Solution :
We know that
![$H_0: \mu_1 = \mu_2=\mu_3$](https://tex.z-dn.net/?f=%24H_0%3A%20%5Cmu_1%20%3D%20%5Cmu_2%3D%5Cmu_3%24)
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
![$\overline X_3 =55.71429$](https://tex.z-dn.net/?f=%24%5Coverline%20X_3%20%3D55.71429%24)
![$s_3^2=36.57143$](https://tex.z-dn.net/?f=%24s_3%5E2%3D36.57143%24)
The grand mean or the overall mean is(GM) :
![$\overline X_{GM}=\frac{\sum \overline X}{N}$](https://tex.z-dn.net/?f=%24%5Coverline%20X_%7BGM%7D%3D%5Cfrac%7B%5Csum%20%5Coverline%20X%7D%7BN%7D%24)
![$=\frac{51+51+...+49+49}{35}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B51%2B51%2B...%2B49%2B49%7D%7B35%7D%24)
= 52.1714
The variance between the groups
![$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$](https://tex.z-dn.net/?f=%24s_B%5E2%3D%5Cfrac%7B%5Csum%20n_i%5Cleft%28%20%5Coverline%20X_i%20-%20%5Coverline%20X_%7BGM%7D%5Cright%29%5E2%7D%7Bk-1%7D%24)
![$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$](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)
![$=\frac{127.1143}{2}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B127.1143%7D%7B2%7D%24)
= 63.55714
The Variance within the groups
![$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$](https://tex.z-dn.net/?f=%24s_W%5E2%3D%5Cfrac%7B%5Csum%28n_i-1%29s_i%5E2%7D%7B%5Csum%28n_i-1%29%7D%24)
![$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B%2814-1%2919.96154%2B%2814-1%2914.68681%2B%287-1%2936.57143%7D%7B%2814-1%29%2B%2814-1%29%2B%287-1%29%7D%24)
![$=\frac{669.8571}{32}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B669.8571%7D%7B32%7D%24)
= 20.93304
The F-test statistics value is :
![$F=\frac{s_B^2}{s_W^2}$](https://tex.z-dn.net/?f=%24F%3D%5Cfrac%7Bs_B%5E2%7D%7Bs_W%5E2%7D%24)
![$=\frac{63.55714}{20.93304}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B63.55714%7D%7B20.93304%7D%24)
= 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