Question

Complete parts​ (a) through​ (h) for the data below.

x- 40, 50, 60, 70, 80

y-62, 58, 55, 47, 33

B) Find the equation of the line containing the points (50, 58) and (80, 33)

y=__x+(__)

D) By hand, determine the least-squares regression line

The equation of the​ least-squares regression line is given by

ModifyingAbove y with caret equals b 1 x plus b 0y=b1x+b0

where b1 equals r times StartFraction s Subscript y Over s Subscript x EndFractionb1=r•sysx

is the slope of the​ least-squares regression line and

b 0 equals y overbar minus b 1 x overbarb0=y−b1x

is the​ y-intercept of the​ least-squares regression line.

First find the correlation​ coefficient, r.

(f) Compute the sum of the squared residuals for the line found in part​ (b).

The residual is given by the formula below.

Residual=observed y−predicted y=y−y^

(g) Compute the sum of the squared residuals for the​ least-squares regression line found in part​(d).

The residual is given by the formula below.

Residual=observed y−predicted y=y−ModifyingAbove y with caret

Answer 1

Solution:-)

R CAUsers Shivam\Desktoploo.R - R Editor x-c (40, 50, 60, 70, 80) RR Console x-c (40, 50, 60, 70, 80) y-c (62, 58, 55, 47, 33) cor (y, x) ## CORRELATION COEFEICIENT yc (62, 58, 55, 47, 33) co 348 CORRELATION COEFFICIENT auRmary (m (V~x) ) plot (y,x, lwd-2) summary (1m (y-x) ) pred<- c(-2.8, 4.0, sum((y-pred) 2) ## Sum of 0.1 2.9, -4.2) squared residual ntformula = y ~ x) vl- 99.6664 -0.83333*58: v2-99.6664 -0.8333 33 el<- 50-yl : e2<- 80-y2 #Residual (el 2 e22) ## Sum of squared residual Residuals: 2 3 45 RR Graphics: Device 2 (ACTIVE) -2.8 0.1 4.0 2.9 -4.2 Coefricients: Estimate Std. Errort value Pr (> | t | 11.62 0.00137 (Intercept) 92,400 7.950 -0.690 0.129 -5.35 0.01277 30 0 0.001 0.01 0.05 .' 0.1 '1 Sianif. codes: Residual standard error: 4.078 on 3 degrees of freedom Adjusted R-squared: Multiple R-squared: 0.9051, 0.8735 F-statistic: 28.62 on 1 and 3 DF, p-value: 0.01277 pred c(-2.8 sum ((y-pred) ^2) 1 13481.1 0.1, 4.0, 2.9, -4.2) # # Sum of squared residual ##using two points 35 45 55 40 50 60 yl 99.6664 -0.83333 58 el<- 50-yl e2<- 80-y2 #Residual (el 2+ e2^2) 63.12564 y2 99.6664 -0.8333 33 Sum of squared residual 08 0L 09 40 50

A) The equation of line passes through two points, (50,58) and (80,33).

y=mx+b, slope is given by

33 58 -0.83333 m = 80 50

Therefore, we have

-0.83333 x 80+8 b99.6664

The equation of the line will be ,

a y99.6664 0.8333

We have computed the following calculations in R code, where on R.H.S. is output and on L.H.S is R code .

D) The equation of the​ least-squares regression line is given by

y= 92.400-0.690

E) The correlation​ coefficient, r =  -0.9513848

(F) The sum of the squared residuals for the line is given by 63.12564

G) The residual is given by the formula below.

Residual=observed y−predicted

-2.8 ,0.1, 4.0, 2.9, -4.2

(g)The sum of the squared residuals for the line is given by  13481.1.