Question

a linear regression was conducted on the following data points.

A(1, 11) B(3, 7) C(5, 2) D(7, 1) E(9, -2)

The residual for which point has the largest absolute value?

Answer 1

looks like you either need to develop the linear regression equation, or use the one they give you if they gave you one, to determine the difference between a stated y, and the lr y value

Answer 2

@ashleybvb

Answer 3

\[Y=\frac{n\sum xy-\sum x\sum y}{n\sum xx-\sum x\sum x}(X-\bar X)+\bar Y\]

Answer 4

I don't understand that equation because I cant afford the calculator

Answer 5

its 5 points ... which can be pretty much done by hand i spose; if organised in the right way

Answer 6

x*x x y x*y
1^2 (1, 11) 1(11)
3^2 (3, 7) 3(7)
5^2 (5, 2) 5(2)
7^2 (7, 1) 7(1)
9^2 (9, -2) 9(-2)
---------------------
totals of columns

Answer 7

when we know the totals of each column, we can then construct the linear regression equation from them

Answer 8

i think its A

Answer 9

so you basically multiply each one and the one that is the highest is the answer?

Answer 10

if they do not give you an equation to work with, then you need to construct one from points

Answer 11

there is no "simple" way to go about it in my view

Answer 12

x*x x y x*y
1^2 (1, 11) 1(11)
3^2 (3, 7) 3(7)
5^2 (5, 2) 5(2)
7^2 (7, 1) 7(1)
9^2 (9, -2) 9(-2)
---------------------
165 25 19 31

n = 5 points, avgX = 25/5 = 5
avgY = 19/5

\[Y=\frac{n\sum xy-\sum x\sum y}{n\sum xx-\sum x\sum x}(X-\bar X)+\bar Y\]
\[Y=\frac{5(31)-(25)(19)}{5(165)-(25)(25)}(X-5)+\frac{19}{5}\]i believe would be the regression equation

Answer 13

to find the difference between Y and a given y point, let X=x and subtract y from both sides

\[Y-y=\frac{5(31)-(25)(19)}{5(165)-(25)(25)}(x-5)+\frac{19}{5}-y\]
and test out each x,y point

Answer 14

10.2 - 11
.8

7 - 7
0

3.8 - 2
1.8

.6 - 1
.4

-2.6 + 2
.6

Answer 15

@amistre64 so the answer would be c(5, 2)

Answer 16

assuming i didnt mess it up along the way, i would say C is the one with the furthest gap, yes

Answer 17

thank you