Question

The data for a college bookstore for 12 semesters is below (X= number of students registered in a course, Y= number of books sold for that course):

X: 36,28,35,39,30,30,31,38,36,38,29,26

Y: 31,29,34,35,29,30,30,38,34,33,29,26

Answer the following questions BY HAND

a) Calculate the least squares regression line for these data.

b) Calculate and explain R^2. Calculate adjusted R^2.

c) Under the simple linear regression model assumptions, what is the unbiased estimator of variance?

d) Calculate the 95% confidence intervals of beta and alpha.

e) Conduct a hypothesis test to verify the number of students registered is a significant predictor for the number of books actually sold for a course at 0.05 significance level.

f) Find the 90% confidence interval for µY |x when x=31, and the prediction interval for a new observation of Y when x=31. Which interval is wider?

Answer 1

	X	Y	X^2	Y^2	XY
	36	31	1296	961	1116
	28	29	784	841	812
	35	34	1225	1156	1190
	39	35	1521	1225	1365
	30	29	900	841	870
	30	30	900	900	900
	31	30	961	900	930
	38	38	1444	1444	1444
	36	34	1296	1156	1224
	38	33	1444	1089	1254
	29	29	841	841	841
	26	26	676	676	676
			0	0	0
			0	0	0
SUM	396	378	13288	12030	12622
n	12		df1	1	k-1
Mean	33	31.5	df2	10	n-k
SSxx	220	Sum(x^2) - ((Sum(x))^2 /n)	SSR	99.56364	slope * Ssxy	MSR	99.56364	SSR/df1
Ssyy	123	Sum(y^2) - ((Sum(y))^2 /n)	SSE	23.43636	SST-SSR	MSE	2.343636	SSE/df2
Ssxy	148	Sum(xy) - (Sum(x)*Sum(y)/n)	SST	123	Ssyy	F	42.48254	MSR/MSE
slope	0.672727	Ssxy/SSxx
intercept	9.3	Mean Y - Mean X * Slope
Se	1.530894	SQRT(SSE/(n-2))
Sb1	0.103213	Se/SQRT(SSxx)
Sb0	3.4346	Se*SQRT(1/n+(Xbar/SSxx))
r	0.8997	Ssxy/SQRT(SSxx*Ssyy)
r^2	0.80946

a)

the least squares regression line for these data

Y = 9.3+0.6727*X

b)

R^2 = 0.8095

80.95% of variation in Y variable is explained by X variable(or regression)

Adj R^2 = 1-[(1-R^2)(N-1)/(n-k)] = 0.7904

If adjusted R^2 nearer to 1 which means regression fits the data

c)

Variance σ^2 = MSE = 2.3436

Variance of β1 = VAR(β1) = σ^2/SSxx = 2.3436/220 = 0.0106 it is nearer to 0

d)

95% CI

Vaiables	Lower 95%	Upper 95%
Intercept	1.647290781	16.95270922
X	0.44275471	0.902699836

95% CI Intercept = (β0 +/- tc * Sb0)

95% C X = (β1 +/- tc * Sb1)

e)

Hypothesis:

H0: β1 = 0

Ha: β1 not = 0

Test:

t stat = b1/Sb1 = 6.5179

P value = 0 < 0.05

the number of students registered is a significant predictor for the number of books actually sold for a course at 0.05 significance level

f)

If x=31

Y = 9.3+0.6727*X = 9.3+0.6727*31 = 30.1545

df = n-k = 10

alpha = 0.1

tc​=1.813 (Use t table)