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The given values are

\(\begin{aligned}{*{20}{c}}{SSA = 35.75}\\{SSB = 861.20}\\{SSAB = 603.51}\\{SSE = 341.82}\end{aligned}\)

Because of the assumption that both factor effects are fixed, let

where the are independent normally distributed random variable with mean 0 and variance\({\sigma ^2}\). The hypotheses of interest are

\({H_{0A}}:{\alpha _1} = {\alpha _2} = \ldots = {\alpha _I} = 0{\rm{\;versus\;}}{H_{aA}}:{\rm{\;at least one\;}}{\alpha _ - }i \ne 0\)

for the the factor B

\({H_{0B}}:{\beta _1} = {\beta _2} = \ldots = {\beta _J} = 0{\rm{\;versus\;}}{H_{aB}}:{\rm{\;at least one\;}}{\beta _ - }j \ne 0{\rm{,\;}}\)

and, for the the interaction

\({H_{0AB}}:{\gamma _{ij}}{\rm{\;for all\;}}i,j{\rm{\;versus\;}}{H_{aAB}}:{\rm{\;at least one\;}}{\gamma _ - }ij \ne 0.{\rm{\;}}\)

Construct ANOVA table in order to test relevant hypotheses. In order to construct an ANOVA table, a couple of values have to be computed. The table is

The degrees of freedom are

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1 = 3 \cdot 6 \cdot 2 - 1 = 35}\\{d{f_E} = IJ(K - 1) = 3 \cdot 6 \cdot (2 - 1) = 18}\end{aligned}\\\begin{aligned}{*{20}{c}}{d{f_A} = I - 1 = 3 - 1 = 2}\\{d{f_B} = J - 1 = 6 - 1 = 5}\end{aligned}\\d{f_{AB}} = (I - 1)(J - 1) = (3 - 1) \cdot (6 - 1) = 10\end{aligned}\)

Fundamental Identity:

\(SST = SSA + SSB + SSAB + SSE\)

By the fundamental identity, the SST is

\(\begin{aligned}{*{20}{c}}{SST = SSA + SSB + SSAB + SSE}\\{ = 35.75 + 861.20 + 603.51 + 341.82 = 1842.28}\end{aligned}\)

The mean squares can now be computed as follows

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{2} \cdot 35.75 = 17.875}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{5} \cdot 861.20 = 172.240}\end{aligned}\\\begin{aligned}{*{20}{c}}{MSAB = \frac{1}{{(I - 1)(J - 1)}} \cdot SSAB = \frac{1}{{10}} \cdot 603.51 = 60.351}\\{MSE = \frac{1}{{IJ(K - 1)}} \cdot SSE = \frac{1}{{18}} \cdot 341.82 = 18.990}\end{aligned}\end{aligned}\)

The test statistic values\(f\)are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSF}} = \frac{{17.875}}{{18.990}} = 0.94}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{172.240}}{{18.990}} = 9.07}\\{{f_{ABS}} = \frac{{MSAB}}{{MSF}} = \frac{{60.351}}{{18.990}} = 3.18.}\end{aligned}\)

The ANOVA table now becomes

There are two ways to make conclusion - using P value and critical value obtained from the table in the appendix.

When testing hypotheses\({H_{0AB}}{\rm{\;versus\;}}{H_{aAB}}\), the test statistic value is

\({f_{AB}} = \frac{{MSAB}}{{MSE}}\)

and the P-value is the area under the\({F_{(I - 1)(J - 1),IJ(K - 1)}}\)curve to the right of the test statistic value\({f_{AB}}\)

Using a software, the P value is

\(P = P\left( {F > {f_{AB}}} \right) = P(F > 3.18) = 0.016,\)

where F has Fisher's distribution with\({\nu _1} = (I - 1)(J - 1) = 10{\rm{\;and\;}}{\nu _2} = IJ(K - 1) = 18\)degrees of freedom. Since

\(P = 0.016 > 0.01 = \alpha \)

do not reject null hypothesis\({H_{0AB}}\)

It appears that there is not interaction between factor A and factor B.

When testing hypotheses\({H_{0A}}{\rm{\;versus\;}}{H_{aA}}\), the test statistic value is

\({f_A} = \frac{{MSA}}{{MSE}}\)

and the P-value is the area under the\({F_{(I - 1)(J - 1),IJ(K - 1)}}\)curve to the right of the test statistic value\({f_A}\)

Using a software, the P value is

\(P = P\left( {F > {f_A}} \right) = P(F > 0.94) = 0.41\)

where F has Fisher's distribution with\({\nu _1} = I - 1 = 2{\rm{\;and\;}}{\nu _2} = IJ(K - 1) = 18\)degrees of freedom. Since

\(P = 0.41 > 0.01 = \alpha \)

do not reject null hypothesis\({H_{0A}}\)

It appears that there is not significant effect due to type of farm.

When testing hypotheses\({H_{0B}}{\rm{\;versus\;}}{H_{aB}}\), the test statistic value is

\({f_B} = \frac{{MSB}}{{MSE}}\)

and the P-value is the area under the\({F_{(I - 1)(J - 1),IJ(K - 1)}}\)curve to the right of the test statistic value\({f_B}\)

Using a software, the P value is

\(P = P\left( {F > {f_B}} \right) = P(F > 9.07) = 0.0002\)

where F has Fisher's distribution with\({\nu _1} = J - 1 = 5{\rm{\;and\;}}{\nu _2} = IJ(K - 1) = 18\)degrees of freedom. Since

\(P = 0.0002 < 0.01 = \alpha \)

reject null hypothesis\({H_{0B}}\)

It appears that there is significant effect due to tractor maintenance method.

The conclusion can be made using the critical values from the table in the appendix.