Question

Refer to the following partial ANOVA results from EXCEL.

source of variation	ss	df	MS	F-sta
among-group	172.740		43.185
within-group	508.972
total

total # of observation: 58

e. the critical value , F x at x=0.05 (3 decimals)

f. at x= 0.05 your conclusion for testing the null hypothesis is that all the groups have the same mean is:

Answer 1

total observation (n) = 58

the necessary calculation table be:-

source of variation	ss	df	MS= ss/df	F-statistic
among-group	172.740	(172.740/43.185) = 4	43.185	=(43.185 / 9.603) = 4.497
within-group	508.972	(57-4) = 53	(508.972/53) =9.603
total		n-1 = (58-1) = 57

e).number of factors (r) = 4+1 = 5

[ because df among group = r-1 = 4]

The degrees of freedom for the numerator are r – 1 = (5-1) = 4

The degrees of freedom for the denominator are (n – r) = ( 58 - 5) = 53

the critical value be:-

[ from f distribution table, for df1 = 4 , df2 = 53 , alpha = 0.05 ]

f).decision:-

so, we reject the null hypothesis.

hence we conclude that,

there is not enough evidence to claim that all the groups have the same mean at 0.05 level of significance.

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