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Given values are

\(\begin{aligned}{*{20}{c}}{SSA = 30.763.0}\\{SSB = 34,185.6}\\{SSE = 97,436.8}\\{SST = 205,966.6}\end{aligned}\)

(a): In order to construct an ANOVA table, a couple of values have to be computed. The table is

The degree of freedom are

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1 = 3 \cdot 4 \cdot 3 - 1 = 35}\\{d{f_E} = IJ(K - 1) = 3 \cdot 4 \cdot (3 - 1) = 24}\end{aligned}\\\begin{aligned}{*{20}{c}}{d{f_A} = I - 1 = 3 - 1 = 2}\\{d{f_B} = J - 1 = 4 - 1 = 3}\end{aligned}\\d{f_{AB}} = (I - 1)(J - 1) = (3 - 1) \cdot (4 - 1) = 6\end{aligned}\).

By the fundamental identity, the SSAB is

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

\(\begin{aligned}{*{20}{c}}{SSAB = SST - SSA - SSB - SSE}\\{ = 205,966.6 - 30,763.0 - 34,185.6 - 97,436.8 = 43,581.2.}\end{aligned}\)

The mean squares now can be computed as,

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{2} \cdot 30,763.0 = 15,381.5}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{3} \cdot 34,185.6 = 11,395.2}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{MSAB = \frac{1}{{(I - 1)(J - 1)}} \cdot SSAB = \frac{1}{6} \cdot 43,581.2 = 7,263.53}\\{MSE = \frac{1}{{IJ(K - 1)}} \cdot SSE = \frac{1}{{24}} \cdot 97,436.8 = 4,059.86}\end{aligned}\)

The test statistic value f are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{15,381.5}}{{4,059.86}} = 3.79}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{11,395.2}}{{4,059.86}} = 2.81}\\{{f_{AB}} = \frac{{MSAB}}{{MSE}} = \frac{{7,263.53}}{{4,059.86}} = 1.79.}\end{aligned}\)

The ANOVA table now becomes

(b) Let

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

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

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

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

There are two ways to make conclusion \(\left. {{F_{(I - 1}} - 1} \right)(.J - 1),IJ\left( {\begin{aligned}{*{20}{c}}{K1}\end{aligned}} \right)\)curve to the right\({f_B}\) - finding critical value(s) or by finding the P value.

From the ANOVA table

\({f_{AB}} = 1.79\)

\({F_{(I - 1)(J - 1),IJ(K - 1)}} = {F_{0.05,6,24}} = 2.51,\)

\({F_{0.05,6,24}} = 2.51 > 1.79 = {f_{AB}},\)

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

thus, there is no interaction between factors.

In order to compute P value you need a software. The P value is

\(\begin{aligned}{*{20}{c}}{P = P\left( {F > {f_{AB}}} \right) = P(F > 1.79) = 0.144.}\\{P = 0.144 > 0.05 = \alpha }\end{aligned}\)

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

(c)Let

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

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

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

There are two ways to make conclusion \(\left. {{F_{(I - 1}} - 1} \right)(.J - 1),IJ\left( {\begin{aligned}{*{20}{c}}{K1}\end{aligned}} \right)\)curve to the right\({f_A}\) - finding critical value(s) or by finding the P value.

From the ANOVA table

\({f_A} = 3.79\)

\({F_{\alpha ,I - 1,IJ(K - 1)}} = {F_{0.05,2,24}} = 3.4\)

\({F_{0.05,2,24}} = 3.4 < 3.79 = {f_A},\)

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

thus, there is no interaction between factors.

In order to compute P value you need a software. The P value is

\(\begin{aligned}{*{20}{c}}{P = P\left( {F > {f_A}} \right) = P(F > 3.79) = 0.037.}\\{P = 0.037 < 0.05 = \alpha ;}\end{aligned}\)

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

The hypotheses of interest are

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

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

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

There are two ways to make conclusion \(\left. {{F_{(I - 1}} - 1} \right)(.J - 1),IJ\left( {\begin{aligned}{*{20}{c}}{K1}\end{aligned}} \right)\)curve to the right\({f_B}\) - finding critical value(s) or by finding the P value.

From the ANOVA table

\({f_B} = 2.81\)

\({F_{\alpha ,J - 1,I.J(K - 1)}} = {F_{0.05,3,24}} = 3.01\)

\({F_{0.05,3,24}} = 3.01 > 2.81 = {f_B},\)

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

thus, there is no interaction between factors.

In order to compute P value you need a software. The P value is

\(\begin{aligned}{*{20}{c}}{P = P\left( {F > {f_B}} \right) = P(F > 2.81) = 0.061.}\\{P = 0.061 > 0.05 = \alpha }\end{aligned}\)

thus, do not reject null hypothesis \({H_{0B}}\).

The rejected hypothesis is \({H_{0A}}\)so it makes sense to use Turkeys method.

In order to compare levels of factor A, first obtain \(\left. {{Q_{\alpha ,I,IJ(K}}1} \right)\). In second step compute value

\(w = {Q_{\alpha ,I,IJ(K - 1)}} \cdot \sqrt {\frac{{MSE}}{{JK}}} \)

In the last step, arrange the sample means \(\left( {{{\bar x}_{i..}}} \right)\)in increasing order and underscore pairs which differs less than w, and identify pairs not underscored by the same line as corresponding to significantly different levels of the given factor.

The sample means are given

\(\begin{aligned}{*{20}{c}}{{{\bar x}_{1..}} = 4010.88}\\{{{\bar x}_{2..}} = 4029.10}\\{{{\bar x}_{3..}} = 3960.02}\end{aligned}\)

From the table in the appendix the value of Q is

\({Q_{\alpha ,I,IJ(K - 1)}} = {Q_{0.05,3,24}} = 3.53\)

The w can be computed as

\(\begin{aligned}{*{20}{c}}{w = {Q_{\alpha ,I,IJ(K - 1)}} \cdot \sqrt {\frac{{MSE}}{{JK}}} = {Q_{0.05,3,24}} \cdot \sqrt {\frac{{MSE}}{{12}}} }\\{ = 3.53 \cdot \sqrt {\frac{{4059.87}}{{12}}} = 64.93.}\end{aligned}\)

From the table at beginning. notice that the ordered means are

\({x_{3..}} < {x_{1..}} < {x_{2..}}\)

thus; compute all differences and see which are smaller than w.

The following are the differences

\(\begin{aligned}{l}{x_{1/.}} - {x_{3..}} = 4010.88 - 3960.02 = 50.86 < w = 64.93\\\begin{aligned}{*{20}{c}}{{x_{2..}} - {x_{3..}} = 4029.1 - 3960.02 = 69.08 > w = 64.93}\\{{x_{2..}} - {x_{1..}} = 4029.1 - 4010.881 = 18.22 < w = 64.93.}\end{aligned}\end{aligned}\)

'The differences marked blue are the ones smaller than w; thus do not, differ a significantly. The other pairs do differ significantly. The pair which statistically differ is

(3,2)

Times 2 and 3 yield significantly different strengths. This can be represented as

\(\begin{aligned}{*{20}{c}}i&3&1&2\\{{{\bar x}_{i,.}}}&{3960.02}&{4010.88}&{4029.1}\end{aligned}\)