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Assume that two and three factor interaction effects are absent!

A Latin square design model equation is given by

\({\rm{\;where\;}}I = J = K = N{\rm{,\;}}\)

and the errors are independent and normally distributed with mean zero and variance \({\sigma ^2}\).The notations for totals and averages are

Obesrved values are denoted as x instead of big X.The notations without line over X are just the sums.

The folowing table needs to completed with corresponding values.

Sum of squares,for a Latin square experiment and the degree of freedom are given by

\(\)

With degrees of freedom respectively,

\(\begin{aligned}{*{20}{c}}{d{f_T} = {N^2} - 1}\\{d{f_A} = N - 1}\\{d{f_B} = N - 1}\\{d{f_C} = N - 1}\\{d{f_E} = (N - 1)(N - 2).}\end{aligned}\)

The sum of measurements obtained when factor A is held at level i are

The Sum of measurements obtained for factor B at level j are

The Sum of measurements obtained for factor C at level k are

The grand sum is

Every sum of square can be computed using the sums above.The SST is

The SSA is

The SSB is

The SSC is

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

By the Fundamental Identity,the error sum of square can be computed as follows

\(\begin{aligned}{*{20}{c}}{SSE = SST - (SSA + SSB + SSC)}\\{ = 8790.98 - 6475.81 - 529.48 - 508.48}\\{ = 1277.21.}\end{aligned}\)

The degrees of freedom are

\(\begin{aligned}{*{20}{c}}{d{f_T} = {N^2} - 1 = 35}\\{d{f_A} = N - 1 = 5}\\{d{f_B} = N - 1 = 5}\\{d{f_C} = N - 1 = 5}\\{d{f_E} = (N - 1)(N - 2) = (6 - 1)(6 - 2) = 20.}\end{aligned}\)

The mean squares are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{d{f_A}}} \times SSA = \frac{1}{5} \times 6475.81 = 1295.16}\\{MSB = \frac{1}{{d{f_B}}} \times SSB = \frac{1}{5} \times 529.48 = 105.90}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{MSC = \frac{1}{{d{f_C}}} \cdot SSC = \frac{1}{5} \cdot 508.48 = 101.70}\\{MSE = \frac{1}{{d{f_E}}} \cdot SSE = \frac{1}{{20}} \cdot 1277.21 = 63.86.}\end{aligned}\)

The corresponding f values are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{1295.16}}{{63.86}} = 20.28}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{105.90}}{{63.86}} = 1.66;}\\{{f_C} = \frac{{MSC}}{{MSE}} = \frac{{101.70}}{{63.86}} = 1.59.}\end{aligned}\)

The P value can be computed using a software.

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 20.28) = 0.00\)

P-value is the area under the \({F_{N - 1,(N - 1)(N - 2)}}{\rm{\;curve to the right of the test statistic value\;}}{f_A}{\rm{.\;}}\)

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 1.66) = 0.19\)

P-value is the area under the \({F_{N - 1,(N - 1)(N - 2))}}{\rm{\;curve to the right of the test statistic value\;}}{f_B}{\rm{.\;}}\)

\({P_C} = P\left( {F > {f_C}} \right) = P(F > 1.59) = 0.21\)

P-value is the area under the \({F_{N - 1,(N - 1)(N - 2)}}{\rm{\;curve to the right of the test statistic value\;}}{f_C}\)

\({F_{0.01,5,20}} = 4.1\)

Finally, ANOVA Table becomes

The Hypothesis of interests are

\(\begin{aligned}{*{20}{c}}{{H_{0A}}:{\alpha _1} = {\alpha _2} = \ldots = {\alpha _I} = 0}&{{\rm{\;versus\;}}}&{{H_{aA}}:{\rm{\;at least one\;}}{\alpha _ - }i \ne 0}\\{{H_{0B}}:{\beta _1} = {\beta _2} = \ldots = {\beta _J} = 0}&{{\rm{\;versus\;}}}&{{H_{aB}}:{\rm{\;at least one\;}}{\beta _ - }j \ne 0}\\{{H_{0C}}:{\delta _1} = {\delta _2} = \ldots = {\delta _K} = 0}&{{\rm{\;versus\;}}}&{{H_{aC}}:{\rm{\;at least one\;}}{\delta _ - }i \ne 0}\end{aligned}\)

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

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

And the p-value is the area under the \({F_{N - 1,(N - 1)(N - 2)}}\)curve to the right of the test statistic value \({f_A}\)

\({\rm{\;When testing hypotheses\;}}{H_{0B}}{\rm{\;versus\;}}{H_a}{B_{{\rm{.\;}}}}\), the test statistics value is

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

And the p-value is the area under the \({F_{N - 1,(N - 1)(N - 2)}}\) curve to the right of the test statistic value \({f_B}\)

\({\rm{\;When testing hypotheses\;}}{H_{0C}}{\rm{\;versus\;}}{H_{aC}}\), the test statistics value is

\({f_C} = \frac{{MSC}}{{MSE}}\)

And the p-value is the area under the \({F_{N - 1,(N - 1)(N - 2)}}\) curve to the right of the test statistic value \({f_C}.\)

For factor A

\(\begin{aligned}{*{20}{c}}{{F_{0.01,5,20}} = 4.1 < 20.28 = {f_A}}\\{{P_A} = 0.00 < 0.01 = \alpha }\end{aligned}\)

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

\({\rm{\;at given significance level\;}}\alpha {\rm{. Factor (main effect)\;}}\)A is statiscally significant.

For factor B

\(\begin{aligned}{*{20}{c}}{{F_{0.01,5,20}} = 4.1 > 1.66 = {f_B}}\\{{P_B} = 0.19 > 0.01 = \alpha }\end{aligned}\)

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

\({\rm{\;at given significance level\;}}\alpha {\rm{. Factor (main effect)\;}}\)B is statiscally significant.

For factor C

\(\begin{aligned}{*{20}{c}}{{F_{0.01,5,20}} = 4.1 > 1.59 = {f_C}}\\{{P_C} = 0.21 > 0.01 = \alpha }\end{aligned}\)

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

\({\rm{\;at given significance level\;}}\alpha {\rm{. Factor (main effect)\;}}\)C is not statiscally significant.

Thus,the shelf space does not affect sales

\(\).