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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

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1 = 2 \cdot 3 \cdot 3 - 1 = 17}\\{d{f_E} = IJ(K - 1) = 2 \cdot 3 \cdot (3 - 1) = 12}\end{aligned}\\\begin{aligned}{*{20}{c}}{d{f_A} = I - 1 = 2 - 1 = 1}\\{d{f_B} = J - 1 = 3 - 1 = 2}\end{aligned}\\d{f_{AB}} = (I - 1)(J - 1) = (2 - 1) \cdot (3 - 1) = 2\end{aligned}\)

Fundamental Identity:

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

By the fundamental identity, the SST is

\(\begin{aligned}{*{20}{c}}{SST = SSA + SSB + SSAB + SSE}\\{ = 2253.44 + 230.81 + 18.58 + 71.87 = 2574.7.}\end{aligned}\)

The mean squares can now be computed as follows

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{1} \cdot 2253.44 = 2253.44}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{2} \cdot 230.81 = 115.41}\end{aligned}\\\begin{aligned}{*{20}{c}}{MSAB = \frac{1}{{(I - 1)(J - 1)}} \cdot SSAB = \frac{1}{2} \cdot 18.58 = 9.29}\\{MSE = \frac{1}{{IJ(K - 1)}} \cdot SSE = \frac{1}{{17}} \cdot 71.87 = 5.99}\end{aligned}\end{aligned}\)

The test statistic values f arc

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.

(a)Let

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

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

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

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

Using a software, the P value is\({f_{AB}}\)

\(P = P\left( {F > {f_{AB}}} \right) = P(F > 1.55) = 0.25,\)

where F has Fisher's distribution with\({\nu _1} = (I - 1)(J - 1) = 4{\rm{\;and\;}}{\nu _2} = IJ(K - 1) = 12\)degrees of freedom. Since

\(P = 0.25 > 0.05 = \alpha \)

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

\(\begin{aligned}{*{20}{c}}{{F_{0.05,1,12}} = 4.75 < 376.2 = {f_A}}\\{{P_A} = 0 < 0.05 = \alpha }\end{aligned}\)

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

at given significance level.

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

\(\begin{aligned}{*{20}{c}}{{F_{0.05,2,12}} = 3.89 < 19.27 = {f_B};}\\{{P_B} = 0 < 0.05 = \alpha ;}\end{aligned}\)

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

at given significance level.

It appears that yield highly depend on both formulation and speed.

(c)The estimates of\({\alpha _i},{\beta _j},{\gamma _{ij}}\)can be computed using the averages of measurements.

Denote with\({\bar X_{i..}}\)the average of measurements obtained when factor A is held at level i

with\({\bar X_{.j.}}\)the average of measurements obtained when factor B is held at level j

with\({\bar X_{ij.}}\)the average of measurements obtained when factor A is held at level i and factor B is held at level j

and with\({\bar X_ \ldots }\)the grand mean

Observed values are denoted with small\(x\)instead of big X. The notations without line over X are just the sums.

The average of measurements obtained when factor A is held at level i are

The average of measurements obtained when factor B is held at level j is

The average of measurements obtained when factor A is held at level i and when factor B is held at level j are

The grand mean is

\({\bar x_ \ldots } = \frac{1}{{2 \cdot 3 \cdot 3}} \cdot (189.7 + 185.1 + \ldots + 161.6 + 170.3) = 175.84\).

\({\rm{\;The estimates can be computed using formulas\;}}\)

\(\begin{aligned}{*{20}{c}}{{{\hat \alpha }_i} = {{\bar x}_{i..}} - {{\bar x}_ \ldots };}\\{{{\hat \beta }_j} = {{\bar x}_{.j.}} - {{\bar x}_ \ldots }}\\{{{\hat \gamma }_{ij}} = {{\bar x}_{ij.}} - \left( {{{\bar x}_ \ldots } + {\alpha _i} + {\beta _j}} \right).}\end{aligned}\)

The estimates of the main effects are

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{{{\hat \alpha }_1} = {{\bar x}_{1..}} - {{\bar x}_ \ldots } = 187.03 - 175.84 = 11.19}\\{{{\hat \alpha }_2} = {{\bar x}_{2..}} - {{\bar x}_ \ldots } = 164.66 - 175.84 = - 11.18}\end{aligned}\\\begin{aligned}{*{20}{c}}{{{\hat \beta }_1} = {{\bar x}_{.1.}} - {{\bar x}_ \ldots } = 177.83 - 175.84 = 1.99}\\{{{\hat \beta }_2} = {{\bar x}_{.2.}} - {{\bar x}_ \ldots } = 170.82 - 175.84 = - 5.02}\\{{{\hat \beta }_3} = {{\bar x}_{.3.}} - {{\bar x}_ \ldots } = 178.88 - 175.84 = 3.04}\end{aligned}\end{aligned}\)

and the estimates of\({\gamma _{ij}}\)are

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{{{\hat \gamma }_{11}} = {{\bar x}_{11.}} - \left( {{{\bar x}_ \ldots } + {\alpha _1} + {\beta _1}} \right) = 0.45}\\{{{\hat \gamma }_{12}} = {{\bar x}_{12.}} - \left( {{{\bar x}_ \ldots } + {\alpha _1} + {\beta _2}} \right) = - 1.41}\\{{{\hat \gamma }_{13}} = {{\bar x}_{13.}} - \left( {{{\bar x}_ \ldots } + {\alpha _1} + {\beta _3}} \right) = 0.96}\end{aligned}\\\begin{aligned}{*{20}{c}}{{{\hat \gamma }_{21}} = {{\bar x}_{21.}} - \left( {{{\bar x}_ \ldots } + {\alpha _2} + {\beta _1}} \right) = - 0.45}\\{{{\hat \gamma }_{22}} = {{\bar x}_{22.}} - \left( {{{\bar x}_ \ldots } + {\alpha _2} + {\beta _2}} \right) = 1.39}\\{{{\hat \gamma }_{23}} = {{\bar x}_{23.}} - \left( {{{\bar x}_ \ldots } + {\alpha _2} + {\beta _3}} \right) = - 0.97}\end{aligned}\end{aligned}\)

The residuals are obviously the same as the given residuals.

First order the residuals:

\( - 3.43, - 3.30, - 2.03, - 1.23, - 1.20, - 1.10, - 0.87, - 0.30, - 0.13,0.07,0.23,0.57,0.63,0.67,1.40,1.97,3.57,4.50.{\rm{\;}}\)Than find corresponding\(Z\)scores:

\( - 1.91, - 1.38, - 1.09, - 0.86, - 0.67, - 0.51, - 0.36, - 0.21, - 0.07,0.07,0.21,0.36,0.51,0.67,0.86,1.09,1.38,1.91\)The following normal probability plots suggest that the assumption of normality is plausible (by looking into the line).