Question

To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, a chemical company obtained the following data on the time (in minutes) needed to mix the material.

Manufacturer
1	2	3
21	29	21
26	25	19
23	31	23
22	27	21

(a)

Use these data to test whether the population mean times for mixing a batch of material differ for the three manufacturers. Use

α = 0.05.

State the null and alternative hypotheses.

H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.

H₀: μ₁ = μ₂ = μ₃
H_a: Not all the population means are equal.    

H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃

H₀: μ₁ ≠ μ₂ ≠ μ₃
H_a: μ₁ = μ₂ = μ₃

H₀: μ₁ = μ₂ = μ₃
H_a: μ₁ ≠ μ₂ ≠ μ₃

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

t stat =

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

p-value =

State your conclusion.

Do not reject H₀. There is not sufficient evidence to conclude that the mean time needed to mix a batch of material is not the same for each manufacturer.

Do not reject H₀. There is sufficient evidence to conclude that the mean time needed to mix a batch of material is not the same for each manufacturer.    

Reject H₀. There is sufficient evidence to conclude that the mean time needed to mix a batch of material is not the same for each manufacturer.

Reject H₀. There is not sufficient evidence to conclude that the mean time needed to mix a batch of material is not the same for each manufacturer.

(b)

At the α = 0.05 level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers 1 and 3.

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

LSD =

Find the pairwise absolute difference between sample means for manufacturers 1 and 3.

x₁ − x_{3 =}

What conclusion can you draw after carrying out this test? = (Choose one)

There is a significant difference between the means for manufacturer 1 and manufacturer 3.

There is not a significant difference between the means for manufacturer 1 and manufacturer 3.   

Answer 1

	Group 1	Group 2	Group 3	Total
Sum	92	112	84	288
Count	4	4	4	12
Mean, Sum/n	23	28	21
Sum of square, Ʃ(xᵢ-x̅)²	14	20	8

a) State the null and alternative hypotheses.

H0: μ1 = μ2 = μ3
Ha: Not all the population means are equal.

Number of treatment, k = 3

Total sample Size, N = 12

df(between) = k-1 = 2

df(within) = N-k = 9

df(total) = N-1 = 11

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = 104

SS(within) = SS1 + SS2 + SS3 = 42

SS(total) = SS(between) + SS(within) = 146

MS(between) = SS(between)/df(between) = 52

MS(within) = SS(within)/df(within) = 4.6667

F = MS(between)/MS(within) = 11.14

p-value = F.DIST.RT(11.1429, 2, 9) = 0.004

Conclusion.

Reject H0. There is sufficient evidence to conclude that the mean time needed to mix a batch of material is not the same for each manufacturer.

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b) At α = 0.05, N-K = 9, t critical value, t_c =T.INV.2T(0.05, 9) = 2.262

LSD = t_c* √(2*MSW*/n) = 2.262√(2*4.6667/4) = 3.46

Pairwise absolute difference between sample means for manufacturers 1 and 3.

x₁ − x₃ = |23 - 21| = 2

Conclusion:

There is not a significant difference between the means for manufacturer 1 and manufacturer 3.