Question

Given: k=5 where n for each group is 12 (ex. k1 n=12, k2 n=12... k5 n=12) so N=60

Sigma squared = 1.54 (Celcius) ^2 =variability within groups

Level of significance= alpha= .05 (5%)

Question: what is the probability of Anova detecting a difference as small as 2.0(Celcius) between population means?      Im just not sure how to do this without any sample data. I calculated v1=4 and v2=55 but i'm not quite sure how to even start this.

Answer 1

required probability=0.000112

here between group df=v1=4 and within gropup df=55 ( it is also known as error df)

SE(difference of two group mean)=sqrt(2*MSE/n)=sqrt(2*1.54/12)=0.5066

here difference of two population mean=2 and

t=differrenc/SE(difference)=2/0.5066=3.95 with error df=55

P(difference of two mean < 2)=P(t<3.95)=0.000112 ( using ms-excel command=TDIST(3.95,55,1))