Can somebody please explain the steps to solve a sum of squares equation in statistics? Thank you.?

Question

I am trying to study for my final, but cannot figure out this method. I don't understand whats behind it. Perhaps someone understands this?! I really appreciate your help.

Answer 1

Yes, I can explain it but you might not want to hear the answer. Which level stats are you doing? Because if it is freshman stats you might just have to use it and not prove it.........

At the heart is solving linear equations: say 2X+3Y=7 and 5X+7Y=-2. You multiply the first by 5 and the second by 2 to get 10X in both: 5{2X+3Y=7}---> 10X+15Y=35 and 2{5X+7Y=-2}--->10X+14Y=-4 then subtract to get Y=39. To solve for Y multiply by 7 and 3....with me so far?

Next up, from the least squares concept [that partial derivatives of sum((m*xi+b-yi)^2,i=1..n) w/r/to m and b have to be zero] you get {wrto m} sum(2*xi*(m*xi+b-yi))=0 which becomes (cancel the extra factor of 2 everywhere)......m*sum(xi^2)+b*sum(xi)-sum(xi*yi)=0 or m*sum(xi^2)+b*sum(xi)=sum(xi*yi). When you differentiate w/r/to b you get sum(2*(m*xi+b-yi))=0 which becomes m*sum(xi)+b*sum(1)-sum(yi) =0 which then becomes m*sum(xi)+b*n=sum(yi)

So our two equations are
m*sum(xi^2)+b*sum(xi) =sum(xi*yi)
and
m*sum(xi)+b*n =sum(yi)

we then do that multiply cancel trick and get the formulas in the book:

m=[n*sum(xi*yi)- sum(xi)*sum(yi)]/ [sum(xi^2)- sum(xi)*sum(xi)]

and b=[sum(xi)*sum(xi*yi)- sum(xi^2)*sum(yi)] /[sum(xi)*sum(xi)- n*sum(xi^2)]

Uggg...hunh?

Answer 2

commencing a sprint late ha? First step is to go over the integer to the different factor of the equation: For 6y - 7 = 34 upload 7 to the two factor or 6y - 7 + 7 = 34 +7 Or 6y = 40-one 2d divide the two factor through variables dissimilar, thus 6 you have 6y/6 = 40-one/6 consequently y = 40-one/6