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	27	19
26	27	19
23	31	23
26	23	23

(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: μ₁ = μ₂ = μ₃
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.     H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃H₀: μ₁ = μ₂ = μ₃
H_a: μ₁ ≠ μ₂ ≠ μ₃H₀: μ₁ ≠ μ₂ ≠ μ₃
H_a: μ₁ = μ₂ = μ₃

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 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 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 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 (supposed to be absolute value)}

=

What conclusion can you draw after carrying out this test?

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

	Manufacturer 1	Manufacturer 2	Manufacturer 3	Total
Sum	96	108	84	288
Count	4	4	4	12
Mean, Sum/n	24	27	21
Sum of square, Ʃ(xᵢ-x̅)²	18	32	16

(a) 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 = 72

SS(within) = SS1 + SS2 + SS3 = 66

SS(total) = SS(between) + SS(within) = 138

MS(between) = SS(between)/df(between) = 36

MS(within) = SS(within)/df(within) = 7.3333

F = MS(between)/MS(within) = 4.9091

p-value = F.DIST.RT(4.9091, 2, 9) = 0.0362

State your conclusion.

Answer b) 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.   

--

(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*7.3333/4) = 4.33

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

xbar1 − xbar3 = 24-21 = 3

What conclusion can you draw after carrying out this test?

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