Question

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

Manufacturer
	1	2	3
	25	33	20
	31	31	19
	29	36	23
	27	32	22

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

Compute the values below (to 2 decimals, if necessary).

Sum of Squares, Treatment
Sum of Squares, Error
Mean Squares, Treatment
Mean Squares, Error

Calculate the value of the test statistic (to 2 decimals).

The -value is - Select your answer

-less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6

What is your conclusion?

- Select your answer

-Conclude the mean time needed to mix a batch of material is not the same for all manufacturersDo not reject the assumption that mean time needed to mix a batch of material is the same for all manufacturersItem 7

b. At the  level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers  and .

Calculate Fisher's LSD Value (to 2 decimals).

What is your conclusion about the mean time for manufacturer  and the mean time for manufacturer ?

- Select your answer

-These manufacturers have different mean timesCannot conclude there is a difference in the mean time for these manufacturersItem 9

Answer 1

	1	2	3
	25	33	20
	31	31	19
	29	36	23
	27	32	22
Total(yi)	112	132	84
Averages y̅i	28	33	21
Treatment Effect	28 - 27.33 = 0.67	33 - 27.33 = 5.67	21 - 27.33 = -6.33

	Sum of Squares	Degree of freedom	Mean Square	F0 = MST / MSE	P value
Treatment	290.68	2	145.34	29.72	< 0.01
Error	44	9	4.89
Total	334.68	11

Overall total = 328
Overall mean Y̅.. = 328 / 12 = 27.33

Y̅ .. is overall mean
y̅i . is treatment mean
SS total = ΣΣ(Yij - & Y̅..)2 = 334.68
SS treatment = ΣΣ(Yij - & y̅i.)2 = 290.68
SS error = Σ(y̅i. - & Y̅..)2 = 44

MS treatment = ΣΣ(Yij - & y̅i.)2 / a - 1 = 145.34
MS error = Σ(y̅i. - & Y̅..)2 / N - a = 4.89

SST = 290.68

SSE = 44.00

MST = 145.34

MSE = 4.89

Test Statistic :-
f = MS treatment / MS error = 29.72

P value < 0.01

Decision based on P value
Reject null hypothesis if P value < α = 0.05
Since P value = 0 < 0.05, hence we reject the null hypothesis
Conclusion :- Treatment means differs

Conclude the mean time needed to mix a batch of material is not the same for all manufacturers

part b)

Fisher's Least significant Difference ( LSD Method )
The pair of means µi. and µj. would be declared significantly different if > LSD
t(α/2 , N-a ) = 2.26 ( Critical value from t table )

LSD = = 3.54 Since the design is balanced n1 = n1 = n3 = n

= | 28 - 33 | = 5
= | 28 - 21 | = 7
= | 33 - 21 | = 12
Since | 28 - 33 | > 3.54, we conclude that the population means µ1. and µ2. differ.
Since | 28 - 21 | > 3.54, we conclude that the population means µ1. and µ3. differ.
Since | 33 - 21 | > 3.54, we conclude that the population means µ2. and µ3. differ.