Unit 7: - Linear Regression and Coefficient of Determination
Guided Exercise 4: Least Squares Line:
The quick Sell car dealership has been using 1-minute spot ads on local tv station. The ads
always occur during the evening hours and advertise different models and price ranges of cars on
the lot that week.
Table 4.8
x
y
6
15
20
31
0
10
14
16
25
28
16
20
28
40
18
25
10
12
8
15
a) Draw a scatter diagram
Scatter diagram and Least-squares line for
table 4-8.
45
40
CARS SOLD
35
30
25
20
15
10
5
0
0
5
10
15
ADS
20
25
30
b) Looking at the equations (3) and (5) pertaining to the least squares line (pages 161 –
162). Two of the quantities that we need to find b are (∑x) and (∑xy). List the others.
Solution:
We also need n, (∑y), (∑x2) and (∑x)2.
c) Completion of table 4.9 (a) on the next page
Solution:
Table 4-9(a)
x
y
x2
xy
6
20
0
14
25
16
28
18
10
8
∑x = 145
15
31
10
16
28
20
40
25
12
15
∑y = 212
36
400
0
196
625
256
784
324
100
64
∑x2 = 2785
90
620
0
224
700
320
1120
450
120
120
∑xy = 3764
d) Computing the sample means x̄ and ȳ
Solution:
x̄ = ∑x = 145
n
10
= 14.5
ȳ = ∑ȳ = 212
n
= 21.2
10
e) Compute a and b for the equation ȳ = a + bx of the least-squares line
Solution:
b = n∑xy – (∑x)(∑y)
n∑x2 – (∑x)2
b = 10(3764) – (145)(212) = 6900
10(2785) – (145)2
6825
= 1.01
a = ȳ - bx
a = 21.2 – 1.01(14.5)
a = 6.56
f) What is the equation of the least-squares a = ȳ - bx
Solution:
Using the values of a and b computed in part (e) or values of and b obtained directly from a
calculator
ȳ = 6.56 + 1.01x
g) Plot the least-squares line on your scatter diagram
Solution:
The least squares line goes through the point (x̄ , ȳ) = (14.5, 21.2). To get another point on
the line, then we select a value of x and compute the corresponding ȳ = 6.56 + 1.01x. For x =
20; we get ȳ = 6.56 + 1.01(20) = 26.8; so the point 920, 26.8) is also on the line. The leastsquares line is shown on the below figure.
Scatter diagram and Least-squares line for
table 4-8.
45
40
CARS SOLD
35
30
25
20
15
10
5
0
0
5
10
15
20
25
30
ADS
h) Read the ȳ value for x = 12 from your graph. Then use the equation of the least squares
line to calculate ȳ when x = 12. How many cars can the manager expect to sel if 12 ads
per week are aired on TV?
Solution:
The graph gives ȳ = 19. From the equation, we get:
ȳ = 6.56 + 1.01x
ȳ = 6.56 +1.01(12)
ȳ = 6.56 + 12.12 = 18.68
i) Interpretation: How reliable do you think the prediction is? Explain (Guided Exercise 5
will show that r = 0.919)
Solution:
The prediction should be fairly reliable. The prediction involves interpolation, and the
scatter diagram shows that the data points are clustered around the least-squares line.
From the next Guided Exercise, we have the information that r is close to 1. Other
variables might affect the value of y for x = 12.
Guided Exercise 5: Coefficient of determination r2
In Guided Exercise 4, we looked at the relationship between x = number of 1-minute spot ads on
TV advertising different models of cars and y = number of cars sold each week by the
sponsoring car dealership.
a) Using the sums found in Guided Exercise 4, compute the sample correlation coefficient r.
n = 10, ∑x = 145, ∑y = 212, ∑x2 = 2785 and ∑xy = 3764. You also need ∑y2 = 5320
r = n ∑xy – (∑x)(∑y)
√n∑x2 – (∑x)2 √n∑y2 – (∑y)2
Solution:
r = 10(3764) – (145)(212)
√10(2785) – (145)2 √10(5320) – (212)2
r = 6900
(82.61)(90.86)
r = 0.919
b) Compute the coefficient of determination r2.
Solution:
r2 = 0.9192
r2 = 0.845
c) Interpretation: What percentage of the variation in the number of car sales can be
explained by the ads and the least-squares line?
Solution:
84.5%
d) Interpretation: What percentage of the variation in the number of car sales is not
explained by the ads and least-squares line?
Solution:
100% - 84.5% = 15.5%
Section 4.2 Problems
#7: Economics: Entry-Level jobs