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
20	28	20
25	25	18
24	32	24
27	27	18

(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: μ₁ = μ₂ = μ₃
H_a: μ₁ ≠ μ₂ ≠ μ₃    H₀: μ₁ ≠ μ₂ ≠ μ₃
H_a: μ₁ = μ₂ = μ₃H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃H₀: 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 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 − x₃

=

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

treatment 1 treatment 2 treatment 3 Total ******** 288 24.0000 grand mean Xgrand- Total=EX = Average=X=EX/n= Total=EX? = s? =(EX?-(2x)2/n:)/(n-1) = 24.0000 2330 112 28.0000 3162 8.6667 20.0000 1624 8.6667 8.0000 Total observation number of groups between groups degree of freedom ofbe Ek-1 = within groups degree of freedom dfwe=N-K = Total degree of freedom dfrotal = treatment 1 treatment 2 treatment 3 total 24.0000 28.0000 20.0000 san 8.667 8.667 8.000 *(x-Xgrand)? 0.00 64.00 64.00 128.0000 (n-1)*s? 26.00 26.00 24.00 76.0000 128.0000 sum of square between groups ESSG = sum of square within groups =SSE = total sum of square =SST = En *(X-Xgrand) = (n-1)*s? SSG+SSE = 76.0000 204.0000 Summary table Source of variation treatments error total MS 64.0000 8.4444 Fp vlaue 7.579 0.012 128.000 76.000 204.000

a)

from above:

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

value of the test statistic F =7.58

p value =0.012

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)

Fisher's (LSD) for group i and j =                  (tN-k)*(sp*√(1/ni+1/nj)           =

4.65

x1-x3

4.00

there is not a significant difference between the means for treatment 1 and 3