5G2 Prof. K. Pearson on Lines and Planes of 



Or, substituting for p from (vii.) and using (ii.) and (iii.) : 

 ItfrxfTxJTxjXu + l 2 (Tx.,(rx u rx. z x u + . . .?«<7ar« + .'•• + lqGx q Xx u rxux (I 



. . (viii.) 



+ ^/« = 0, 



is the type equation. 



Now (vi.) may be written : — 



U = n{l 1 2 a\ + / 2 2 crV + ..". + l q *o* Xq + 2W,a* 1 o* il rj, i » 1 



+ . . . + 2/ ? _i/ 5 0-a,v y _iO".r ( // , .r 7 _i^[ 



(viii.) bis 



Multiplying each type equation by its corresponding l Uf 

 adding together and remembering (iv.), we find 



5s + Q =0> or Q=-U«, 

 n n 



where U»» is the minimum value of U. 



Now let 2 2 be the mean square of the residuals, or 



2 2 = 



_ S(Z 1 ^7 1 + l 2 X. 2 + . . . + IqXq — pf 



Then 



Q 



= -V 



and a physical meaning has been given to Q, ^ — Q/n is the 

 " mean square residual," — /. e., the quantity, the square of 

 which is the mean square of the residuals. 



The type equation (viii.) may now be written : 



llCTx^xuTx^Xu + h^x./Txtirx^Xu + . . • + lu((T x u — £ 2 ) 



+ l Q <rx q crx lt rx q xu = 0. . (ix.) 



We can eliminate the Z's and dividing out row and column 

 of resulting determinant by the corresponding <r, we have: 



1- 



1" 



Tx-lX.-,, rx \ x 3 



1 % X x Xq 



^> l3 



1- 



$2 



r X. z X^ • • . Tx 2 Xq 



r x,x v 



rx q X oJ 



Tx n X 



9^3? 



1 



Xq 



=0,. 



Ix.) 



as a determinants! equation to find Z 2 . We must choose the 

 least root of this equation, for the mean square residual must 



