91影视

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 - Medium at level corresponding level

\(\begin{aligned}{*{20}{c}}{{x_{1.}} = 0.2433 + 0.383 + 0.5625 + 0.7258 = 1.9146}\\{{x_{2.}} = 0.159 + 0.2649 + 0.3609 + 0.4773 = 1.2621}\end{aligned}\)

Values of

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

Are given by

\(\begin{aligned}{*{20}{c}}{{{\bar x}_{1.}} = \frac{1}{4} \cdot 1.9146 = 0.47865}\\{{{\bar x}_{2.}} = \frac{1}{4} \cdot 1.2621 = 0.31553.}\end{aligned}\)

The average of measurements for factor B- Working current, at level corresponding level

\(\begin{aligned}{*{20}{c}}{{x_{.1}} = 0.2433 + 0.159 = 0.4023}\\{{x_{.2}} = 0.383 + 0.2649 = 0.6479}\\{{x_{.3}} = 0.5625 + 0.3609 = 0.9234}\\{{x_{.4}} = 0.7258 + 0.4773 = 1.2031.}\end{aligned}\)

Values of

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

Are given by

\(\begin{aligned}{*{20}{c}}{{{\bar x}_{.1}} = \frac{1}{2} \cdot 0.4023 = 0.20115}\\{{{\bar x}_{.2}} = \frac{1}{2} \cdot 0.6479 = 0.32395}\\{{{\bar x}_{.3}} = \frac{1}{2} \cdot 0.9234 = 0.46170}\\{{{\bar x}_{.4}} = \frac{1}{2} \cdot 1.2031 = 0.60155}\end{aligned}\)

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

\(\begin{aligned}{*{20}{c}}{ = \left( {{{60.2433}^2} + {{0.383}^2} + \ldots + {{0.3609}^2} + {{0.4773}^2}} \right) - \frac{1}{{2 \cdot 4}} \cdot {{3.17670}^2}}\\{ = 1.502593 - 1.261428}\\{ = 0.241165}\end{aligned}\)

SSA can be computed as

SSB can be computed as

\(SST = SSA + SSB + SSE\)

By the fundamental identity, the SSE can be computed as

\(SSE = SST - SSA - SSB = 0.241165 - 0.053220 - 0.179441 = 0.008504\)

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

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJ - 1 = 2 \cdot 4 - 1 = 7}\\{d{f_A} = I - 1 = 2 - 1 = 1}\\{d{f_B} = J - 1 = 4 - 1 = 3}\\{d{f_E} = (I - 1)(J - 1) = (2 - 1) \cdot (4 - 1) = 3.}\end{aligned}\)

The mean square are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{1} \cdot 0.053220 = 0.053220}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{3} \cdot 0.179441 = 0.059814}\\{MSE = \frac{1}{{(I - 1)(J - 1)}} \cdot SSE = \frac{1}{3} \cdot 0.008504 = 0.002835.}\end{aligned}\)

The value of test statistics are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{0.053220}}{{0.002835}} = 18.77;}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{0.059814}}{{0.002835}} = 21.10.}\end{aligned}\)

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

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 18.77) = 0.02\)

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

\({P_A} = 0.02 < 0.05 = \alpha \)

Do not reject hypothesis H_0A

at given significance level.

The P_B value when testing hypotheses H_OB versus H_aB is

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 21.10) = 0.20,\)

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

\({P_B} = 0.02 < 0.05 = \alpha \)

reject hypothesis H_0B

at given significance level.

using the fact that

\({F_{\alpha ,I - 1,(I - 1)(J - 1)}} = {F_{0.05,1,3}} = 10.13,\)

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

\({F_{0.05,1,3}} = 10.13 < 18.77 = {f_A}\)

reject hypothesis H_0A

at given significance level.

Also, using the fact that

\({F_{\alpha ,J - 1,(I - 1)(J - 1)}} = {F_{0.05,3,3}} = 9.28\)

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

\({F_{0.05,3,3}} = 9.28 < 21.10 = {f_B}\)

reject hypothesis H_0B

at given significance level.

Finally, the result can be represented in a table

B)-

In (a) the null hypothesis H_0B was rejected. It makes sense to use Turkey's procedure to identify significance difference between the levels of the factor B.

In order to compare levels of factor B, first obtain\({Q_{\alpha ,J,(I - 1)(.J - 1)}}\). In second step compute value

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

In the last step, arrange the sample means 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.

Given in the exercise,

\({Q_{\alpha ,J,(I - 1)(J - 1)}} = {Q_{0.05,4,3}} = 6.825.\)

The value of w is

\(\begin{aligned}{*{20}{c}}{w = {Q_{\alpha ,J,(I - 1)(J - 1)}} \cdot \sqrt {\frac{{MSE}}{I}} = {Q_{0.05,4,3}} \cdot \sqrt {\frac{{MSE}}{2}} }\\{ = 6.825 \cdot \sqrt {\frac{{0.002835}}{2}} = 0.257.}\end{aligned}\)

The following table can help with the values required.

From the table, notice that the ordered means are

\(x.1 < x.2 < x.3 < x.4\)

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

The following are the differences

\(\begin{aligned}{*{20}{c}}{{x_{.2}} - {x_{.1}} = 0.323950 - 0.201150 = 0.122800 < w = 0.257;}\\{{x_{.3}} - {x_{.1}} = 0.461700 - 0.201150 = 0.260550 > w = 0.257;}\\{{x_{.4}} - {x_{.1}} = 0.601550 - 0.201150 = 0.400400 > w = 0.257;}\\{{x_{.3}} - {x_{.2}} = 0.461700 - 0.323950 = 0.137750 < w = 0.257;}\\{{x_{.4}} - {x_{.2}} = 0.601550 - 0.323950 = 0.277600 > w = 0.257;}\\{{x_{.4}} - {x_{.3}} = 0.601550 - 0.461700 = 0.139850 > w = 0.257.}\end{aligned}\)

The differences are the ones smaller than w; thus, do not differ a significantly. The other pair do differ significantly. The pair which differ are

(1,3), (1,3), (2,4).