Consider the following four simples A, B, C, D. Are there significant different between the mean of the treatment of the samples on the basis of the 5% of level of significant?

• Null Hypothesis ($H_0$): There is no significant difference between the means of the four samples. $$H_0: \mu_A = \mu_B = \mu_C = \mu_D$$
• Alternative Hypothesis ($H_1$): At least one sample mean is significantly different from the others. $$H_1: \text{At least one } \mu_j \text{ is different}$$

The significance level is $\alpha = 0.05$.

Let $n_j$ be the number of observations in sample $j$, $\sum x_j$ be the sum of observations, and $\sum x_j^2$ be the sum of squared observations.

• Sample A: $n_A = 5$ $\sum x_A = 9+7+12+14+13 = 55$ $\bar{x}_A = 11$ $\sum x_A^2 = 9^2+7^2+12^2+14^2+13^2 = 81+49+144+196+169 = 639$
• Sample B: $n_B = 6$ $\sum x_B = 18+12+15+13+16+16 = 90$ $\bar{x}_B = 15$ $\sum x_B^2 = 18^2+12^2+15^2+13^2+16^2+16^2 = 324+144+225+169+256+256 = 1374$
• Sample C: $n_C = 6$ $\sum x_C = 10+9+11+10+7+10 = 57$ $\bar{x}_C = 9.5$ $\sum x_C^2 = 10^2+9^2+11^2+10^2+7^2+10^2 = 100+81+121+100+49+100 = 551$
• Sample D: $n_D = 6$ $\sum x_D = 13+8+9+12+11+10 = 63$ $\bar{x}_D = 10.5$ $\sum x_D^2 = 13^2+8^2+9^2+12^2+11^2+10^2 = 169+64+81+144+121+100 = 679$

Total number of observations $N = n_A+n_B+n_C+n_D = 5+6+6+6 = 23$.
Grand Sum $\sum X = 55+90+57+63 = 265$.
Grand Sum of Squares $\sum X^2 = 639+1374+551+679 = 3243$.
Correction Factor $CF = \frac{(\sum X)^2}{N} = \frac{265^2}{23} = \frac{70225}{23} \approx 3053.2609$.

• Sum of Squares Total (SST): $$SST = \sum X^2 - CF = 3243 - 3053.2609 = 189.7391$$
• Sum of Squares Between Treatments (SSTreatment): $$SSTreatment = \sum_{j=1}^{k} \frac{(\sum x_j)^2}{n_j} - CF$$ $$SSTreatment = \left( \frac{55^2}{5} + \frac{90^2}{6} + \frac{57^2}{6} + \frac{63^2}{6} \right) - 3053.2609$$ $$SSTreatment = \left( \frac{3025}{5} + \frac{8100}{6} + \frac{3249}{6} + \frac{3969}{6} \right) - 3053.2609$$ $$SSTreatment = (605 + 1350 + 541.5 + 661.5) - 3053.2609$$ $$SSTreatment = 3158 - 3053.2609 = 104.7391$$
• Sum of Squares Within Treatments (SSError): $$SSError = SST - SSTreatment = 189.7391 - 104.7391 = 85.0000$$

• df for Treatments ($df_1 = k-1$): $4-1 = 3$
• df for Error ($df_2 = N-k$): $23-4 = 19$
• df Total ($N-1$): $23-1 = 22$

• Mean Square for Treatments (MSTreatment): $$MSTreatment = \frac{SSTreatment}{df_1} = \frac{104.7391}{3} \approx 34.9130$$
• Mean Square for Error (MSError): $$MSError = \frac{SSError}{df_2} = \frac{85.0000}{19} \approx 4.4737$$

For $\alpha = 0.05$, $df_1 = 3$, and $df_2 = 19$, the critical F-value from the F-distribution table is approximately $F_{critical} = 3.127$.

Since the calculated F-statistic (7.8041) is greater than the critical F-value (3.127), we reject the null hypothesis ($H_0$).

There is sufficient evidence to conclude that there are significant differences between the means of the treatments (samples A, B, C, D) at the 5% level of significance.

The final answer is $\boxed{\text{Yes, there are significant differences between the means of the treatments.}}$