Statistics For Business & Economics solution Chapter 14 – 14 – 1 © 2017 Cengage Learning. All Rights – Studocu

Mục Lục

14 – 1

© 2017 Cengage Learning. All Rights Reserved.

Chapter 14

Simple Linear Regression

Learning Objectives

  1. Understand how regression analysis can be used to develop an equation that estimates

mathematically how two variables are related.

  1. Understand the differences between the regression model, the regression equation, and the estimated

regression equation.

  1. Know how to fit an estimated regression equation to a set of sample data based upon the least-

squares method.

  1. Be able to determine how good a fit is provided by the estimated regression equation and compute

the sample correlation coefficient from the regression analysis output.

  1. Understand the assumptions necessary for statistical inference and be able to test for a significant

relationship.

  1. Know how to develop confidence interval estimates of y given a specific value of x in both the case

of a mean value of y and an individual value of y.

  1. Learn how to use a residual plot to make a judgement as to the validity of the regression

assumptions.

  1. Know the definition of the following terms:

independent and dependent variable

simple linear regression

regression model

regression equation and estimated regression equation

scatter diagram

coefficient of determination

standard error of the estimate

confidence interval

prediction interval

residual plot

Chapter 14

14 – 2

© 2017 Cengage Learning. All Rights Reserved.

Solutions:

1 a.

b. There appears to be a positive linear relationship between x and y.

c. Many different straight lines can be drawn to provide a linear approximation of the

relationship between x and y; in part (d) we will determine the equation of a straight line

that “best” represents the relationship according to the least squares criterion.

d.

15 40
3 8
55

ii
xy
xy
nn


 

2
()( )26 ()1 0
ii i
    xxyy xx

1

2

()( ) 26
2.
() 10

ii

i

xxyy
b
xx

 



bybx
01

 8263 02(.)().

yx ˆ0 2.

e. ˆ y  0 2(4) 10.

0
2
4
6
8
10
12
14
16
0123456

y

x

Chapter 14

14 – 4

© 2017 Cengage Learning. All Rights Reserved.

  1. a.

b.

50 83
10 16.
55

ii
xy
xy
nn


 

2
()( )171 ()1 90
ii i
    xxyy xx

1 2

()( ) 171
0.
() 190

ii

i

xxyy
b
xx

 



01
bybx   16 (0)(10)7.

yx ˆ7 0.

c. ˆ y  7 0(6) 13

0
5
10
15
20
25
30
0 5 10 15 20 25

y

x

Simple Linear Regression

14 – 5

© 2017 Cengage Learning. All Rights Reserved.

  1. a.

b. There appears to be a positive linear relationship between the percentage of women working in the

five companies ( x ) and the percentage of management jobs held by women in that company ( y )

c. Many different straight lines can be drawn to provide a linear approximation of the

relationship between x and y ; in part (d) we will determine the equation of a straight line

that “best” represents the relationship according to the least squares criterion.

d.

300 215
60 43
55

xyii
xy
nn


 

2
   ( xxyyii )( ) 624 ( xxi ) 480

1 2

()( ) 624
1.
( ) 480

ii

i

xxyy
b
xx

 



bybx 01   43 1(60) 35

yx ˆ 35 1.

e. yx ˆ   35 1 35 1(60)43%

0
10
20
30
40
50
60
70
40 45 50 55 60 65 70 75

% Management

% Working

Simple Linear Regression

14 – 7

© 2017 Cengage Learning. All Rights Reserved.

  1. a.

b. The scatter diagram indicates a positive linear relationship between x = average number of passing

yards per attempt and y = the percentage of games won by the team.

c. xxn   ii / 680 / 10 6  yyn  / 464 / 1046.

2
( )( ) 121 ( ) 7.
ii i
  xxyy   xx

1 2

( )( ) 121.
17.
( ) 7.

ii

i

xxyy
b
xx

 



bybx 01   46 (17)(6)70.

yx ˆ70 17

d. The slope of the estimated regression line is approximately 17. So, for every increase of one yard

in the average number of passes per attempt, the percentage of games won by the team increases by

17%.

e. With an average number of passing yards per attempt of 6, the predicted percentage of games won

is y ˆ= -70 + 17(6) = 36%. With a record of 7 wins and 9 loses, the percentage of wins that

the Kansas City Chiefs won is 43 or approximately 44%. Considering the small data size, the

prediction made using the estimated regression equation is not too bad.

0
10
20
30
40
50
60
70
80
90
456789

Win%

Yds/Att

Chapter 14

14 – 8

© 2017 Cengage Learning. All Rights Reserved.

  1. a.

b. Let x = years of experience and y = annual sales ($1000s)

70 1080
7 108
10 10

xyii
xy
nn


  

2
   () xxyyii ( )568 ()1 xxi 42

12

()( ) 568
4
() 142

ii

i

xxyy
b
xx

 



bybx 01     108 ()()4 7 80

yx 80 4

c. yx  80 4 80 4 9() 116 or $116,

50
60
70
80
90
100
110
120
130
140
150
02468101214

Annual Sales ($1000s)

Years of Experience

Chapter 14

14 – 10

© 2017 Cengage Learning. All Rights Reserved.

  1. a.

b. The scatter diagram indicates a positive linear relationship between x = cars in service (1000s) and y

= annual revenue ($millions).

c. / 43 / 6 7 / 462 / 6 77
xxn   ii  yyn   

2
( )( ) 734 ( ) 56.
ii i
  xxyy   xx

1

2

( )( ) 734.
12.
( ) 56.

ii

i

xxyy
b
xx

 



bybx 01   77 (12)(7)17.

yx ˆ17 12

d. For every additional 1000 cars placed in service annual revenue will increase by 12 ($millions)

or $12,966,000. Therefor every additional car placed in service will increase annual revenue by

$12,966.

e. yx ˆ17 12 17 12(11) 125 

A prediction of annual revenue for Fox Rent A Car is approximately $126 million.

0
20
40
60
80
100
120
140
160
0 2 4 6 8 10 12 14

Annual Revenue ($millions)

Cars in Service (1000s)

Simple Linear Regression

14 – 11

© 2017 Cengage Learning. All Rights Reserved.

10. a.

b. The scatter diagram indicates a positive linear relationship between x = percentage increase in the

stock price and y = percentage gain in options value. In other words, options values increase as stock

prices increase.

c. / 2939 / 10 293 / 6301 / 10 630.
xxn   ii  yyn   

2
 ( xxyyii )( ) 314,501 (  xxi ) 115,842.

12

( )( ) 314,501.
2.
( ) 115,842.

ii

i

xxyy
b
xx

 



01
bybx   630 (2)(293)167.

yx ˆ167 2

d. The slope of the estimated regression line is approximately 2. So, for every percentage increase in

the price of the stock the options value increases by 2%.

e. The rewards for the CEO do appear to be based upon performance increases in the stock value.

While the rewards may seem excessive, the executive is being rewarded for his/her role in increasing

the value of the company. This is why such compensation schemes are devised for CEOs by boards

of directors. A compensation scheme where an executive got a big salary increase when the

company stock went down would be bad. And, if the stock price for a company had gone down

during the periods in question, the value of the CEOs options would also go down.

0
200
400
600
800
1000
1200
1400
0 100 200 300 400 500 600

% Gain in Options Value

% Increase in Stock Price

Simple Linear Regression

14 – 13

© 2017 Cengage Learning. All Rights Reserved.

12. a.

b. The scatter diagram indicates a positive linear relationship between x = hotel room rate and the

amount spent on entertainment.

c. / 945 / 9 105 / 1134 / 9 126
xxn   ii  yyn   

2
   ( xxyyii )( ) 4237 ( xxi ) 4100

12

( )( ) 4237
1.
( ) 4100

ii

i

xxyy
b
xx

 



01
bybx   126 (1)(105) 17

yx ˆ17 1.

d. With a value of x = $128, the predicted value of y for Chicago is

yx ˆ 17 1 17 1(128) 150

Note: In The Wall Street Journal article the entertainment expense for Chicago was $146. Thus, the

estimated regression equation provided a good estimate of entertainment expenses for Chicago.

70
90
110
130
150
170
190
70 90 110 130 150 170

Entertainment ($)

Hotel Room Rate ($)

Chapter 14

14 – 14

© 2017 Cengage Learning. All Rights Reserved.

  1. a.

b. Let x = adjusted gross income and y = reasonable amount of itemized deductions

399 97.
57 13.
77

xyii
xy
nn


  

2
 ( xxyyii )( ) 1233 (  xxi ) 7648

12

()( )1233.
0.
() 7648

ii

i

xxyy
b
xx

 



bybx 01  13 (0)(57) 4.

yx 468 016..

c. yx  4 68 0 16.. ..(4 68 0 1652.). 5 13 08or approximately $13,080.

The agent’s request for an audit appears to be justified.

0.
5.
10.
15.
20.
25.
30.
0 20 40 60 80 100 120 140.

Reasonable Amount of Itemized

Deductions ($1000s)

Adjusted Gross Income ($1000s)

Chapter 14

14 – 16

© 2017 Cengage Learning. All Rights Reserved.

  1. a. The estimated regression equation and the mean for the dependent variable are:

ˆ yxi  68 3 y 35

The sum of squares due to error and the total sum of squares are

22
SSE ( ˆ) 230 SST ( ) 1850
   yyii    yyi

Thus, SSR = SST – SSE = 1850 – 230 = 1620

b. r
2
= SSR/SST = 1620/1850 =.

The least squares line provided an excellent fit; 87% of the variability in y has been explained by

the estimated regression equation.

c. rxy .876 .

Note: the sign for r is negative because the slope of the estimated regression equation is negative.

( b 1 = -3)

  1. The estimated regression equation and the mean for the dependent variable are:

ˆ yxi 7 .9 y 16.

The sum of squares due to error and the total sum of squares are

22
SSE ( ˆ) 127 SST ( ) 281.
ii i
   yy    yy

Thus, SSR = SST – SSE = 281 – 127 = 153.

r

2
= SSR/SST = 153.9/281 =.

We see that 54% of the variability in y has been explained by the least squares line.

.547.

xy
r 

  1. a. / 600 / 6 100 / 330 / 6 55
    ii
    xxn    yyn   

22
SST = ( ) 1800 SSE = ( ˆ) 287.
  yyii   yyi

SSR = SST – SSR = 1800 – 287 = 1512.

b.

2 SSR 1512.
.
SST 1800

r  

c.

2
rr  .84.

Simple Linear Regression

14 – 17

© 2017 Cengage Learning. All Rights Reserved.

  1. a. The estimated regression equation and the mean for the dependent variable are:

y ˆ= 80 + 4 x y = 108

The sum of squares due to error and the total sum of squares are

22
SSE ( ˆ) 170 SST ( ) 2442
   yyii    yyi

Thus, SSR = SST – SSE = 2442 – 170 = 2272

b. r

2
= SSR/SST = 2272/2442 =.

We see that 93% of the variability in y has been explained by the least squares line.

c. .93.
xy
r 

  1. a. / 160 / 10 16 / 55,500 / 10 5550
    ii
    xxn    yyn   

2
 ( xxyyii )( ) 31, 284 (  xxi ) 21.

12

()( )31,
1439
() 2 1.

ii

i

xxyy
b
xx

  



01
bybx   5550 ( 1439)(16)28,

yx ˆ28, 574 1439

b. SST = 52,120,800 SSE = 7,102,922.

SSR = SST – SSR = 52,120,800 – 7,102,922 = 45,017,

2
r = SSR/SST = 45,017,877/52,120,800 =.

The estimated regression equation provided a very good fit.

c. yx ˆ 28,574 1439 28,574 1439(15) 6989

Thus, an estimate of the price for a bike that weighs 15 pounds is $6989.

  1. a.
3450 33, 700
575 5616.
66

ii
xy
xy
nn


   

2
 ( xxyyii )( ) 712, 500 (  xxi ) 93, 750

12

()( )712, 500
7.
() 93, 750

ii

i

xxyy
b
xx

 



bybx 01  5616 (7)(575) 1246 

yx 1246 67 7 6..

Simple Linear Regression

14 – 19

© 2017 Cengage Learning. All Rights Reserved.

e. MSR = SSR / 1 = 67.

F = MSR / MSE = 67 / 4 = 16.

Using F table (1 degree of freedom numerator and 3 denominator), p -value is between .025 and.

Using Excel or Minitab, the p -value corresponding to F = 16 is .0272.

Because p -value, we reject H 0 :  1 = 0

Source

of Variation

Sum

of Squares

Degrees

of Freedom

Mean

Square
F
p -value

Regression 67 1 67 16.

Error 12 3 4.

Total 80 4

  1. a. s
    2
    = MSE = SSE/( n – 2) = 230/3 = 76.

b. s  MSE 76 8.

c.

2
()1 80
i
  xx

12

8.
0.
() 180

b

i

s
s

xx




d.

1

1 3
4.
.

b

b
t
s

 

Using t table (3 degrees of freedom), area in tail is less than .01; p -value is less than.

Using Excel or Minitab, the p -value corresponding to t = -4 is .0193.

Because p -value, we reject H 0 :  1 = 0

e. MSR = SSR/1 = 1620

F = MSR/MSE = 1620/76 = 21.

Using F table (1 degree of freedom numerator and 3 denominator), p -value is less than.

Using Excel or Minitab, the p -value corresponding to F = 21 is .0193.

Because p -value, we reject H 0 :  1 = 0

Source

of Variation

Sum

of Squares

Degrees

of Freedom

Mean

Square
F
p -value

Regression 1620 1 1620 21.

Error 230 3 76.

Total 1850 4

Chapter 14

14 – 20

© 2017 Cengage Learning. All Rights Reserved.

  1. a. s

2
= MSE = SSE/( n – 2) = 127/3 = 42.

s  MSE 42 6.

b.

2
 ()1 xxi 90

12

6.
0.
() 190

b

i

s
s

xx




1

1

.
1.
.

b

b
t
s

 

Using t table (3 degrees of freedom), area in tail is between .05 and.

p -value is between .10 and.

Using Excel or Minitab, the p -value corresponding to t = 1 is .1530.

Because p -value >, we cannot reject H 0 :  1 = 0; x and y do not appear to be related.

c. MSR = SSR/1 = 153 /1 = 153.

F = MSR/MSE = 153.9/42 = 3.

Using F table (1 degree of freedom numerator and 3 denominator), p -value is greater than.

Using Excel or Minitab, the p -value corresponding to F = 3 is .1530.

Because p -value >, we cannot reject H 0 :  1 = 0; x and y do not appear to be related.
  1. a. In the statement of exercise 18, y ˆ= 23 + .318 x

In solving exercise 18, we found SSE = 287.

2
sn MSE = SSE/( -2) =287 / 4 71.

s  MSE 71 8.

2
()1 xx 4,

12

8.
.
()14, 950

b

s
s

xx



 

1

1

.
4.
.

b

b
t
s

 

Using t table (4 degrees of freedom), area in tail is between .005 and.

p -value is between .01 and.