Question

To test for any significant difference in the number of hours between breakdowns for four machines, the following data were obtained.

Machine 1	Machine 2	Machine 3	Machine 4
6.4	8.8	10.9	9.7
7.9	7.6	10.3	12.6
5.5	9.5	9.6	11.8
7.5	10.3	10.2	10.7
8.4	9.3	9.0	11.0
7.5	10.3	8.8	11.4

a) Use Fisher's LSD procedure to test for the equality of the means for machines 2 and 4. Use a 0.05 level of significance.

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

LSD =

b) Find the pairwise absolute difference between sample means for machines 2 and 4.

x₂ − x₄ =

Answer 1

	Mean	n	Std. Dev
Machine 1	7.20	6	1.062
Machine 2	9.30	6	1.018
Machine 3	9.80	6	0.812
Machine 4	11.20	6	0.990
Total	9.38	24	1.726

ANOVA table
Source	SS	df	MS	F	p-value
Treatment	49.485	3	16.4950	17.34	8.71E-06
Error	19.020	20	0.9510
Total	68.505	23

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 ) = 2.086 ( Critical value from t table )

LSD = = 1.17 Since the design is balanced n1 = n1 = n3 = n

Part b)

Since | 9.3 - 11.2 | > 1.1745, we conclude that the population means µ2. and µ4. differ.