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	21
26	27	19
24	30	22
25	32	26

Use Fisher's LSD procedure to develop a 95% confidence interval estimate (in minutes) of the difference between the means for manufacturer 1 and manufacturer 2. (Round your answers to two decimal places.)

min to  min

Answer 1

Performing the analysis using SPSS

Descriptives
Observation
	N	Mean	Std. Deviation	Std. Error	95% Confidence Interval for Mean		Minimum	Maximum
					Lower Bound	Upper Bound
1.00	4	24.0000	2.16025	1.08012	20.5626	27.4374	21.00	26.00
2.00	4	29.0000	2.44949	1.22474	25.1023	32.8977	27.00	32.00
3.00	4	22.0000	2.94392	1.47196	17.3156	26.6844	19.00	26.00
Total	12	25.0000	3.83761	1.10782	22.5617	27.4383	19.00	32.00

Test of Homogeneity of Variances
Observation
Levene Statistic	df1	df2	Sig.
.176	2	9	.841

Hypothesis:

H0: the mean time needed to mix a batch of material is the same for machines produced by three manufacturers

H1: There is significant difference in the mean time needed to mix a batch of material for machines produced by three manufacturers.

Level of significance: 0.05

Conclusion Since p value =0.010 is less than 0.05 , hence we reject H0 and conclude that there is significant difference in the mean time needed to mix a batch of material for machines produced by three manufacturers

ANOVA
Observation
	Sum of Squares	df	Mean Square	F	Sig.
Between Groups	104.000	2	52.000	8.069	.010
Within Groups	58.000	9	6.444
Total	162.000	11

95% confidence interval estimate (in minutes) of the difference between the means for manufacturer 1 and manufacturer 2.

  (-9.0607   -.9393)

Multiple Comparisons
Dependent Variable: Observation LSD
(I) Manufecturer	(J) Manufecturer	Mean Difference (I-J)	Std. Error	Sig.	95% Confidence Interval
					Lower Bound	Upper Bound
1.00	2.00	-5.00000*	1.79505	.021	-9.0607	-.9393
	3.00	2.00000	1.79505	.294	-2.0607	6.0607
2.00	1.00	5.00000*	1.79505	.021	.9393	9.0607
	3.00	7.00000*	1.79505	.004	2.9393	11.0607
3.00	1.00	-2.00000	1.79505	.294	-6.0607	2.0607
	2.00	-7.00000*	1.79505	.004	-11.0607	-2.9393
*. The mean difference is significant at the 0.05 level.