Show that Total SS = SST+SSB+SSE .

Question

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Answer 1

Question

Chapter 13, Problem 88SE

To determine

Show that $Total SS = SST+SSB+SSE$ .

Expert Solution & Answer

Explanation of Solution

The formulas for sum of squares are as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯)2SST =b∑i=1k(Y¯i.−Y¯)2SSB =k∑j=1b(Y¯.j−Y¯)2SSE=∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j−Y¯)2$ .

The total sum of squares is calculated is as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯i.+Y¯i.−Y¯.j+Y¯.j−Y¯+Y¯−Y¯)2=∑i=1k∑j=1b((Y¯i.−Y¯)+(Y¯.j−Y¯)+Yij−Y¯i.−Y¯.j+Y¯)2=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]$

Consider

$∑j=1b(Y¯.j−Y¯)=[∑j=1b1k(∑i=1kYij)−b×∑i=1k∑j=1bYijbk] =[1k∑i=1k∑j=1bYij−1k∑i=1k∑j=1bYij]=0$

$∑i=1k(Y¯.i−Y¯)=[∑i=1k1b(∑j=1bYij)−k×∑i=1k∑j=1bYijbk] =[1b∑i=1k∑j=1bYij−1b∑i=1k∑j=1bYij]=0$

$∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)=∑i=1k(Yij−1b∑j=1bYij−1k∑i=1kYij−1bk∑i=1k∑j=1bYij)=∑i=1kYij−1b∑i=1k∑j=1bYij−kk∑i=1kYij+1b∑i=1k∑j=1bYij=0$

Thus, the required TSS is as follows:

$TSS =[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k(Y¯i.−Y¯)∑i=1k(Y¯.j−Y¯)+2∑j=1b(Y¯i.−Y¯)∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)+ 2∑j=1k(Y¯.j−Y¯)∑i=1b(Yij−Y¯i.−Y¯.j+Y¯)]=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+0+0+0=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2 =b∑i=1k(Y¯i.−Y¯)2+k∑j=1k(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2=SST+SSB+SSE$

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Answer 2

Question

Chapter 13, Problem 88SE

To determine

Show that $Total SS = SST+SSB+SSE$ .

Expert Solution & Answer

Explanation of Solution

The formulas for sum of squares are as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯)2SST =b∑i=1k(Y¯i.−Y¯)2SSB =k∑j=1b(Y¯.j−Y¯)2SSE=∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j−Y¯)2$ .

The total sum of squares is calculated is as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯i.+Y¯i.−Y¯.j+Y¯.j−Y¯+Y¯−Y¯)2=∑i=1k∑j=1b((Y¯i.−Y¯)+(Y¯.j−Y¯)+Yij−Y¯i.−Y¯.j+Y¯)2=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]$

Consider

$∑j=1b(Y¯.j−Y¯)=[∑j=1b1k(∑i=1kYij)−b×∑i=1k∑j=1bYijbk] =[1k∑i=1k∑j=1bYij−1k∑i=1k∑j=1bYij]=0$

$∑i=1k(Y¯.i−Y¯)=[∑i=1k1b(∑j=1bYij)−k×∑i=1k∑j=1bYijbk] =[1b∑i=1k∑j=1bYij−1b∑i=1k∑j=1bYij]=0$

$∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)=∑i=1k(Yij−1b∑j=1bYij−1k∑i=1kYij−1bk∑i=1k∑j=1bYij)=∑i=1kYij−1b∑i=1k∑j=1bYij−kk∑i=1kYij+1b∑i=1k∑j=1bYij=0$

Thus, the required TSS is as follows:

$TSS =[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k(Y¯i.−Y¯)∑i=1k(Y¯.j−Y¯)+2∑j=1b(Y¯i.−Y¯)∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)+ 2∑j=1k(Y¯.j−Y¯)∑i=1b(Yij−Y¯i.−Y¯.j+Y¯)]=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+0+0+0=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2 =b∑i=1k(Y¯i.−Y¯)2+k∑j=1k(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2=SST+SSB+SSE$

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Answer 3

Question

Chapter 13, Problem 88SE

To determine

Show that $Total SS = SST+SSB+SSE$ .

Expert Solution & Answer

Explanation of Solution

The formulas for sum of squares are as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯)2SST =b∑i=1k(Y¯i.−Y¯)2SSB =k∑j=1b(Y¯.j−Y¯)2SSE=∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j−Y¯)2$ .

The total sum of squares is calculated is as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯i.+Y¯i.−Y¯.j+Y¯.j−Y¯+Y¯−Y¯)2=∑i=1k∑j=1b((Y¯i.−Y¯)+(Y¯.j−Y¯)+Yij−Y¯i.−Y¯.j+Y¯)2=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]$

Consider

$∑j=1b(Y¯.j−Y¯)=[∑j=1b1k(∑i=1kYij)−b×∑i=1k∑j=1bYijbk] =[1k∑i=1k∑j=1bYij−1k∑i=1k∑j=1bYij]=0$

$∑i=1k(Y¯.i−Y¯)=[∑i=1k1b(∑j=1bYij)−k×∑i=1k∑j=1bYijbk] =[1b∑i=1k∑j=1bYij−1b∑i=1k∑j=1bYij]=0$

$∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)=∑i=1k(Yij−1b∑j=1bYij−1k∑i=1kYij−1bk∑i=1k∑j=1bYij)=∑i=1kYij−1b∑i=1k∑j=1bYij−kk∑i=1kYij+1b∑i=1k∑j=1bYij=0$

Thus, the required TSS is as follows:

$TSS =[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k(Y¯i.−Y¯)∑i=1k(Y¯.j−Y¯)+2∑j=1b(Y¯i.−Y¯)∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)+ 2∑j=1k(Y¯.j−Y¯)∑i=1b(Yij−Y¯i.−Y¯.j+Y¯)]=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+0+0+0=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2 =b∑i=1k(Y¯i.−Y¯)2+k∑j=1k(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2=SST+SSB+SSE$

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Answer 4

Question

Chapter 13, Problem 88SE

To determine

Show that $Total SS = SST+SSB+SSE$ .

Expert Solution & Answer

Explanation of Solution

The formulas for sum of squares are as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯)2SST =b∑i=1k(Y¯i.−Y¯)2SSB =k∑j=1b(Y¯.j−Y¯)2SSE=∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j−Y¯)2$ .

The total sum of squares is calculated is as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯i.+Y¯i.−Y¯.j+Y¯.j−Y¯+Y¯−Y¯)2=∑i=1k∑j=1b((Y¯i.−Y¯)+(Y¯.j−Y¯)+Yij−Y¯i.−Y¯.j+Y¯)2=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]$

Consider

$∑j=1b(Y¯.j−Y¯)=[∑j=1b1k(∑i=1kYij)−b×∑i=1k∑j=1bYijbk] =[1k∑i=1k∑j=1bYij−1k∑i=1k∑j=1bYij]=0$

$∑i=1k(Y¯.i−Y¯)=[∑i=1k1b(∑j=1bYij)−k×∑i=1k∑j=1bYijbk] =[1b∑i=1k∑j=1bYij−1b∑i=1k∑j=1bYij]=0$

$∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)=∑i=1k(Yij−1b∑j=1bYij−1k∑i=1kYij−1bk∑i=1k∑j=1bYij)=∑i=1kYij−1b∑i=1k∑j=1bYij−kk∑i=1kYij+1b∑i=1k∑j=1bYij=0$

Thus, the required TSS is as follows:

$TSS =[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k(Y¯i.−Y¯)∑i=1k(Y¯.j−Y¯)+2∑j=1b(Y¯i.−Y¯)∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)+ 2∑j=1k(Y¯.j−Y¯)∑i=1b(Yij−Y¯i.−Y¯.j+Y¯)]=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+0+0+0=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2 =b∑i=1k(Y¯i.−Y¯)2+k∑j=1k(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2=SST+SSB+SSE$

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Answer 5

Chapter 13, Problem 88SE

To determine

Show that $Total SS = SST+SSB+SSE$ .

Expert Solution & Answer

Explanation of Solution

The formulas for sum of squares are as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯)2SST =b∑i=1k(Y¯i.−Y¯)2SSB =k∑j=1b(Y¯.j−Y¯)2SSE=∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j−Y¯)2$ .

The total sum of squares is calculated is as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯i.+Y¯i.−Y¯.j+Y¯.j−Y¯+Y¯−Y¯)2=∑i=1k∑j=1b((Y¯i.−Y¯)+(Y¯.j−Y¯)+Yij−Y¯i.−Y¯.j+Y¯)2=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]$

Consider

$∑j=1b(Y¯.j−Y¯)=[∑j=1b1k(∑i=1kYij)−b×∑i=1k∑j=1bYijbk] =[1k∑i=1k∑j=1bYij−1k∑i=1k∑j=1bYij]=0$

$∑i=1k(Y¯.i−Y¯)=[∑i=1k1b(∑j=1bYij)−k×∑i=1k∑j=1bYijbk] =[1b∑i=1k∑j=1bYij−1b∑i=1k∑j=1bYij]=0$

$∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)=∑i=1k(Yij−1b∑j=1bYij−1k∑i=1kYij−1bk∑i=1k∑j=1bYij)=∑i=1kYij−1b∑i=1k∑j=1bYij−kk∑i=1kYij+1b∑i=1k∑j=1bYij=0$

Thus, the required TSS is as follows:

$TSS =[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k(Y¯i.−Y¯)∑i=1k(Y¯.j−Y¯)+2∑j=1b(Y¯i.−Y¯)∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)+ 2∑j=1k(Y¯.j−Y¯)∑i=1b(Yij−Y¯i.−Y¯.j+Y¯)]=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+0+0+0=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2 =b∑i=1k(Y¯i.−Y¯)2+k∑j=1k(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2=SST+SSB+SSE$

Answer 6

Chapter 13, Problem 88SE

To determine

Show that $Total SS = SST+SSB+SSE$ .

Expert Solution & Answer

Explanation of Solution

The formulas for sum of squares are as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯)2SST =b∑i=1k(Y¯i.−Y¯)2SSB =k∑j=1b(Y¯.j−Y¯)2SSE=∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j−Y¯)2$ .

The total sum of squares is calculated is as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯i.+Y¯i.−Y¯.j+Y¯.j−Y¯+Y¯−Y¯)2=∑i=1k∑j=1b((Y¯i.−Y¯)+(Y¯.j−Y¯)+Yij−Y¯i.−Y¯.j+Y¯)2=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]$

Consider

$∑j=1b(Y¯.j−Y¯)=[∑j=1b1k(∑i=1kYij)−b×∑i=1k∑j=1bYijbk] =[1k∑i=1k∑j=1bYij−1k∑i=1k∑j=1bYij]=0$

$∑i=1k(Y¯.i−Y¯)=[∑i=1k1b(∑j=1bYij)−k×∑i=1k∑j=1bYijbk] =[1b∑i=1k∑j=1bYij−1b∑i=1k∑j=1bYij]=0$

$∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)=∑i=1k(Yij−1b∑j=1bYij−1k∑i=1kYij−1bk∑i=1k∑j=1bYij)=∑i=1kYij−1b∑i=1k∑j=1bYij−kk∑i=1kYij+1b∑i=1k∑j=1bYij=0$

Thus, the required TSS is as follows:

$TSS =[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k(Y¯i.−Y¯)∑i=1k(Y¯.j−Y¯)+2∑j=1b(Y¯i.−Y¯)∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)+ 2∑j=1k(Y¯.j−Y¯)∑i=1b(Yij−Y¯i.−Y¯.j+Y¯)]=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+0+0+0=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2 =b∑i=1k(Y¯i.−Y¯)2+k∑j=1k(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2=SST+SSB+SSE$

Answer 7

Chapter 13, Problem 88SE

To determine

Show that $Total SS = SST+SSB+SSE$ .

Answer 8

Expert Solution & Answer

Explanation of Solution

The formulas for sum of squares are as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯)2SST =b∑i=1k(Y¯i.−Y¯)2SSB =k∑j=1b(Y¯.j−Y¯)2SSE=∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j−Y¯)2$ .

The total sum of squares is calculated is as follows:

$TSS =∑i=1k∑j=1b(Yij−Y¯i.+Y¯i.−Y¯.j+Y¯.j−Y¯+Y¯−Y¯)2=∑i=1k∑j=1b((Y¯i.−Y¯)+(Y¯.j−Y¯)+Yij−Y¯i.−Y¯.j+Y¯)2=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]$

Consider

$∑j=1b(Y¯.j−Y¯)=[∑j=1b1k(∑i=1kYij)−b×∑i=1k∑j=1bYijbk] =[1k∑i=1k∑j=1bYij−1k∑i=1k∑j=1bYij]=0$

$∑i=1k(Y¯.i−Y¯)=[∑i=1k1b(∑j=1bYij)−k×∑i=1k∑j=1bYijbk] =[1b∑i=1k∑j=1bYij−1b∑i=1k∑j=1bYij]=0$

$∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)=∑i=1k(Yij−1b∑j=1bYij−1k∑i=1kYij−1bk∑i=1k∑j=1bYij)=∑i=1kYij−1b∑i=1k∑j=1bYij−kk∑i=1kYij+1b∑i=1k∑j=1bYij=0$

Thus, the required TSS is as follows:

$TSS =[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k∑j=1b(Y¯i.−Y¯)(Y¯.j−Y¯)+2∑i=1k∑j=1b(Y¯i.−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)+2∑i=1k∑j=1b(Y¯.j−Y¯)(Yij−Y¯i.−Y¯.j+Y¯)]=[∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+2∑i=1k(Y¯i.−Y¯)∑i=1k(Y¯.j−Y¯)+2∑j=1b(Y¯i.−Y¯)∑i=1k(Yij−Y¯i.−Y¯.j+Y¯)+ 2∑j=1k(Y¯.j−Y¯)∑i=1b(Yij−Y¯i.−Y¯.j+Y¯)]=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2+0+0+0=∑i=1k∑j=1b(Y¯i.−Y¯)2+∑i=1k∑j=1b(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2 =b∑i=1k(Y¯i.−Y¯)2+k∑j=1k(Y¯.j−Y¯)2+∑i=1k∑j=1b(Yij−Y¯i.−Y¯.j+Y¯)2=SST+SSB+SSE$