Question

Consider the data.

xi	1	2	3	4	5
yi	3	7	5	11	14

The estimated regression equation for these data is

ŷ = 0.20 + 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.)

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.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.     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.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

a)

x	y	Ŷ	residual,ei=y-yhat	SSE = (Y-Ŷ)²	SST=(Y-Ȳ)²	SSR= Σ(Ŷ - Ȳ)²
1	3	0.000	3.000	9.00	25	0.00
2	7	0.000	7.000	49.00	1	0.00
3	5	0.000	5.000	25.00	9	0.00
4	11	0.000	11.000	121.00	9	0.00
5	14	0.000	14.000	196.00	36	0.00

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    12.4

SSR = 67.6

SST = 80.0

B)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	15	40	10	80.0	26.00
mean	3.00	8.00	SSxx	SSyy	SSxy

R² =    (Sxy)²/(Sx.Sy) =    0.845

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)

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.9192

Please revert back in case of any doubt.

Please upvote. Thanks in advance.