Question

Consider the data in the table collected from three independent populations.		Sample 1	Sample 2	Sample 3
		4	3	4
​a) Calculate the total sum of squares​ (SST) and partition the SST into its two​ components, the sum of squares between​ (SSB) and the sum of squares within​ (SSW).		2	2	2
		9	1	1
				5

​b) Use these values to construct a​ one-way ANOVA table.

​c) Using alphaαequals=0.05​,

what conclusions can be made concerning the population​ means?

Answer 1

For the given data using Anova Single Factor in Excel we get output as

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Sample 1	3	15	5	13
Sample 2	3	6	2	1
Sample 3	4	12	3	3.333333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	14.1	2	7.05	1.298684	0.331368	4.737414
Within Groups	38	7	5.428571
Total	52.1	9

( a ) From the above output

SST = 52.1

SST = SSB + SSW

= 14.1 + 38

= 52.1

( b ) one-way ANOVA table.

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	14.1	2	7.05	1.298684	0.331368	4.737414
Within Groups	38	7	5.428571
Total	52.1	9

( c ) P value < l.o.s

0.3314 > 0.05

So fail to reject H0

At α = 0.05 l.o..s there is enough evidence that all population means are equal