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Let factor A be spray volume, factor B bell speed, and factor C brand. From the given table of signs you can create the following table in standard order

Where the - representsand + represents 鈥�2鈥�

An efficient method for hand computation due to Yates is: write total of 8 cells totals in the standard order. Add three more columns to the table of 8 cells. The entries in the columns represent sum of particular entries of the previous columns - first four are sum of 1 and 2 ; 3 and 4 ; 5 and 6 ; 7 and 8 ; and the last four are differences between 2 and 1 ; 4 and 3 ; 6 and 5 ; 7 and 8. The table becomes:

To understand the table above, see the following details:

First (second) column consist of corresponding sums of particular factors at corresponding factor levels, e.g.\({x_{1111}}\)is the sum when the factor Ais held at level 1,factor B is held at level 1, and factor Cis held at level 1. Third column is constructed as follows:

First rowis sum of \({x_{111}}\)and \({x_{211}}\)

\({x_{111}}\)+\({x_{211}}\)=76+53=129;

Second rowis sum of \({x_{121}}\)and\({x_{221}}\)

\({x_{121}}\)+\({x_{221}}\)=62+98=160;

Third row is sum of \({x_{112}}\)and\({x_{212}}\)

\({x_{112}}\)+\({x_{212}}\)=88+55=143;

Fourth row is sum of \({x_{122}}\)and\({x_{222}}\)

\({x_{122}}\)+ \({x_{222}}\)=59+101=160;

Fifth rowis difference of \({x_{211.}}{\rm{\;and\;}}{x_{111.}}\)

\({x_{211.}} + {x_{111.}} = 53 - 76 = - 23\);

Sixth row is difference of \({x_{221.}}{\rm{\;and\;}}{x_{121}}\)

\({x_{221.}} + {x_{121.}} = 98 - 62 = 36\);

Seventh row is difference of \({x_{212}}.{\rm{\;and\;}}{x_{112}}\)

\({x_{212.}} + {x_{112.}} = 55 - 88 = - 33\);

Sixth row is difference of \({x_{222.}}{\rm{\;and\;}}{x_{122}}\)

\({x_{222.}} + {x_{122.}} = 101 - 59 = 42\);

In a similar manner obtain fourth and fifth column. The formula to obtain the fifth formula is given in the table.

The computed sum of squares, from the table, corresponding to the letters in the first column, are

SSA=30.25

SSB=144.00

SSAB=1122.25

SSC=12.25

SSAC=1

SSBC=12.25

SSABC=16

The ANOVA table is

There are missing two sum of squares, SSE and SST. The degrees of freedom are simply computed because each factor has only two levels.

The SST is

\( = {40^2} + {36^2} + \ldots + {49^2} - \frac{1}{{2 \cdot 2 \cdot 2 \cdot 2}} \cdot (40 + 36 + \ldots + 49)\)

\( = 1374\)

Fundamental identity:

\(SST = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABC + SSE\)

\({\rm{\;By the fundamental identity, value of the missing values,\;}}SSABC{\rm{, is\;}}\)

\(SSE = SST - (SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABC)\)

\( = 1374 - 30.25 - 144 - 12.25 - \)

\( - 1122.25 - 1 - 12.25 - 16\)

=36

The degree of freedom are

\(d{f_T} = IJKL - 1 = 2 \cdot 2 \cdot 2 \cdot 2 - 1 = 15\)

\(d{f_E} = IJK(L - 1) = 2 \cdot 2 \cdot 2 \cdot (2 - 1) = 8\)

\(d{f_A} = I - 1 = 2 - 1 = 1\)

\(d{f_B} = J - 1 = 2 - 1 = 1\)

\(d{f_C} = K - 1 = 2 - 1 = 1\)

\(d{f_{AB}} = (I - 1)(J - 1) = (2 - 1) \cdot (2 - 1) = 1\)

\(d{f_{AC}} = (I - 1)(K - 1) = (2 - 1) \cdot (2 - 1) = 1\).

\(d{f_{BC}} = (J - 1)(K - 1) = (2 - 1) \cdot (2 - 1) = 1\)

\(d{f_{ABC}} = (I - 1)(J - 1)(K - 1) = (2 - 1) \cdot (2 - 1) \cdot (2 - 1) = 1\)

The mean squares can be computed using sum of squares as follows

\(MSA = \frac{1}{{d{f_A}}} \cdot SSA = \frac{1}{1} \cdot 30.25 = 30.25\)

\(MSB = \frac{1}{{d{f_B}}} \cdot SSB = \frac{1}{1} \cdot 144 = 144\)

\(MSC = \frac{1}{{d{f_C}}} \cdot SSC = \frac{1}{1} \cdot 12.25 = 12.25\)

\(MSAB = \frac{1}{{d{f_{AB}}}} \cdot SSAB = \frac{1}{1} \cdot 1122.25 = 1122.25\)

\(MSAC = \frac{1}{{d{f_{AC}}}} \cdot SSAC = \frac{1}{1} \cdot 1 = 1\)

\(MSBC = \frac{1}{{d{f_{BC}}}} \cdot SSBC = \frac{1}{1} \cdot 12.25 = 12.25\)

\(MSABC = \frac{1}{{d{f_{ABN}}}} \cdot SSABC = \frac{1}{1} \cdot 16 = 16\)

\(MSE = \frac{1}{{d{f_E}}} \cdot SSE = \frac{1}{8} \cdot 36 = 4.5\)

\({\rm{\;The corresponding\;}}f{\rm{\;values are\;}}\)

\({f_A} = \frac{{MSA}}{{MSE}} = \frac{{30.25}}{{4.5}} = 6.72\)

\({f_B} = \frac{{MSB}}{{MSE}} = \frac{{144}}{{4.5}} = 32.00\)

\({f_C} = \frac{{MSC}}{{MSE}} = \frac{{12.25}}{{4.5}} = 2.72\)

\({f_{AB}} = \frac{{MSAB}}{{MSE}} = \frac{{1122.25}}{{4.5}} = 249.39\)

\({f_{AC}} = \frac{{MSAC}}{{MSE}} = \frac{1}{{4.5}} = 0.22\)

\({f_{BC}} = \frac{{MSBC}}{{MSE}} = \frac{{12.25}}{{4.5}} = 2.72\)

\({f_{ABC}} = \frac{{MSABC}}{{MSE}} = \frac{{16}}{{4.5}} = 3.56\)

The P values can be computed using a software. The following holds:

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 25) = 0.001\)

P-value is the area under the \({F_{I - 1,IJK(L - 1)}}\)curve to the right of the test static value\({f_A}\)

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 32.00) = 0.00\)

P-value is the area under the\({F_{J - 1,IJK(L - 1)}}\)curve to the right of the test static value \({f_B}\)

\({P_C} = P\left( {F > {f_C}} \right) = P(F > 2.72) = 0.14\)

P-value is the area under the \({H_K} - 1,IJK(L - 1)\)curve to the right of the test static value\({f_C}\)

\({P_{AB}} = P\left( {F > {f_{AB}}} \right) = P(F > 249.39) = 0.00\)

P-value is the area under the \({F_{(I - 1)(J - 1),IJK(L - 1)}}\) curve to the right of the test static value\({f_{AB}}\)

\({P_{AC}} = P\left( {F > {f_{AC}}} \right) = P(F > 0.22) = 0.65\)

P-value is the area under the \({F_{(I - 1)(K - 1),IJK(L - 1)}}\) curve to the right of the test static value\({f_{AC}}\)

\({P_{BC}} = P\left( {F > {f_{BC}}} \right) = P(F > 2.72) = 0.14\)

P-value is the area under the \({F_{(J - 1)(K - 1),IJK(L - 1)}}\)curve to the right of the test static value\({f_{BC}}\)

\({P_{ABC}} = P\left( {F > {f_{ABC}}} \right) = P(F > 3.56) = 0.10\)

P-value is the area under the\({F_{(I - 1)(J - 1)(K - 1),IJK(L - 1)}}\) curve to the right of the test static value\({f_{ABC}}\)

The only value left is the critical value:

\({F_{0.05,1,8}} = 5.32\)

The ANOVA table is

\({\rm{\;The fixed effects model for three - factor, for\;}}I = J = K = 2{\rm{, ANOVA is\;}}\)

\({\rm{\;for\;}}i = 1,2, \ldots ,I;j = 1,2, \ldots ,J;k = 1,2, \ldots ,K;l = 1,2, \ldots ,L,{\rm{\;and\;}}\)\({L_{ijk}} = L\)are no. of observation made at the factor of A at the level of i, factor Bat level j, factor C at level k, where the are independent normally distributed random variable with mean 0 and variance\({\sigma ^2}\). Notion \({\mu _{ijk}}\)stands for

\({\mu _{ijk}} = \mu + {\alpha _i} + {\beta _j} + {\delta _k} + \gamma _{ij}^{AB} + \gamma _{ij}^{AC} + \gamma _{ij}^{BC} + {\gamma _{ijk}}\)

The relevant hypotheses are

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

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

\({H_{0C}}:{\delta _1} = {\delta _2} = \ldots = {\delta _K} = 0{\rm{\;versus\;}}{H_{aC}}:{\rm{\;at least one\;}}{\delta _ - }i \ne 0\)

And for the interaction

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

\({H_{0AC}}:\gamma _{ij}^{AC} = 0{\rm{\;for all\;}}i,j{\rm{\;versus\;}}{H_{aAC}}:{\rm{\;at least one\;}}{\gamma _ - }i{j^{AC}} \ne 0\)\({H_{0BC}}:\gamma _{ij}^{BC} = 0{\rm{\;for all\;}}i,j{\rm{\;versus\;}}{H_{aBC}}:{\rm{\;at least one\;}}{\gamma _ - }i{j^{BC}} \ne 0\)

And for the interaction ABC

\({H_{0ABC}}:{\gamma _{ijk}} = 0{\rm{\;for all\;}}i,j,k{\rm{\;versus\;}}{H_{aABC}}:{\rm{\;at least one\;}}{\gamma _ - }ijk \ne 0\)

When testing hypotheses\({H_{0{\rm{A}}}}\)versus \({H_{aA}}\)the test static value is \({f_A} = \frac{{MSA}}{{MSE}}\)

P-value is the area under the \({F_{I - 1,IJK(L - 1)}}\) curve to the right of the test static value\({f_A}\)

When testing hypotheses\({H_{0B}}\)versus \({H_{aB}}\)the test static value is \({f_B} = \frac{{MSB}}{{MSE}}\)

P-value is the area under the\({F_{J - 1,IJK(L - 1)}}\) curve to the right of the test static value \({f_B}\)

When testing hypotheses\({H_{0C}}\)versus\({H_{aC}}\) the test static value is \({f_C} = \frac{{MSC}}{{MSE}}\)

P-value is the area under the \({F_{K - 1,IJK(L - 1)}}\) curve to the right of the test static value\({f_C}\)

From the ANOVA table you can take the required values.

\({P_A} = 0.03 < 0.05 = \alpha \)

Reject null hypotheses \({H_{0A}}\)

At given significance level\(\alpha \) Factor (main effect) A is statistically significant.

For factor B

\({P_B} = 0.00 < 0.05 = \alpha \)

Reject null hypothesis \({H_{0B}}\)

At given significance level\(\alpha \) Factor (main effect) B is statistically significant.

For factor C

\({P_C} = 0.14 > 0.05 = \alpha \)

Do not Reject null hypothesis\({H_{0C}}\)

At given significance level\(\alpha \) Factor (main effect) C is statistically significant.

When testing hypotheses\({H_{0AB}}\)versus \({H_{aAB}}\) the test static value is\({f_{AB}} = \frac{{MSAB}}{{MSE}}\)

And the P-value is the area under the\({F_{(I - 1)(J - 1),IJK(L - 1)}}\) curve to the right of the test statistic value\({f_{AB}}\)

When testing hypotheses \({{\rm{H}}_{0{\rm{AC}}}}\)versus \({H_{aAC}}\) the test static value is \({f_{AC}} = \frac{{MSAC}}{{MSE}}\)

And the P-value is the area under the\({F_{(I - 1)(K - 1),IJK(L - 1)}}\) curve to the right of the test statistic value\({f_{AC}}\)

When testing hypotheses \({H_{0BC}}\) versus \({H_{aBC}}\) the test static value is\({f_{BC}} = \frac{{MSBC}}{{MSE}}\)

And the P-value is the area under the \({F_{(J - 1)(K - 1),IJK(L - 1)}}\) curve to the right of the test statistic value\({f_{BC}}\)

When testing hypotheses \({H_{0ABC}}\)versus\({H_{aABC}}\) the test static value is\({f_{ABC}} = \frac{{MSABC}}{{MSE}}\)

And the P-value is the area under the\({F_{(I - 1)(J - 1)(K - 1),IJK(L - 1)}}\) curve to the right of the test statistic value\({f_{ABC}}\)

From the ANOVA table you can take the required values.

For interaction AB

\({P_{AB}} = 0.00 < 0.05 = \alpha \)

Reject null hypothesis\({H_{0AB}}\)

at given significance level. Interaction AB is statistically significant.

For interaction AC

\({P_{AC}}0.65 > 0.05 = \alpha \)

Do not Reject null hypothesis\({H_{0AC}}\)

at given significance level. Interaction AB is statistically significant.

For interaction BC

\({P_{BC}} = 0.14 > 0.05 = \alpha \)

Do not reject null hypothesis\({H_{0BC}}\)

at given significance level. Interaction BC is not statistically significant.

For interaction ABC

\({P_{ABC}} = 0.10 > 0.05 = \alpha \)

Do not reject null hypothesis\({H_{0ABC}}\)

at given significance level. Interaction $A B C$, the three way interaction, is not statistically significant.

It appears that main effects A and B as well as interaction AB are statistically significant. No other main effects or interactions are statistically significant.

Thus It appears that main effects A and B as well as interaction AB are statistically significant. No other main effects or interactions are statistically significant.