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Chapter 10: Q28E (page 434)

For a single-factor ANOVA with sample sizes\(J\left( {i = 1,2,....I} \right)\,\)show that\(SSTr = \Sigma {J_i}{\left( {{{\mathop X\limits^\_ }_i}\, - \,\mathop X\limits^\_ ...} \right)^2} = \Sigma {J_i}\mathop X\limits^\_ _i^2 - \mathop X\limits^\_ _{..}^2\)where\(n = \Sigma {J_i}\).

Short Answer

Expert verified

use the definition of SSTr to prove the equality.

Step by step solution

01

Step 1: Single-factor ANOVA

Given usual notations, sample sizes

\({J_{i,\,\,}}i = 1,2, \ldots ,l\,\,\,\,\,\,\,I\,\,\,and\,\,\,n = \sum\limits_{i = 1}^I {{J_i},} \)

And the fact that

\(SSTr = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} \, \left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)} ^2} = \sum\limits_{i = 1}^I {\frac{1}{{{J_i}}}\,} x_{_{i\,.}}^2 - \frac{1}{n}x_{ \cdot \cdot }^2\,\)

The following is true

\(\begin{aligned}{l}SSTr = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} \, \left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)} ^2} = \sum\limits_{i = 1}^I {\left( {\sum\limits_{j = 1}^{{J_i}} {\,{{\left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)}^2}} } \right)\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_{i = 1}^I \, {J_i}\,.\,{\left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)^2}\end{aligned}\)

Which proves the first equality

02

Solution for SSTr

Also, the following is true

\(\begin{aligned}{l}SSTr = \sum\limits_{i = 1}^I \, {J_i}\,.\,{\left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_{i = 1}^I \, {J_i}\mathop X\limits^\_ _i^2 - 2.\mathop {X..}\limits^\_ \,\,\sum\limits_{i = 1}^I \, {J_i}{\mathop X\limits^\_ _i}.\, + \mathop X\limits^\_ _{..}^2\,\sum\limits_{i = 1}^I \, {J_i}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_{i = 1}^I \, {J_i}\mathop X\limits^\_ _i^2 - 2.\mathop {X..}\limits^\_ \,X..\, + \,n\,\mathop X\limits^\_ _{..}^2\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_{i = 1}^I \, {J_i}\mathop X\limits^\_ _i^2 - 2n\,\mathop X\limits^\_ _{..}^2 + n\,\mathop X\limits^\_ _{..}^2\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_{i = 1}^I \, {J_i}\mathop X\limits^\_ _i^2 - n\,\mathop X\limits^\_ _{..}^2,\end{aligned}\)

Which proves the second equality. Remember the following things that were used:

\(\begin{aligned}{l}{x_i} = \sum\limits_{j = 1}^{{J_j}} {{x_{i\,j}}} \,;\\{x_{ \cdot \cdot }} = \sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} {\,{x_{i\,j}}} \,} .\end{aligned}\)

Here, the final result is using the definition of SSTr to prove the equality.

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