In a completely randomized design 12 experimental units were

In a completely randomized design, 12 experimental units were for the first treatment 15 for the second treatment and 20 for the third treatment. Complete the following analysis of variance

Solution

as mentioned in the question the treatment has three factors- first treatment, second treatment, third treatment

so number of factors=k=3

now there are 12 experimental units for the first treatment

15 experimental units for the second treatment

20 experimental units in the third treatment

so total number of experimental units=n=15+12+20=47

we are to test at 5% level of significance whether there exist a significant difference between the treatments.

let a1 be the additional effect of the first treatment, a2 be the additional effect of the second treatment, a3 be the additional effect of the third treatment.

so here the null hypothesis is H0: a1=a2=a3=0 [no significant difference between the treatments] vs H1: there exist significant difference between the treatments

now sum of squares of treatment+sum of squares of error=total sum of squares

now it is given that sum of squares of treatment=1200   and total sum of squares=1800

so sum of suqares of error=1800-1200=600

now degrees of freedom of treatment is k-1=3-1=2

degrees of freedom of error is n-k=47-3=44

so total degrees of freedom =44+2=46

hence mean square of treatment=MST=sum of squares of treatment/degrees of freedom of treatment=1200/2=600

mean square of error=MSE=sum of squares of error/degrees of freedom of error=600/44=13.636=13.64 [rounded to two decimal]

hence the F ratio is defined as F=MST/MSE=600/13.64=43.988=43.99 [rounded to two decimal]

hence the analysis of variance table is

source of variation     sum of squares      degrees of freedom     mean square      F

treatment                       1200                           2                          600            43.99

error                               600                            44                        13.64

total                              1800                            46

now under H0   F follows an F distribution with dfs k-1=2 and n-k=44

at 5% level of significance we reject H0 if p value<levek of significance=0.05

here a right tailed test is used.

so p value=p=P[F_2,44>43.99]=0.000 [answer]

so p<0.05

hence H0 is rejected and the conlusion is that there exist significant difference between the treatments [answer]