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Denote The total sum of squares And treatment sum of squares And Error sum of squares.

Denote with,

\(\begin{aligned}{l}{x_{i.}}\sum\limits_{j = 1}^J {{x_{ij.}}} \\{x_{..}}\sum\limits_{i = 1}^J {\sum\limits_{j = 1}^J {{x_{ij.}}} } \end{aligned}\)

The total sum of squares

\(\left( {SST} \right),\)

And treatment sum of squares

\(\left( {SSTr} \right)\),

And Error sum of squares

\({\rm{ }}\left( {SSE} \right)\)are given by

\(SST = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {{x_{ij}} - \overline x ..} \right)}^2}} = } \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{x^2}_{ij} - \frac{1}{{I\,\,.\,J}}} {x^2}_{..;}} \)

\(SSTr = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {\overline {{x_{i.}}} - \overline x ..} \right)}^2}} = \frac{1}{J}.} \sum\limits_{j = 1}^J {{x^2}_{i.} - \frac{1}{{I\,\,.\,J}}{x^2}_{..;}} \)

\(SSE = \sum\limits_{i = 1}^I {{{\sum\limits_{j = 1}^J {\left( {{x_{ij}} - \overline {{x_{i.}}} } \right)} }^2}} .\)

The mean squares are

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.SSTr;\\MSE = \frac{1}{{I.\left( {J - 1} \right)}}.SSE.\end{aligned}\)

F is ratio of the two mean

\(F = \frac{{MSTr}}{{MSE}}.\)

Compute all value:

\(\begin{aligned}{l}I = 3;\\J = 5;\\\overline {{X_{1.}}} = 10;\\\overline {{X_{2.}}} = 15;\\\overline {{X_{3.}}} = 20.\end{aligned}\)

The degree f freedom of the F Statistic is tested using single Factor ANOVA are

\(\begin{aligned}{l}I - 1 = 2\\I(J - 1) = 12\end{aligned}\)

First compute SSTr

\(\begin{aligned}{l}SSTr = \frac{1}{J}.\sum\limits_{i = 1}^I {{x^2}_{i.} - \frac{1}{{I\,\,.\,J}}{x^2}..} \\ = \frac{1}{J}\left( {{x_{1.}}^2 + {x_{2.}}^2 + {x_{3.}}^2} \right) + \frac{1}{{I\,\,.\,J}}{x^2}..\end{aligned}\)

\(\begin{aligned}{l} = \frac{1}{J} + \left( {{J^2}.{x_{1.}}^2 + {J^2}.{x_{2.}}^2 + {J^2}.{x_{3.}}^2} \right) + \frac{1}{{I\,\,.\,J}}{\left( {{x_{1.}} + {x_{2.}} + {x_{3.}}} \right)^2}\\ = \frac{1}{5} + \left( {{5^2}{{.10}^2} + {5^2}{{.12}^2} + {5^2}{{.20}^2}} \right) + \frac{1}{{3\,.\,5}}\left( {5.10 + 5.15 + 5.20} \right)\end{aligned}\)

\(\begin{aligned}{l} = \frac{1}{5}\left( {2500 + 5625 + 10,000} \right) - 3375\\ = 250\end{aligned}\)

Now, the MSTr is

\(MSTr = \frac{1}{{I - 1}}.SSTr = \frac{1}{2}.250 = 125.\)

The f value is

\(f = \frac{{MSTr}}{{MSE}} = \frac{{125}}{{SSE/12}} = \frac{{1500}}{{SSE}}\)

First condition is

\(f > {F_{0.05,2,12}}\)

Where \({F_{0.05,2,12}} = 3.89\)which can be obtained using the table in the appendix; thus

\(\frac{{1500}}{{SSE}} > 3.89\)

or equally

\(SSE < 385.6\)

However, be careful, this is not the only condition! Remember the following:

The T Method for Identifying Significantly Different 渭i 鈥檚

Find value \({Q_{\alpha .I,I(J - 1)}}\)at the Table A.10. in the appendix of the book for given 伪 .

Compute and a list the sample means in increasing order. Calculate

\(w = {Q_{\alpha ,}}I,{I_{\left( {j - 1} \right)}}.\sqrt {\frac{{MSE}}{J}} \)

And underline pairs of the sample means that differ by less than . The pair of sample which are not underscored by the same line corresponding of population or treatment means that they are significantly different.

From this, another condition can be obtain so when the Turkey's procedure is applied none of the averages are statistically different from one another. From the table

\({Q_{0.05,3,12}} = 3.77\)

hence, another condition for SSE can be obtained from

\(\begin{aligned}{l}w = {Q_{\alpha ,I,I\left( {J - 1} \right)}}.\sqrt {\frac{{MSE}}{J}} \\ = 3.77.\,\sqrt {\frac{{SSE/\left( {12} \right)}}{5}} \\ = 3.77.\,\sqrt {\frac{{SSE}}{{60}}} \\ = 0.4867.\sqrt {SSE.} \end{aligned}\)

It is required that all mean differences are smaller than W it is enough to take maximum value of the mean differences and make it bigger than that value. The maximum difference is

\({\overline x _{3.\,\,\,\,}} - {\overline x _{1.\,\,\,}} = 20 - 10 = 10\)

Therefore, the second condition is

\(w > 10\)

or equally

\(0.4867.\sqrt {SSE} > 10\)

or finally

\(SSE > 422.16\)

The answer can be any real number SSE which satisfies the following

\(422.16 < SSE < 385.6.\)

Hence,

\(422.16 < SSE < 385.6.\)