Question

2) Use the least square regressing to fit a curve on the form: y = a + bx’ suitable for this data X 0 1 2 3 4 5 у 1.0 3.0 15

Please can you solve it on a paper

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Answer #1

We know that for a polynomial equation of the form

The polynomial regression model Yi Bo + Bixi + B2x} + ... + Bmx + εi (i = 1, 2, ..., n) =

Then the model can be written as a system of linear equations as follows

[1 Yi 21 x E1 Y2 1 22 c! m 2 . Bo Bi B2 E2 חוז. 1 13 Y3 + = 3 E3 ... Lyn ] [i 2 27 Lßm Len.

So we have

Bo 0 y ya YS & Ez Ez ey Yo E6 15 20 घ Ez 3 Eu Eig B3 140 250 127 64 125 رمی کے 1= Bot & 32 Bot Bz+E2 15: BoT883 + Ez 20= $+27

HT (45.457. 2:05 22 Bo = -5.457. ß3= 2.6522 so regression equation is gewon by -5.457 +2:0522.23

So the R2 can be computed as follows

R2 = Sum of (ypredicted - ymean)2 / Sum of (y-ymean)2

= 50864/69126/52061.5 = .977

х у 0 1 1 3 2 15 (X - Mean(x)) (Y - Mean(Y)) (X - Mean(X)) * (Y - Mean(Y)) -2.5 -70.5 176.25 -1.5 -68.5 102.75 -0.5 -56.5 28.

Correlation coefficient is calculated as follows

the formula known as the Pearson correlation coefficient, shown below: р = (–a)(*_*)()

So for our data we find that corelation coefficient = 1/5 * 0* 0 = 0

X у 0 1 1 3 2 15 (X - Mean(X)) (X - Mean(x))/stdev(x) -2.5 -1.33630621 -1.5 -0.801783726 -0.5 -0.267261242 0.5 0.267261242 1.

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