91Ó°ÊÓ

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 given values are

The SST is

The SSA is

The SSB is

The SSC is

Fundamental Identity

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

By the Fundamental Identity,

\(\begin{aligned}{*{20}{c}}{SSE = SST - (SSA + SSB + SSC)}\\{ = 168.07 - 67.32 - 51.06 - 5.43}\\{ = 44.26}\end{aligned}\)

The degrees of freedom are

\(\begin{aligned}{*{20}{c}}{d{f_T} = {N^2} - 1 = 48}\\{d{f_A} = N - 1 = 6}\\{d{f_B} = N - 1 = 6}\\{d{f_C} = N - 1 = 6}\\{d{f_E} = (N - 1)(N - 2) = (7 - 1)(7 - 2) = 30}\end{aligned}\)

The mean squares are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{d{f_A}}} \cdot SSA = \frac{1}{6} \cdot 67.32 = 11.02}\\{MSB = \frac{1}{{d{f_B}}} \cdot SSB = \frac{1}{6} \cdot 51.06 = 8.51}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{MSC = \frac{1}{{d{f_C}}} \cdot SSC = \frac{1}{6} \cdot 5.43 = 0.91}\\{MSE = \frac{1}{{d{f_E}}} \cdot SSE = \frac{1}{{30}} \cdot 44.26 = 1.48.}\end{aligned}\)

The corresponding f values are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{11.02}}{{1.48}} = 7.6}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{8.51}}{{1.48}} = 5.8}\\{{f_C} = \frac{{MSC}}{{MSE}} = \frac{{0.91}}{{1.48}} = 0.61.}\end{aligned}\)

The P value can be computed using a software.

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 7.6) = 0.00\)

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

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 5.8) = 0.00\)

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

\({P_C} = P\left( {F > {f_C}} \right) = P(F > 0.61) = 0.72\)

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

\({F_{0.05,6,30}} = 2.42\)

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

\(\begin{aligned}{*{20}{r}}{{F_{0.05,6,30}} = 2.42 < 7.6 = {f_A}}\\{{P_A} = 0.00 < 0.05 = \alpha }\end{aligned}\)

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

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

For factor B

\(\begin{aligned}{*{20}{r}}{{F_{0.05,6,30}} = 2.42 < 5.8 = {f_B}}\\{{P_B} = 0.034 < 0.05 = \alpha }\end{aligned}\)

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

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

For factor C

\(\begin{aligned}{*{20}{c}}{{F_{0.05,6,30}} = 2.42 > 0.91 = {f_C}}\\{{P_C} = 0.72}\end{aligned}\)

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

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

Thus, it appears that the main effect C,the heat treatment had no effect on ranging.