Question

Consider the data.

xi	1	2	3	4	5
yi	4	7	5	11	15

The estimated regression equation for these data is  ŷ = 0.60 + 2.60x.

(a)Compute SSE, SST, and SSR using equations SSE = Σ(y_i − ŷ_i)², SST = Σ(y_i − y)², and SSR = Σ(ŷ_i − y)².

SSE=SST=SSR=

(b) Compute the coefficient of determination r².

r² =

Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.)

A) The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squares line.

B) The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares line.   

C) The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares line.

D) The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squares line.

(c)Compute the sample correlation coefficient. (Round your answer to three decimal places.)

Answer 1

The given data is:

x	y
1	4
2	7
3	5
4	11
5	15

Using Excel<data<megastat<correlation/regression<regression

Regression Analysis
	r²	0.813
	r	0.901
	Std. Error	2.280
	n	5
	k	1
	Dep. Var.	y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	67.6000	1	67.6000	13.00	.0366
Residual	15.6000	3	5.2000
Total	83.2000	4
Regression output					confidence interval
variables	coefficients	std. error	t (df=3)	p-value	95% lower	95% upper
Intercept	0.6000	2.39	0.25	0.82	-7.01	8.21
x	2.6000	0.7211	3.606	.0366	0.3051	4.8949

a)

SSE= 15.6

SST= 83.2

SSR= 67.6

b)

r^2=0.813

The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares line

c) r= 0.901

Please do the comment for any doubt or clarification. Please upvote if this helps you out. Thank You!