Question

Apply the technique of ANOVA to the following data showing the yields
of 3 varieties of a crop each from 4 blocks, and test whether the mean
yields of the varieties are equal or not. Also test the equality of the block
means.
Varieties Blocks

I II III IV
A 4 8 6 8
B 5 5 7 8
C 6 7 9 5

Answer 1

The following table is obtained:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & A & B & C \\ \hline & 4 & 5 & 6 \\ & 8 & 5 & 7 \\ & 6 & 7 & 9 \\ & 8 & 8 & 5 \\ \hdashline Sum= & 26 & 25 & 27 \\ \hdashline Average= & 6.5 & 6.25 & 6.75 \\ \hdashline \sum_iX_{ij}^2= & 180 & 163 & 191 \\ \hdashline St.Dev.= & 1.915 & 1.5 & 1.708 \\ \hdashline SS= & 11 & 6.75 & 8.75 \\ \hdashline n= & 4 & 4 & 4 \\ \hdashline \end{array}$$

The total sample size is $N=12.$ Therefore, the total degrees of freedom are:

$$df=12-1=11$$

The between-groups degrees of freedom are $df_{between}=3-1=2,$ and the within-groups degrees of freedom are:

$$df_{within}=df_{total}-df_{between}=11-2=9$$

We need to compute the total sum of values and the grand mean. The following is obtained

$$\sum_{\mathclap{ i, j}} X_{ij}=25+26+27=78$$

Also, the sum of squared values is

$$\sum_{\mathclap{ i, j}} X_{ij}^2=180+163+191=534$$

The total sum of squares is computed as follows

$$SS_{total}=\sum_{\mathclap{ i, j}} X_{ij}^2-\dfrac{1}{N}(\sum_{\mathclap{ i, j}} X_{ij})^2=$$

$$=534-\dfrac{(78)^2}{12}=27$$

The within sum of squares is computed as shown in the calculation below:

$$SS_{within}=\sum SS_{withingroups}=11+6.75+8.75=26.5$$

The between sum of squares is computed directly as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

$$MS_{between}=\dfrac{SS_{between}}{df_{between}}=\dfrac{0.5}{2}=0.25$$

$$MS_{within}=\dfrac{SS_{within}}{df_{within}}=\dfrac{26.5}{9}=2.944$$

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{0.25}{2.944}=0.085$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2=\mu_3$

$H_1:$ Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

Based on the information provided, the significance level is $\alpha=0.05,$ and the degrees of freedom are $df_1=2$ and $df_2=9,$ therefore, the rejection region for this F-test is $R=\{F:F>F_c=4.2565\}$

Test Statistics

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{0.25}{2.944}=0.085$$

Since it is observed that $F=0.085<4.2565=F_c,$ it is then concluded that the null hypothesis is not rejected.  Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the $\alpha=0.05$ significance level.

Using the P-value approach:

The p-value is $p=0.919241,$ and since $p=0.919241>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the $\alpha=0.05$ significance level.

The following table is obtained:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c: c} & I & II & III & IV \\ \hline & 4 & 8 & 6 & 8 \\ & 5 & 5 & 7 & 8 \\ & 6 & 7 & 9 & 5 \\ \hdashline Sum= & 15 & 20 & 22 & 21 \\ \hdashline Average= & 5 & 6.667 & 7.333 & 7 \\ \hdashline \sum_iX_{ij}^2= & 77 & 138 & 166 & 153 \\ \hdashline St.Dev.= & 1 & 1.528 & 1.528 & 1.732 \\ \hdashline SS= & 2 & 4.667 & 4.667 & 6 \\ \hdashline n= & 3 & 3 & 3 & 3 \\ \hdashline \end{array}$$

The total sample size is $N=12.$ Therefore, the total degrees of freedom are:

$$df=12-1=11$$

The between-groups degrees of freedom are $df_{between}=4-1=3,$ and the within-groups degrees of freedom are:

$$df_{within}=df_{total}-df_{between}=11-3=8$$

We need to compute the total sum of values and the grand mean. The following is obtained

$$\sum_{\mathclap{ i, j}} X_{ij}=15+20+22+21=78$$

Also, the sum of squared values is

$$\sum_{\mathclap{ i, j}} X_{ij}^2=77+138+166+153=534$$

The total sum of squares is computed as follows

$$SS_{total}=\sum_{\mathclap{ i, j}} X_{ij}^2-\dfrac{1}{N}(\sum_{\mathclap{ i, j}} X_{ij})^2=$$

$$=534-\dfrac{(78)^2}{12}=27$$

The within sum of squares is computed as shown in the calculation below:

$$SS_{within}=\sum SS_{withingroups}=$$ $$=2+4.667+4.667+6=17.333$$

The between sum of squares is computed directly as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

$$MS_{between}=\dfrac{SS_{between}}{df_{between}}=\dfrac{9.667}{3}=3.222$$

$$MS_{within}=\dfrac{SS_{within}}{df_{within}}=\dfrac{17.333}{8}=2.167$$

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{3.222}{2.167}=1.487$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2=\mu_3=\mu_4$

$H_1:$ Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

Based on the information provided, the significance level is $\alpha=0.05,$ and the degrees of freedom are $df_1=3$ and $df_2=8,$ therefore, the rejection region for this F-test is $R=\{F:F>F_c=4.0662\}$

Test Statistics

$$F=\dfrac{MS_{between}}{MS_{within}}=\dfrac{3.222}{2.167}=1.487$$

Since it is observed that $F=1.487<4.0662=F_c,$ it is then concluded that the null hypothesis is not rejected.  Therefore, there is not enough evidence to claim that not all 4 population means are equal, at the $\alpha=0.05$ significance level.

Using the P-value approach:

The p-value is $p=0.290001,$ and since $p=0.290001>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that not all 4 population means are equal, at the $\alpha=0.05$ significance level.