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Let

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

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

For the factor B

\({H_{0B}}:\sigma _B^2 = 0\)versus \({H_{aB}}:\sigma _B^2 > 0\)

\({H_{0G}}:\sigma _G^2 = 0\)versus \({H_{aG}}:\sigma _G^2 > 0\)

Sum of squares are given by

With degrees of freedom respectively,

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1}\\{d{f_E} = IJ(K - 1)}\\{d{f_A} = I - 1}\\{d{f_B} = J - 1}\\{d{f_{AB}} = (I - 1)(J - 1).}\end{aligned}\)

The given data can be represented in a following table

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

The grand sum is

\({x_ \ldots } = 1847 + 1942 + \ldots + 1891 + 1756 = 27,479\)

The sum of squares can now be computed. This is a little bit trickier because not all the data points are given; however, it is possible to compute using the given values.

The SST is

\(\begin{aligned}{*{20}{c}}{ = \frac{1}{{5 \cdot 3}} \cdot \left( {{{9410}^2} + {{8835}^2} + {{9234}^2}} \right) - \frac{1}{{3 \cdot 5 \cdot 3}} \cdot 27,{{479}^2}}\\{ = 16,791,472.067 - 16,779,898.689}\\{ = 11,573.378}\end{aligned}\)

The SSA is

\(\begin{aligned}{*{20}{c}}{ = \frac{1}{{5 \cdot 3}} \cdot \left( {{{9410}^2} + {{8835}^2} + {{9234}^2}} \right) - \frac{1}{{3 \cdot 5 \cdot 3}} \cdot 27,{{479}^2}}\\{ = 16,791,472.067 - 16,779,898.689}\\{ = 11,573.378.}\end{aligned}\)

The SSB is

\(\begin{aligned}{*{20}{c}}{ = \frac{1}{{3 \cdot 3}} \cdot \left( {{{5432}^2} + {{5684}^2} + {{5619}^2} + {{5567}^2} + {{5177}^2}} \right) - \frac{1}{{3 \cdot 5 \cdot 3}} \cdot 27,{{479}^2}}\\{ = 16,797,828.778 - 16,779,898.689}\\{ = 17,930.089}\end{aligned}\)

The SSE is

\(\begin{aligned}{*{20}{c}}{ = 16,815,853 - \frac{1}{3} \cdot 50,443,409}\\{ = 1383.333}\end{aligned}\)

By the fundamental identity the SSAB

\(\begin{aligned}{*{20}{c}}{SSAB = SST - SSA - SSB - SSE}\\{ = 35,954.311 - 11,573.378 - 17,930.089 - 1383.333}\\{ = 5067.511.}\end{aligned}\)

The degrees of freedom are

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1 = 3 \cdot 5 \cdot 2 - 1 = 44}\\{d{f_E} = IJ(K - 1) = 3 \cdot 5 \cdot (3 - 1) = 30}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{d{f_A} = I - 1 = 3 - 1 = 2}\\{d{f_B} = J - 1 = 5 - 1 = 4}\\{d{f_{AB}} = (I - 1)(J - 1) = (3 - 1) \cdot (5 - 1) = 8}\end{aligned}\)

The mean squares are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{2} \cdot 11,573.378 = 5786.689}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{4} \cdot 17,930.089 = 4482.522}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{MSAB = \frac{1}{{(I - 1)(J - 1)}} \cdot SSAB = \frac{1}{8} \cdot 5067.511 = 633.439}\\{MSE = \frac{1}{{IJ(K - 1)}} \cdot SSE = \frac{1}{{30}} \cdot 1383.333 = 46.111}\end{aligned}\)

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

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

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

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

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

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

When testing hypotheses \({H_{0G}}\)versus \({H_{aG}}\)the test statistic value is

\({f_G} = \frac{{MSAB}}{{MSE}}\)

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

Hence, the values f values are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSAB}} = \frac{{5786.689}}{{633.439}} = 9.141}\\{{f_B} = \frac{{MSB}}{{MSAB}} = \frac{{4482.522}}{{633.439}} = 7.076}\\{{f_G} = \frac{{MSAB}}{{MSE}} = \frac{{633.439}}{{46.111}} = 13.737}\end{aligned}\)

The critical values are, respectively,

\(\begin{aligned}{*{20}{c}}{{F_{\alpha ,I - 1,(I - 1)(J - 1)}} = {F_{0.01,2,8}} = 8.65}\\{{F_{\alpha ,J - 1,(I - 1)(J - 1)}} = {F_{0.01,4,8}} = 7.01}\\{{F_{\alpha ,(I - 1)(J - 1),IJ(K - 1)}} = {F_{0.01,8,30}} = 3.17,}\end{aligned}\)

which were computed from the table in the appendix.

The P values are the mentioned areas under corresponding F curve, their values are, respectively,

\(\begin{aligned}{*{20}{c}}{{P_A} = 0.009}\\{{P_B} = 0.0097}\\{{P_G} = 0.00}\end{aligned}\)

which were computed using a software.

The ANOVO table is

Always test hypothesis\({H_{0G}}\). Since the ANOVA table has been already computed, there is no need for that because the values have been computed already.

When testing\({H_{0A}}\)versus\({H_{aA}}\)because

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,8}} = 8.65 < 9.141 = {f_A};}\\{{P_A} = 0.009 < 0.01 = \alpha ;}\end{aligned}\)

reject null hypothesis $H_{0 A}$

at given significance level.

When testing\({H_{0A}}\)versus\({H_{0B}}\)because

\(\begin{aligned}{*{20}{c}}{{F_{0.01,4,8}} = 7.01 < 7.076 = {f_B}}\\{{P_B} = 0.0097 < 0.01 = \alpha }\end{aligned}\)

reject null hypothesis $H_{0 B}$

at given significance level.

When testing \({H_{0G}}\) versus \({H_{aG}}\) because