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The fundamental identity defines the SST as:

\(SST = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABC + SSE\)

The sum of squares of each gives the mean square.

The fixed effect model for three-factor ANOVA is defined as:

\({X_{ijkl}} = \mu + {\alpha _i} + {\beta _j} + {\delta _k} + \gamma _{ij}^{AB} + \gamma _{ij}^{AC} + \gamma _{ij}^{BC} + {\gamma _{ijk}} + {{\rm{\varepsilon }}_{ij}}\)

where the values of\(i = 1,2,...,I\), the values of\(j = 1,2,...,J\), the values of \(k = 1,2,...,K\) and the values of \(l = 1,2,...,L\) .

\({L_{ijk}} = L\)are the numbers of observation made with factor \(A\) at level \(i\) , factor \(B\) at level \(j\) and the factor \(C\) at level \(k\) , where \({{\rm{\varepsilon }}_{ij}}\) are independent normally distributed random variable wih mean \(0\) and the variance is \({\sigma ^2}\) .

Here, \({\mu _{ijk}}\) stands for \({\mu _{ijk}} = \mu + {\alpha _i} + {\beta _j} + {\delta _k} + \gamma _{ij}^{AB} + \gamma _{ij}^{AC} + \gamma _{ij}^{BC} + {\gamma _{ijk}}\) .

Tabulating the data mentioned,

\(\begin{aligned}{*{20}{c}}{ Source of Variation }&{ SS }&{ df }&{ MS }&{ f }&{ P value }&{ F crit }\\{ Factor A }&{SSA}&{I - 1}&{MSA}&{{f_A}}&{{P_A}}&{{F_{Acit}}}\\{ Factor B }&{SSB}&{J - 1}&{MSB}&{{f_B}}&{{P_B}}&{{F_{Bcrit}}}\\{ Factor C }&{SSC}&{K - 1}&{MSC}&{{f_C}}&{{P_C}}&{{F_{Ccrit}}}\\{ Interaction AB }&{SSAB}&{(I - 1)(J - 1)}&{MSAB}&{{f_{AB}}}&{{P_{AB}}}&{{F_{ABcrit}}}\\{ Interaction AC }&{SSAC}&{(I - 1)(K - 1)}&{MSAC}&{{f_{AC}}}&{{P_{AC}}}&{{F_{AC rit }}}\\{ Interaction BC }&{SSBC}&{(J - 1)(K - 1)}&{MSBC}&{{f_{BC}}}&{{P_{BC}}}&{{F_{BC it }}}\\{ Interaction ABC }&{SSABC}&{(I - 1)(J - 1)(K - 1)}&{MSABC}&{{f_{ABC}}}&{{P_{ABC}}}&{{F_{ABCcrit}}}\\{ Error }&{SSE}&{IJK(L - 1)}&{MSE}&{}&{}&{}\\{ Total }&{SST}&{IJKL - 1}&{}&{}&{}&{}\\{}&{}&{}&{}&{}&{}&{}\end{aligned}\)

This shows a lot of data as missing.

Determining the missing values,

From the Fundamental Identity,

\(SST = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABC + SSE\)

Substituting the mentioned values in the table,

\(\begin{aligned}{c}SST = 39.171 + 0.665 + 21.508 + 1.43 + 15.953\\ + 1.382 + 9.016 + 115.820\\ = 204.947\end{aligned}\)

Determining the degrees of freedom as per mentioned data in the table,

\(\begin{aligned}{c}d{f_T} = IJKL - 1\\ = 4 \times 3 \times 6 \times 3 - 1\\ = 215\\d{f_E} = IJK(L - 1)\\ = 4 \times 3 \times 6 \times (3 - 1)\\ = 144\end{aligned}\)

\(\begin{aligned}{c}d{f_A} = I - 1\\ = 4 - 1\\ = 3\\d{f_B} = J - 1\\ = 3 - 1\\ = 2\end{aligned}\)

\(\begin{aligned}{c}d{f_C} = K - 1\\ = 6 - 1\\ = 5\\d{f_{AB}} = (I - 1)(J - 1)\\ = (4 - 1) \times (3 - 1)\\ = 6\end{aligned}\)

\(\begin{aligned}{c}d{f_{AC}} = (I - 1)(K - 1)\\ = (4 - 1) \times (6 - 1)\\ = 15\\d{f_{BC}} = (J - 1)(K - 1)\\ = (3 - 1) \times (6 - 1)\\ = 10\\d{f_{ABC}} = (I - 1)(J - 1)(K - 1)\\ = (4 - 1) \times (3 - 1) \times (6 - 1)\\ = 30\end{aligned}\)

Determining the mean squares by finding the sum of squares,

\(\begin{aligned}{c}MSA = \frac{1}{{d{f_A}}} \times SSA\\ = \frac{1}{3} \times 39.171\\ = 13.057\\MSB = \frac{1}{{d{f_B}}} \times SSB\\ = \frac{1}{2} \times 0.665\\ = 0.3325\end{aligned}\)

\(\begin{aligned}{c}MSC = \frac{1}{{d{f_C}}} \times SSC\\ = \frac{1}{5} \times 21.508\\ = 4.3016\\MSAB = \frac{1}{{d{f_{AB}}}} \times SSAB\\ = \frac{1}{6} \times 1.432\\ = 0.2387\end{aligned}\)

\(\begin{aligned}{c}MSAC = \frac{1}{{d{f_{AC}}}} \times SSAC\\ = \frac{1}{{15}} \times 15.953\\ = 1.0635\\MSBC = \frac{1}{{d{f_{BC}}}} \times SSBC\\ = \frac{1}{{10}} \times 1.382\\ = 0.1382\end{aligned}\)

\(\begin{aligned}{c}MSABC = \frac{1}{{d{f_{ABC}}}} \times SSABC\\ = \frac{1}{{30}} \times 9.016\\ = 0.3005\\MSE = \frac{1}{{d{f_E}}} \times SSE\\ = \frac{1}{{144}} \times 115.820\\ = 0.8043\end{aligned}\)

Determining the function values,

\(\begin{aligned}{c}{f_A} = \frac{{MSA}}{{MSE}}\\ = \frac{{13.057}}{{0.8043}}\\ = 16.23\\{f_B} = \frac{{MSB}}{{MSE}}\\ = \frac{{0.3325}}{{0.8043}}\\ = 0.41\end{aligned}\)

\(\begin{aligned}{c}{f_C} = \frac{{MSC}}{{MSE}}\\ = \frac{{4.3016}}{{0.8043}}\\ = 5.35\\{f_{AB}} = \frac{{MSAB}}{{MSE}}\\ = \frac{{0.2387}}{{08043}}\\ = 0.30\end{aligned}\)

\(\begin{aligned}{c}{f_{AC}} = \frac{{MSAC}}{{MSE}}\\ = \frac{{1.0635}}{{0.8043}}\\ = 1.32\\{f_{BC}} = \frac{{MSBC}}{{MSE}}\\ = \frac{{0.1382}}{{0.8043}}\\ = 0.17\\{f_{ABC}} = \frac{{MSABC}}{{MSE}}\\ = \frac{{0.3005}}{{0.8043}}\\ = 0.37\end{aligned}\)

Determining P-values using a software,

\(\begin{aligned}{c}{P_A} = P\left( {F > {f_A}} \right)\\ = P(F > 16.23)\\ = 0.00\\{P_B} = P\left( {F > {f_B}} \right)\\ = P(F > 0.41)\\ = 0.66\end{aligned}\)

\(\begin{aligned}{c}{P_C} = P\left( {F > {f_C}} \right)\\ = P(F > 5.35)\\ = 0.00\\{P_{AB}} = P\left( {F > {f_{AB}}} \right)\\ = P(F > 0.30)\\ = 0.94\end{aligned}\)

\(\begin{aligned}{c}{P_{AC}} = P\left( {F > {f_{AC}}} \right)\\ = P(F > 1.32)\\ = 0.20\\{P_{BC}} = P\left( {F > {f_{BC}}} \right)\\ = P(F > 0.17)\\ = 1.00\\{P_{ABC}} = P\left( {F > {f_{ABC}}} \right)\\ = P(F > 0.37)\\ = 1.00\end{aligned}\)

The values to be filled are critical values as:

\(\begin{aligned}{*{20}{c}}{{F_{0.01,3,144}} = 3.92}\\{{F_{0.01,2,144}} = 4.76}\\{{F_{0.01,5,144}} = 3.15}\\{{F_{0.01,6,144}} = 2.93}\\{{F_{0.01,15,144}} = 2.17}\\{{F_{0.01,10,144}} = 2.45}\\{{F_{0.01,30,144}} = 1.83}\end{aligned}\)

The ANOVA table will be:

\(\begin{aligned}{*{20}{c}}{ Source of Variation }&{ SS }&{ df }&{ MS }&{ f }&{ P value }&{ F crit }\\{ Factor A }&{39.171}&3&{13.057}&{16.23}&{0.00}&{3.92}\\{ Factor B }&{0.665}&2&{0.3325}&{0.41}&{0.66}&{4.76}\\{ lector C }&{21.508}&5&{4.3016}&{5.35}&{0.00}&{3.15}\\{ Intcraction AB }&{1.432}&6&{0.2387}&{0.3}&{0.91}&{2.93}\\{ Interaction AC }&{15.953}&{15}&{1.0635}&{1.32}&{0.20}&{2.17}\\{ Interaction BC }&{1.382}&{10}&{0.1382}&{0.17}&{1.00}&{2.45}\\{ Interaction ABC }&{9.016}&{30}&{0.3005}&{0.37}&{1.00}&{1.83}\\{ Error }&{115.820}&{144}&{0.8043}&{}&{}&{}\\{ Total }&{204.947}&{215}&{}&{}&{}&{}\\{}&{}&{}&{}&{}&{}&{}\end{aligned}\)

According to the definition of fixed effects model for three-factor ANOVA,

The relevant hypotheses will be:

\(\begin{aligned}{*{20}{c}}{{H_{0A}}:{\alpha _1} = {\alpha _2} = \ldots = {\alpha _I} = 0}&{ versus }&{{H_{aA}}: at least one {\alpha _ - }i \ne 0}\\{{H_{0B}}:{\beta _1} = {\beta _2} = \ldots = {\beta _J} = 0}&{ versus }&{{H_{aB}}: at least one {\beta _ - }j \ne 0}\\{{H_{0C}}:{\delta _1} = {\delta _2} = \ldots = {\delta _K} = 0}&{ versus }&{{H_{aC}}: at least one {\delta _ - }i \ne 0}\end{aligned}\)

This is for the interactions:

\(\begin{aligned}{*{20}{l}}{{H_{0AB}}:\gamma _{ij}^{AB} = 0 for all i,j}&{ versus }&{{H_{aAB}}: at least one {\gamma _ - }i{j^{AB}} \ne 0}\\{{H_{0AC}}:\gamma _{ij}^{AC} = 0 for all i,j}&{ versus }&{{H_{aAC}}: at least one {\gamma _ - }i{j^{AC}} \ne 0}\\{{H_{0BC}}:\gamma _{ij}^{BC} = 0 for all i,j}&{ versus }&{{H_{aBC}}: at least one {\gamma _ - }i{j^{BC}} \ne 0}\end{aligned}\)

And for the interaction ABC,

\({H_{0ABC}}:{\gamma _{ijk}} = 0 for all i,j,k versus {H_{aABC}}: at least one {\gamma _ - }ijk \ne 0\)

From the hypotheses tests performed,

The test static values will be:

\({f_A} = \frac{{MSA}}{{MSE}}\)and the\(P\)- value is the area under the\({F_{I - 1,I,J,K(L - 1)}}\)curve to the right of the test static value\({f_A}\).

Similarly for the other test static values,

\(\begin{aligned}{l}{f_B} = \frac{{MSB}}{{MSE}}\\{f_C} = \frac{{MSC}}{{MSE}}\\{f_{AB}} = \frac{{MSAB}}{{MSE}}\\{f_{AC}} = \frac{{MSAC}}{{MSE}}\\{f_{BC}} = \frac{{MSBC}}{{MSE}}\\{f_{ABC}} = \frac{{MSABC}}{{MSE}}\end{aligned}\)

Considering the factor \(A\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,3,144}} = 3.92 < 16.23 = {f_A};}\\{{P_A} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

So, the null hypothesis\({H_{0A}}\) is rejected at any given significance level of\(\alpha \), the factor\(A\)is statistically significant.

Considering the factor \(B\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,144}} = 4.76 > 0.41 = {f_B};}\\{{P_B} = 0.66 > 0.01 = \alpha ,}\end{aligned}\)

So, the null hypothesis \({H_{0B}}\)is not rejected at any significance level of\(\alpha \), the factor\(B\)is not statistically significant.

Considering the factor \(C\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,5,144}} = 3.15 < 5.35 = {f_C};}\\{{P_C} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

So, the null hypothesis \({H_{0C}}\) is rejected at any given significance level of \(\alpha \), the factor \(C\)is statistically significant

Considering the interaction \(AB\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,6,144}} = 2.93 > 0.3 = {f_{AB}};}\\{{P_{AB}} = 0.94 > 0.01 = \alpha ,}\end{aligned}\)

So, the null hypothesis \({H_{0AB}}\)is not rejected at any significance level of\(\alpha \), the factor\(AB\)is not statistically significant.

Considering the interaction \(AC\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,15,144}} = 2.17 > 1.32 = {f_{AC}};}\\{{P_{AC}} = 0.20 > 0.01 = \alpha ,}\end{aligned}\)

So, the null hypothesis \({H_{0AC}}\)is not rejected at any significance level of\(\alpha \), the factor\(AC\)is not statistically significant.

Considering the interaction \(BC\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,20,144}} = 2.45 > 0.17 = {f_{BC}};}\\{{P_{BC}} = 1.00 > 0.01 = \alpha ,}\end{aligned}\)

So, the null hypothesis \({H_{0BC}}\)is not rejected at any significance level of\(\alpha \), the factor\(BC\)is not statistically significant.

Considering the interaction \(ABC\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,30,144}} = 1.83 > 0.37 = {f_{ABC}};}\\{{P_{ABC}} = 1.00 > 0.01 = \alpha ,}\end{aligned}\)

So, the null hypothesis \({H_{0ABC}}\)is not rejected at any significance level of\(\alpha \), the factor\(ABC\)is not statistically significant.

This shows that the laundry treatment and fabric type are statistically significant main effects.