Solution :
We know that

At least one mean is different form the others (claim)
We need to find the critical values.
We know k = 3 , N = 35, α = 0.05
d.f.N = k - 1
= 3 - 1 = 2
d.f.D = N - k
= 35 - 3 = 32
SO the critical value is 3.295
The mean and the variance of each sample :
Goust Jet red Cloudtran


The grand mean or the overall mean is(GM) :


= 52.1714
The variance between the groups

![$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B%5Cleft%5B14%2850.5-52.1714%29%5E2%2B14%2852.07143-52.1714%29%5E2%2B7%2855.71426-52.1714%29%5E2%5Cright%5D%7D%7B3-1%7D%24)

= 63.55714
The Variance within the groups



= 20.93304
The F-test statistics value is :


= 3.036212
Now since the 3.036 < 3.295, we do not reject the null hypothesis.
So there is no sufficient evidence to support the claim that there is a difference among the means.
The ANOVA table is :
Source Sum of squares d.f Mean square F
Between 127.1143 2 63.55714 3.036212
Within 669.8571 32 20.93304
Total 796.9714 34