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a)-

Let

Where the following holds

and where the are independent normally distributed random variable with mean 0 and varianceThe hypotheses of interest are

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

And for the factor B

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

The following table summarizes values which will be required to compute the test statistic value.

Values of I and J are

\(I = 4\)- number of rows;

J = 3 鈥� number of columns.

Before continuing, denote withthe average of measurements obtained when factor A is held at level i.

Withthe average of measurements obtained when factor B is held at level j

And with.. the grand mean

Observed values are denoted with small instead of big X. The notations without line over X are just the sums.

Compute them one by one. Starting with average of measurements for factor A at level corresponding level

\(\begin{aligned}{*{20}{c}}{{x_{1.}} = 64 + 49 + 50 = 163}\\{{x_{2.}} = 53 + 51 + 48 = 152}\\{{x_{3.}} = 47 + 45 + 50 = 142}\\{{x_{4.}} = 51 + 43 + 52 = 146}\end{aligned}\)

Values of

\({\bar x_{i.}} = \frac{1}{J} \cdot {x_i}\)

Are given by

\(\begin{aligned}{*{20}{c}}{{{\bar x}_{1.}} = \frac{1}{3} \cdot 163 = 54.33}\\{{{\bar x}_{2.}} = \frac{1}{3} \cdot 152 = 50.67}\\{{{\bar x}_{3.}} = \frac{1}{3} \cdot 142 = 47.33}\\{{{\bar x}_{4.}} = \frac{1}{3} \cdot 146 = 48.67}\end{aligned}\)

The average of measurements for factor B at level corresponding level

\(\begin{aligned}{*{20}{c}}{{x_{.1}} = 64 + 53 + 47 + 51 = 215}\\{{x_{.2}} = 49 + 51 + 45 + 43 = 188}\\{{x_{.3}} = 50 + 48 + 50 + 52 = 200.}\end{aligned}\)

Values of

\({\bar x_{ \cdot j}} = \frac{1}{I} \cdot {x_{ \cdot j}}\)

Are given by

\(\begin{aligned}{*{20}{c}}{{{\bar x}_{.1}} = \frac{1}{4} \cdot 215 = 53.75;}\\{{{\bar x}_{.2}} = \frac{1}{4} \cdot 188 = 47;}\\{{{\bar x}_{.3}} = \frac{1}{4} \cdot 200 = 50.}\end{aligned}\)

The grand sum is

The grand mean is

The sum of squares are given by

with degrees of freedom respectively,

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

SST can be computed as

SSA can be computed as

\(\begin{aligned}{*{20}{c}}{ = \frac{1}{3} \cdot \left( {{{163}^2} + {{152}^2} + {{142}^2} + {{146}^2}} \right) - \frac{1}{{4 \cdot 3}} \cdot {{603}^2}}\\{ = 30,384.33 - 30,300.75}\\{ = 83.58}\end{aligned}\)

\(SST = SSA + SSB + SSE\)

By the fundamental identity, the SSE can be computed as

SSE= SST 鈥� SSA 鈥� SSB =298.25 - 83.58 - 91.5 = 123.17.

When testing hypotheses H_0A versus H_aA, the test statistic value is

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

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_A.

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

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

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

The degrees of freedom are

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJ - 1 = 4 \cdot 3 - 1 = 11}\\{d{f_A} = I - 1 = 4 - 1 = 3}\\{d{f_B} = J - 1 = 3 - 1 = 2;}\\{d{f_E} = (I - 1)(J - 1) = (4 - 1) \cdot (3 - 1) = 6.}\end{aligned}\)

The mean square are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{3} \cdot 83.58 = 27.86}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{2} \cdot 91.5 = 45.75}\\{MSE = \frac{1}{{(I - 1)(J - 1)}} \cdot SSE = \frac{1}{6} \cdot 123.17 = 20.53.}\end{aligned}\)

The value of test statistics are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{27.86}}{{20.53}} = 1.36}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{45.75}}{{20.53}} = 2.23.}\end{aligned}\)

There are two ways to make conclusion. Using the table in the appendix or compute P value using a software.

The P_A value when testing hypotheses H_OA versus H_aA is

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 1.36) = 0.34\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(I - 1 = 3{\rm{\;and\;}}(I - 1)(J - 1) = 6\)because

\({P_A} = 0.34 > 0.05 = \alpha \)

do not reject hypothesis H_0A

at given significance level.

The P_B value when testing hypotheses H_OB versus H_aB is

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 2.23) = 0.19\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(J - 1 = 2{\rm{\;and\;}}(I - 1)(J - 1) = 6.{\rm{ Because\;}}\)

\({P_B} = 0.19 > 0.05 = \alpha \)

do not reject hypothesis H_0B

at given significance level.

using the fact that

\({F_{\alpha ,I - 1,(I - 1)(J - 1)}} = {F_{0.05,3,6}} = 4.76,\)

which was obtained from the table in the appendix, and the fact that

\({F_{0.05,3,6}} = 4.76 > 1.36 = {f_A}\)

do not reject hypothesis H_0A

at given significance level.

Also, using the fact that

\({F_{\alpha ,J - 1,(I - 1)(J - 1)}} = {F_{0.05,2,6}} = 5.14,\)

which was obtained from the table in the appendix, and the fact that

\({F_{0.05,2,6}} = 5.14 > 2.23 = {f_B}\)

do not reject hypothesis H_0B

at given significance level.

Finally, the result can be represented in a table

B)-

The estimates can be computed using the sample means; thus, depending on what value is estimated, the formula is given below

\(\begin{aligned}{*{20}{c}}{\hat \mu = {{\bar x}_{..}} = 50.25}\\{{{\hat \alpha }_1} = {{\bar x}_{1.}} - {{\bar x}_{..}} = 54.33 - 50.25 = 4.08}\\{{{\hat \alpha }_2} = {{\bar x}_{2.}} - {{\bar x}_{..}} = 50.67 - 50.25 = 0.42;}\\{{{\hat \alpha }_3} = {{\bar x}_{3.}} - {{\bar x}_{..}} = 47.33 - 50.25 = - 2.92;}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{{\hat \alpha }_4} = {{\bar x}_{4.}} - {{\bar x}_{..}} = 48.67 - 50.25 = - 1.58;}\\{{{\hat \beta }_1} = {{\bar x}_{.1}} - {{\bar x}_{..}} = 53.75 - 50.25 = 3.5}\\{{{\hat \beta }_2} = {{\bar x}_{.1}} - {{\bar x}_{..}} = 53.75 - 50.25 = - 3.25;}\\{{{\hat \beta }_3} = {{\bar x}_{.1}} - {{\bar x}_{..}} = 53.75 - 50.25 = - 0.25.}\end{aligned}\)