Question

1) Present the analysis of variance (ANOVA) table in standard format.

2) Report the result of the test, including an appropriate critical value for the test.

3) Comment on the validity of test result from question

4) Append a relevant boxplot and a means plot. Use these plots to interpret the results in (2) and (3).

The data set:

"Group" "Y"
"A" -2.379
"A" 0.879
"A" -4.805
"A" -4.879
"A" -6.177
"A" 3.568
"A" -0.181
"A" 1.438
"A" -3.016
"A" -2.304
"B" 1.226
"B" 0.281
"B" 4.517
"B" 0.064
"B" 2.78
"B" 0.765
"C" 1.045
"C" -3.7
"C" -1.904
"C" -3.814
"C" -2.77
"C" 0.022
"C" -5.203
"C" -1.521
"D" -2.704
"D" 0.616
"D" -4.831
"D" 1.292
"D" -2.937
"D" -1.766
"D" -0.49
"D" 6.499
"D" -11.003
"D" -8.869
"D" -2.268
"E" -7.252
"E" -3.603
"E" -4.424
"E" -7.569
"E" -13.572
"E" 2.913
"E" 4.061
"F" 2.771
"F" 0.912
"F" 7.875
"F" 8.463
"F" 5.563
"F" 6.285
"F" 6.816

Answer 1

(1)

One factor ANOVA
	Mean	n	Std. Dev
	-1.7856	10	3.14170	A
	1.6055	6	1.72263	B
	-2.2306	8	2.07730	C
	-2.4055	11	4.77636	D
	-4.2066	7	6.16096	E
	5.5264	7	2.74664	F
	-0.8835	49	4.75491	Total
ANOVA table
Source	SS	df	MS	F	p-value
Treatment	450.21930	5	90.043860	6.10	.0002
Error	635.02148	43	14.767941
Total	1,085.24078	48

(b) Critical value for α = 0.05 and df = (5, 43) is F = 2.4322

Since 6.10 > 2.4322, we reject Ho and conclude that at least two of the six groups have significantly different means

(c) We assumed the following so that our model is valid:

Independence of cases – this is an assumption of the model that simplifies the statistical analysis.

Normality – the distributions of the residuals are normal.

Equality (or "homogeneity") of variances

(d)

We see from the plot that the following groups may have significantly different means {A and F, B and D, B and E, C and F, D and F, E and F}