	111.2

the value of the test statistic is given by:

(c)

By Technology,

p - value = 0.6489

(d)

From the above data. ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p - value
Between treatments	25.2	2	12.6	12.6/28.1=0.45	0.6489
Within treaments	337.2	12	28.1
Total	362.4	14

(e)

Correct option:

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