91影视

The given values are

\(\begin{aligned}{*{20}{c}}{SSA = 22,941.80;}\\{SSB = 22,765.53;}\\{SSE = 15,253.50,}\\{SST = 64,954.70.}\end{aligned}\)

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

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

for the the factor B

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

and, for the the interaction

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

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

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1 = 3 \cdot 5 \cdot 2 - 1 = 29}\\{d{f_E} = TJ(K - 1) = 3 \cdot 5 \cdot (2 - 1) = 15}\\{d{f_A} = T - 1 = 3 - 1 = 2}\\{d{f_B} = J - 1 = 5 - 1 = 4}\\{d{f_{AB}} = (I - 1)(J - 1) = (3 - 1) \cdot (5 - 1) = 8.}\end{aligned}\).

Fundamental Identity :

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

By the fundamental identity, the SSAB is

\(\begin{aligned}{*{20}{c}}{SSAB = SST - SSA - SSB - SSE}\\{ = 64,954.70 - 22,941.80 - 22,765.53 - 15,253.50 = 3993.87.}\end{aligned}\)

The mean squares can now be computed as follows

\(MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{2} \cdot 22,941.80 = 11,470.90\)

\(MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{4} \cdot 22,765.53 = 5691.38\)

\(\begin{aligned}{*{20}{c}}{MSAB = \frac{1}{{(I - 1)(J - 1)}} \cdot SSAB = \frac{1}{8} \cdot 3993.87 = 499.23}\\{MSE = \frac{1}{{IJ(K - 1)}} \cdot SSE = \frac{1}{{15}} \cdot 15,253.50 = 1016.9}\end{aligned}\)

The test statistic values f are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{11.470.90}}{{1016.9}} = 22.98}\\{{f_{1S}} = \frac{{MSE}}{{MSF}} = \frac{{5691.38}}{{1016.9}} = 11.40}\\{{f_{AB}} = \frac{{MSAB}}{{MSE}} = \frac{{499.23}}{{1016.9}} = 0.49.}\end{aligned}\)

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 \({H_{0AB}}\)versus \({H_{0AB}}\), 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

\(P = P\left( {F > {f_{AB}}} \right) = P(F > 0.49) = 0.845\)

where F has Fisher's distribution with \({\nu _1} = (I - 1)(J - 1) = 8{\rm{\;and\;}}{\nu _2} = IJ(K - 1) = 15\) degrees of freedom. Since

\(P = 0.845 > 0.05 = \alpha \)

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

It appears that there is no interaction between factor A and factor B.

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

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

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

Using a software, the P value is

\(P = P\left( {F > {f_A}} \right) = P(F > 22.98) = 0.00,\)

Where F has Fisher's distribution with \({\nu _1} = I - 1 = 2{\rm{\;and\;}}{\nu _2} = IJ(K - 1) = 15\) degrees of freedom. Since

\(P = 0.00 < 0.05 = \alpha \)

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

It appears that there is significant effect due to factor A.

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

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

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

Using a software, the P value is

\(P = P\left( {F > {f_B}} \right) = P(F > 11.4) = 0.0\)

where F has Fisher's distribution with \({\nu _1} = (J - 1) = 4{\rm{\;and\;}}{\nu _2} = IJ(K - 1) = 15\) degrees of freedom. Since

\(P = 0.00 < 0.05 = \alpha \)

do not reject null hypothesis

\({H_{0AB}}\)

It appears that there is significant effect due to factor B.

The other way to make conclusion about the hypotheses is by finding the critical values from the table in the appendix

\(\begin{aligned}{*{20}{c}}{{F_{0.05,2,15}}}\\{{F_{0.05,4,15}}}\\{{F_{0.05,8,15}}}\end{aligned}\)