Problem 1029

Solution. To simplify the calculation, we subtract the overall mean $\bar{x}=29$ from each observed value $x_{i}$, i.e., we transition to reduced values: $y_{i j}=x_{i j}-29$. For example, $y_{11}=x_{11}-29=38-29=9 ; y_{21}=x_{21}-29=36-29=7$ and so on.

We will construct the calculation table 45.

Table

45

| Test Number $\boldsymbol{c}$ | Factor Levels | | | | | | Column Totals |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | $F_{1}$ | | $F_{2}$ | | $F_{3}$ | | |
| | $y_{i 1}$ | $y_{i 1}^{2}$ | $y_{i 2}$ | $y_{i-2}^{2}$ | $y_{i s}$ | $\boldsymbol{y}_{i 3}^{2}$ | |
| 1 2 3 4 | 9 7 6 2 | 81 49 36 4 | -9 -5 -3 1 | 81 25 9 1 | -8 -7 2 5 | 64 49 4 25 | |
| $Q_{j}=\sum y_{i j}^{2}$ $T_{j}=\Sigma \boldsymbol{T}_{i j}^{\dot{Q}}$ $\mathbf{T}_{i j}$ | 24 576 | 170 | -16 256 | 116 | -8 64 | 142 | $\Sigma Q_{j}=428$ $\Sigma T_{j}=0$ $\Sigma T_{j}^{2}=896$ |

Using the summary column of Table 45, we will find the total and factor sums of squares of deviations, considering that the number of factor levels $p=3$, and the number of trials at each level $q=4$:

$$\begin{gathered} S_{06 \mathrm{~m}}=\sum_{i=1}^{p} Q_{j}-\left[\sum_{j=1}^{p} T_{j}\right]^{2} /(p q)=428-0=428 \\ S_{\text {factor }}=\left[\sum_{j=1}^{p} T_{j}^{2}\right] / q-\left[\sum_{j=1}^{p} T_{j}\right]^{2} / p q=896 / 4-0=224 \end{gathered}$$

We will find the residual sum of squares of deviations:

$$S_{\text {residual }}=S_{\text {total }}-S_{\text {factor }}=428-224=204$$

We will find the factor variance; for this, we divide $S_{\text {factor }}$ by the number of degrees of freedom $p-1=3-1=2$:

$$s_{\text {factor }}^{2}=S_{\text {factor }} /(p-1)=224 / 2=112 .$$

We will find the residual variance; for this, we divide $S_{\text {residual }}$ by the number of degrees of freedom $p(q-1)=3(4-1)=9$:

$$s_{\text {residual }}^{2}=S_{\text {residual }} / p(q-1)=204 / 9=22.67$$

We will compare the factor and residual variances using the Fisher-Snedecor criterion (see Chapter XIII, § 2). For this, we first find the observed value of the criterion:

$$F_{\text {obs }}=s_{\text {factor }}^{2} / s_{\text {residual }}^{2}=112 / 22.67=4.94$$

Considering that the number of degrees of freedom of the numerator $k_{1}=2$, and the denominator $k_{2}=9$, and that the significance level $\alpha=0.05$, we find the critical point from the table in Appendix 7:

$$F_{\text {crit }}(0.05 ; 2 ; 9)=4.26$$

Since $F_{\text {obs }}>F_{\text {crit }}$, we reject the null hypothesis of equality of group means. In other words, the group means differ significantly "overall."