91影视

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,

\(\begin{aligned}{*{20}{c}}{ Source of Variation }&{SS}&{ df }&{ MS }&{ f }&{ P value }&{ F crit }\\{ Factor A }&{4414.658}&2&{2207.329}&{2259.29}&{0.00}&{5.25}\\{ Factor B }&{47.255}&1&{47.255}&{48.37}&{0.00}&{7.40}\\{ Factor C }&{983.566}&2&{491.783}&{503.36}&{0.00}&{5.25}\\{ Factor D }&{0.044}&1&{0.044}&{0.05}&{0.82}&{7.40}\\{ AB }&{30.606}&2&{15.303}&{15.66}&{0.00}&{5.25}\\{ AC }&{1101.784}&4&{275.446}&{281.93}&{0.00}&{3.89}\\{ AD }&{0.940}&2&{0.47}&{0.48}&{0.62}&{5.25}\\{ BC }&{4.282}&2&{2.141}&{2.19}&{0.13}&{5.25}\\{ BD }&{0.273}&1&{0.273}&{0.28}&{0.60}&{7.40}\\{ CD }&{0.494}&2&{0.247}&{0.25}&{0.78}&{5.25}\\{ ABC }&{14.856}&4&{3.714}&{3.8}&{0.01}&{3.89}\\{ ABD }&{8.144}&2&{4.072}&{4.17}&{0.02}&{5.25}\\{ ACD }&{3.068}&4&{0.767}&{0.79}&{0.54}&{3.89}\\{ BCD }&{0.560}&2&{0.28}&{0.29}&{0.75}&{5.25}\\{ ABCD }&{1.388}&4&{0.347}&{0.355}&{0.84}&{3.89}\\{ Error }&{35.172}&{36}&{0.977}&{}&{}&{}\\{ Total }&{6647.090}&{71}&{}&{}&{}&{}\\{}&{}&{}&{}&{}&{}&{}\end{aligned}\)

This shows a lot of data as missing.

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

\(\begin{aligned}{c}d{f_T} = IJKML - 1\\ = 3 \times 2 \times 3 \times 2 \times 2 - 1\\ = 71\\d{f_E} = IJKM(L - 1)\\ = 3 \times 2 \times 3 \times 2 \times (2 - 1)\\ = 36\end{aligned}\)

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

\(\begin{aligned}{c}d{f_C} = K - 1\\ = 3 - 1\\ = 2\\d{f_D} = M - 1\\ = 2 - 1\\ = 1\\d{f_{AB}} = (I - 1)(J - 1)\\ = (3 - 1) \times (2 - 1)\\ = 2\end{aligned}\)

\(\begin{aligned}{c}d{f_{AC}} = (I - 1)(K - 1)\\ = (3 - 1) \times (3 - 1)\\ = 4\\d{f_{AD}} = (I - 1)(M - 1)\\ = (3 - 1) \times (2 - 1)\\ = 2\\d{f_{BC}} = (J - 1)(K - 1)\\ = (3 - 1) \times (3 - 1)\\ = 4\end{aligned}\)

Determining for other data as well,

\(\begin{aligned}{c}d{f_{BD}} = (J - 1)(M - 1)\\ = (2 - 1) \times (2 - 1)\\ = 1\\d{f_{CD}} = (K - 1)(M - 1)\\ = (3 - 1) \times (2 - 1)\\ = 2\\d{f_{ABC}} = (I - 1)(J - 1)(K - 1)\\ = (3 - 1) \times (2 - 1) \times (3 - 1)\\ = 4\end{aligned}\)

\(\begin{aligned}{c}d{f_{ABD}} = (I - 1)(J - 1)(M - 1)\\ = (3 - 1) \times (2 - 1) \times (2 - 1)\\ = 2\\d{f_{ACD}} = (I - 1)(K - 1)(M - 1)\\ = (3 - 1) \times (3 - 1) \times (2 - 1)\\ = 4\end{aligned}\)

\(\begin{aligned}{c}d{f_{BCD}} = (J - 1)(K - 1)(M - 1)\\ = (2 - 1) \times (3 - 1) \times (2 - 1)\\ = 2\\d{f_{ABCD}} = (I - 1)(J - 1)(K - 1)(M - 1)\\ = (3 - 1) \times (2 - 1) \times (3 - 1) \times (2 - 1)\\ = 4\end{aligned}\)

Determining the mean squares by finding the sum of squares,

\(\begin{aligned}{*{20}{c}}{SSA = d{f_A} \times MSA = 2 \times 2207.329 = 4414.658}\\{SSB = d{f_B} \times MSB = 1 \times 47.255 = 47.255}\\{SSC = d{f_C} \times MSC = 2 \times 491.783 = 983.566;}\\{SSD = d{f_D} \times MSD = 1 \times 0.044 = 0.044}\\{SSAB = d{f_{AB}} \times MSAB = 2 \times 15.303 = 30.606;}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{SSAC = d{f_{AC}} \times MSAC = 4 \times 275.446 = 1101.784}\\{SSAD = d{f_{AD}} \times MSAD = 2 \times 0.47 = 0.940}\\{SSBC = d{f_{BC}} \times MSBC = 2 \times 2.141 = 4.282}\\{SSBD = d{f_{BD}} \times MSBD = 1 \times 0.273 = 0.273}\\{SSCD = d{f_{CD}} \times MSCD = 2 \times 0.247 = 0.494}\end{aligned}\)
\(\) \(\begin{aligned}{*{20}{c}}{SSABC = d{f_{ABC}} \times MSABC = 4 \times 3.714 = 14.856}\\{SSABD = d{f_{ABD}} \times MSABD = 2 \times 4.072 = 8.144}\\{SSACD = d{f_{ACD}} \times MSACD = 4 \times 0.767 = 3.068}\\{SSBCD = d{f_{BCD}} \times MSBCD = 2 \times 0.28 = 0.560}\\\begin{aligned}{l}SSE = d{f_E} \times MSE = 36 \times 0.911 = 32.796\\SST = d{f_T} \times MST = 71 \times 93.621 = 6647.090\end{aligned}\end{aligned}\)

Determining the missing values,

From the Fundamental Identity,

\(SSABCD = SST - SSA - SSB - \ldots - SSBCD - SSE\)

Substituting the mentioned values in the table,

\(\begin{aligned}{*{20}{c}}{SSABCD = SST - SSA - SSB - \ldots - SSBCD - SSE}\\{ = 6647.090 - 4414.658 - 47.255 - \ldots - 0.560 - 32.796 = 1.388.}\end{aligned}\)

So, the value obtained is:

\(MSABCD = \frac{1}{{d{f_{ABCD}}}} \times SSABCD = \frac{1}{4} \times 1.388 = 0.347.\)

Determining the function values,

\(\begin{aligned}{c}{f_A} = \frac{{MSA}}{{MSE}} = \frac{{2207.329}}{{0.977}} = 2259.29\\{f_B} = \frac{{MSB}}{{MSE}} = \frac{{47.255}}{{0.977}} = 48.37\end{aligned}\)

\(\begin{aligned}{c}{f_C} = \frac{{MSC}}{{MSE}} = \frac{{491.783}}{{0.977}} = 503.36\\{f_D} = \frac{{MSD}}{{MSE}} = \frac{{0.044}}{{0.977}} = 0.05\\{f_{AB}} = \frac{{MSAB}}{{MSE}} = \frac{{15.303}}{{0.977}} = 15.66\end{aligned}\)

\(\begin{aligned}{c}{f_{AC}} = \frac{{MSAC}}{{MSE}} = \frac{{275.446}}{{0.977}} = 281.93\\{f_{AD}} = \frac{{MSAD}}{{MSE}} = \frac{{0.47}}{{0.977}} = 0.48\end{aligned}\)

Determining the other function values,

\(\begin{aligned}{c}{f_{BC}} = \frac{{MSBC}}{{MSE}} = \frac{{2.141}}{{0.977}} = 2.19\\{f_{BD}} = \frac{{MSBD}}{{MSE}} = \frac{{0.273}}{{0.977}} = 0.28\\{f_{CD}} = \frac{{MSCD}}{{MSE}} = \frac{{0.247}}{{0.977}} = 0.25\\{f_{ABC}} = \frac{{MSABC}}{{MSE}} = \frac{{3.714}}{{0.977}} = 3.8\end{aligned}\)

\(\begin{aligned}{c}{f_{ABD}} = \frac{{MSABD}}{{MSE}} = \frac{{4.072}}{{0.977}} = 4.17\\{f_{ACD}} = \frac{{MSACD}}{{MSE}} = \frac{{0.767}}{{0.977}} = 0.79\\{f_{BCD}} = \frac{{MSBCD}}{{MSE}} = \frac{{0.28}}{{0.977}} = 0.29\\{f_{ABCD}} = \frac{{MSABCD}}{{MSE}} = \frac{{0.347}}{{0.977}} = 0.355\end{aligned}\)

Determining P-values using a software,

\(\begin{aligned}{c}{P_A} = P\left( {F > {f_A}} \right) = P(F > 2259.29) = 0.00\\{P_B} = P\left( {F > {f_B}} \right) = P(F > 48.37) = 0.00\\{P_C} = P\left( {F > {f_C}} \right) = P(F > 503.36) = 0.00\\{P_D} = P\left( {F > {f_D}} \right) = P(F > 0.05) = 0.82\end{aligned}\)

\(\begin{aligned}{c}{P_{AB}} = P\left( {F > {f_{AB}}} \right) = P(F > 15.66) = 0.00\\{P_{AC}} = P\left( {F > {f_{AC}}} \right) = P(F > 281.93) = 0.00\\{P_{AD}} = P\left( {F > {f_{AD}}} \right) = P(F > 0.48) = 0.62\\{P_{BC}} = P\left( {F > {f_{BC}}} \right) = P(F > 2.19) = 0.13\end{aligned}\)

\(\begin{aligned}{l}{P_{BD}} = P\left( {F > {f_{BD}}} \right) = P(F > 0.28) = 0.60\\{P_{CD}} = P\left( {F > {f_{CD}}} \right) = P(F > 0.25) = 0.78\\{P_{ABC}} = P\left( {F > {f_{ABC}}} \right) = P(F > 3.8) = 0.01\\{P_{ABD}} = P\left( {F > {f_{ABD}}} \right) = P(F > 4.17) = 0.02\end{aligned}\)

\(\begin{aligned}{l}{P_{ACD}} = P\left( {F > {f_{ACD}}} \right) = P(F > 0.79) = 0.54\\{P_{BCD}} = P\left( {F > {f_{BCD}}} \right) = P(F > 0.29) = 0.75\\{P_{ABCD}} = P\left( {F > {f_{ABCD}}} \right) = P(F > 0.355) = 0.84\end{aligned}\)

The values to be filled are critical values as:

\(\begin{aligned}{l}{F_{0.01,2,36}} = 5.25\\{F_{0.01,1,36}} = 7.40\\{F_{0.01,4,36}} = 3.89\end{aligned}\)

The ANOVA table will be:

\(\begin{aligned}{*{20}{c}}{ Source of Variation }&{ SS }&{ df }&{ MS }&{ f }&{ P value }&{ F crit }\\{ Factor A }&{4414.658}&2&{2207.329}&{2259.29}&{0.00}&{5.25}\\{ Factor B }&{47.255}&1&{47.255}&{48.37}&{0.00}&{7.40}\\{ Factor C }&{983.566}&2&{491.783}&{503.36}&{0.00}&{5.25}\\{ Factor D }&{0.044}&1&{0.044}&{0.05}&{0.82}&{7.40}\\{ AB }&{30.606}&2&{15.303}&{15.66}&{0.00}&{5.25}\\{ AC }&{1101.784}&4&{275.446}&{281.93}&{0.00}&{3.89}\\{ AD }&{0.940}&2&{0.47}&{0.48}&{0.62}&{5.25}\\{ BC }&{4.282}&2&{2.141}&{2.19}&{0.13}&{5.25}\\{ BD }&{0.273}&1&{0.273}&{0.28}&{0.60}&{7.40}\\{ CD }&{0.494}&2&{0.247}&{0.25}&{0.78}&{5.25}\\{ ABC }&{14.856}&4&{3.714}&{3.8}&{0.01}&{3.89}\\{ ABD }&{8.144}&2&{4.072}&{4.17}&{0.02}&{5.25}\\{ ACD }&{3.068}&4&{0.767}&{0.79}&{0.54}&{3.89}\\{ BCD }&{0.560}&2&{0.28}&{0.29}&{0.75}&{5.25}\\{ ABCD }&{1.388}&4&{0.347}&{0.355}&{0.84}&{3.89}\\{ Error }&{35.172}&{36}&{0.977}&{}&{}&{}\\{ Total }&{6647.090}&{71}&{}&{}&{}&{}\\{}&{}&{}&{}&{}&{}&{}\end{aligned}\)

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

Considering the factor \(A\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 < 2259.29 = {f_A};}\\{{P_A} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

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

Considering the factor \(A\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 < 2259.29 = {f_A};}\\{{P_A} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

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

Considering the factor \(B\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,1,36}} = 7.40 < 48.37 = {f_B};}\\{{P_B} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

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

Considering the factor \(A\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 < 503.36 = {f_C};}\\{{P_C} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

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

Considering the factor \(D\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,1,36}} = 7.4 > 0.05 = {f_D};}\\{{P_A} = 0.82 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(AB\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 < 15.66 = {f_{AB}};}\\{{P_{AB}} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(AC\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,4,36}} = 3.89 < 281.93 = {f_{AC}};}\\{{P_{AC}} = 0.00 < 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(AD\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 > 0.48 = {f_{AD}};}\\{{P_{AD}} = 0.62 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(BC\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 > 0.48 = {f_{BC}};}\\{{P_{BC}} = 0.13 > 0.01 = \alpha ,}\end{aligned}\)

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 \(BD\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,1,36}} = 7.40 > 0.28 = {f_{BD}};}\\{{P_{BD}} = 0.60 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(CD\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 > 0.25 = {f_{CD}};}\\{{P_{CD}} = 0.78 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(ABC\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,4,36}} = 3.89 > 3.8 = {f_{ABC}};}\\{{P_{ABC}} = 0.011 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(ABD\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 > 4.17 = {f_{ABD}};}\\{{P_{ABD}} = 0.02 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(ACD\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,4,36}} = 3.89 > 0.79 = {f_{ACD}};}\\{{P_{ACD}} = 0.54 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(BCD\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,36}} = 5.25 > 0.29 = {f_{BCD}};}\\{{P_{BCD}} = 0.75 > 0.01 = \alpha ,}\end{aligned}\)

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

Considering the interaction \(ABCD\),

\(\begin{aligned}{*{20}{c}}{{F_{0.01,4,36}} = 3.89 > 0.355 = {f_{ABCD}};}\\{{P_{ABCD}} = 0.84 > 0.01 = \alpha ,}\end{aligned}\)

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

This shows that the main effects A,B and C are statistically significant, and that the interactions AB and AC are statistically significant.