Question

In an effort to stabilize the process, there has been some discussion about 3 machines
that appear to be causing variation in the mean part thicknesses (measured
prior to plating). It appears that these 3 machines may be producing parts with different average
material thickness even though they are supposed to be producing parts with the same average thickness.
It was suggested that a One-Way ANOVA be used to compare the mean thicknesses
produced by these 3 machines. You will be checking for significant differences between
the mean material thicknesses produced by these 3 machines. Use a 5% level of significance.
Data:
	Machine 1	Machine 2	Machine 3
	0.546	0.573	0.573
	0.526	0.592	0.570
	0.587	0.571	0.527
	0.563	0.556	0.572
Question:
Determine degrees of freedom (df) total AND df factor

I got 11 for df-total and 3 for df-factor

Answer 1

df(Total) = 11

df(factor) = 3-1 = 2

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Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Machine 1	4	2.222	0.5555	0.00067
Machine 2	4	2.292	0.573	0.000218
Machine 3	4	2.242	0.5605	0.0005
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	0.00065	2	0.000325	0.70245	0.52062	4.256495
Within Groups	0.004164	9	0.000463
Total	0.004814	11

Conclusion= There is no significant difference between the mean of machines (Reason:- P-value is > 0.05 so, we fail to reject Ho)