Question

The following data are from a completely randomized design. Treatment Treatment Treatment 32 30 30 26 32 30 35 38 37 38 42 38 6.5 45 45 47 49 46 Sample mean Sample variance At the α-.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error

Answer 1

Here we have data:

A	B	C
32	45	35
30	44	38
30	45	37
26	47	38
32	49	42
30	46	38
6	4	6.5

Here we calculate by Excel

Excel output

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Column 1	7	186	26.57143	86.28571
Column 2	7	280	40	254.6667
Column 3	7	234.5	33.5	146.0833
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	631.3571	2	315.6786	1.944489	0.171958	3.554557
Within Groups	2922.214	18	162.3452
Total	3553.571	20

Sum of square treatment = 631.4

Sum of square of error = 2922.2

Mean square of treatment = 315.7

Mean square of error = 162.3

Here we have not sufficient evidence to reject the null hypothesis, because F-observed value (1.94) is less than F-critical value(3.55)

Conclusion: here we can say three treatment is equal.