Statistics for Business and Economics chapter 14 Simple Linear Regression

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.

2.

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

3.

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

4.

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.

5.

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

6.

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.

7.

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

8.

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

14 ­ 1

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Chapter 14
Solutions:

1

a.
16
14
12
y

10
8
6
4
2
0
0

1

2

3

4

5

6

x

 

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.

x

xi 15

3
n
5

y

( xi  x )( yi  y )  26
b1 

yi 40

8

n
5
( xi  x ) 2  10

( xi  x )( yi  y ) 26
 2.6
( xi  x ) 2
10

b0  y  b1 x 8  (2.6)(3) 0.2
y�0.2  2.6 x
e.

y�0.2  2.6(4) 10.6

14 ­ 2

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Simple Linear Regression
2.

a.

b.

There appears to be a negative 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.

x

xi 55

 11
n
5

y

( xi  x )( yi  y )  540
b1 

yi 175

 35
n
5
( xi  x ) 2  180

( xi  x )( yi  y ) 540


 3
( xi  x ) 2
180

b0  y  b1 x  35  (3)(11)  68
yˆ  68  3 x
e.

yˆ  68  3(10)  38

14 ­ 3

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Chapter 14
3.

a.

b.

x

xi 50

 10
n
5

y

( xi  x )( yi  y )  171
b1 

yi 83

 16.6
n
5
( xi  x ) 2  190

( xi  x )( yi  y ) 171

 0.9
( xi  x ) 2
190

b0  y  b1 x  16.6  (0.9)(10)  7.6
yˆ  7.6  0.9 x
c.

yˆ  7.6  0.9(6)  13

14 ­ 4

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Simple Linear Regression
4. a.

135
130

Weight

125
120
115
110
105
100
61

62

63

64

65

66

67

68

69

Height

b.

There appears to be a positive linear relationship between x = height and y = weight.

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.

x

xi 325

 65
n
5

( xi  x )( yi  y )  110
b1 

y

yi 585

 117
n
5

( xi  x ) 2  20

( xi  x )( yi  y ) 110

5.5
( xi  x ) 2
20

b0  y  b1 x 117  (5.5)(65)  240.5
y  240.5  55
. x
e.

y  240.5  55
. x  240.5  55
. (63) 106 pounds

14 ­ 5

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Chapter 14

5.

a.

b.

There appears to be a positive relationship between price and rating. The sign that says “Quality: 
You Get What You Pay For” does fairly reflect the price­quality relationship for ellipticals.

c.

Let x = price ($) and y = rating.
x

xi 1500

 1875
n
8

y

( xi  x )( yi  y )  68,900
b1 

yi 592

 74
n
8

( xi  x )2  8,155,000

( xi  x )( yi  y )
68,900

 .008449
( xi  x ) 2
8,155,000

b0  y  b1 x  74  (.008449)(1875)  58.158
yˆ  58.158  .008449 x
d.

yˆ  58.158  .008449 x  58.158  .008449(1500)  70.83 or approximately 71

14 ­ 6

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Simple Linear Regression
6.

a.

b.

There appears to be a negative linear relationship between x = miles and y = sales price.
If the car has higher miles, the sales price tends to be lower.

c.

x

xi 874

 87.4
n
10

y

( xi  x )( yi  y )  135.66
b1 

yi 66.4

 6.64
n
10

( xi  x )2  5152.4

( xi  x )( yi  y ) 135.66

 .02633
( xi  x ) 2
5152.40

b0  y  b1 x  6.64  (.02633)(87.4)  8.9412
yˆ  8.9412  .02633 x
d.

The slope of the estimated regression equation is -.02633. Thus, a one unit increase in the value of
x will result in a decrease in the estimated value of y equal to .02633. Because the data were
recorded in thousands, every additional 1000 miles on the car’s odometer will result in a $26.33
decrease in the estimated price.

e.

yˆ  8.9412  .02633(100)  6.3 or $6300

14 ­ 7

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Chapter 14
a.
150
140
130
Annual Sales ($1000s)

7.

120
110

100
90
80
70
60
50
0

2

4

6

8

10

12

14

Years of Experience

b.

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

yi 1080


 108
n
10

( xi  x )( yi  y )  568

( xi  x ) 2  142

x

xi 70

7
n
10

b1 

( xi  x )( yi  y ) 568

4
( xi  x ) 2
142

b0  y  b1 x 108  (4)(7) 80
y 80  4 x
c.

y 80  4 x 80  4(9) 116 or $116,000

14 ­ 8

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Simple Linear Regression
8.

a.

b.

The scatter diagram and the slope of the estimated regression equation indicate a negative linear
relationship between x = temperature rating and y = price. Thus, it appears that sleeping bags with
a lower temperature rating cost more than sleeping bags with a higher temperature rating. In other
words, it costs more to stay warmer.

c.

x  xi / n  209 /11  19
( xi  x )( yi  y )  10,090
b1 

y  yi / n  2849 /11  259
( xi  x ) 2  1912

( xi  x )( yi  y ) 10, 090


 5.2772
( xi  x ) 2
1912

b0  y  b1 x  259  (5.2772)(19)  359.2668
yˆ  359.2668  5.2772 x
d.

yˆ  359.2668  5.2772 x  359.2668  5.2772(20)  253.72
Thus, the estimate of the price of sleeping bag with a temperature rating of 20 is approximately
$254.

14 ­ 9

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Chapter 14
9.

a.

b.

There appears to be a positive linear relationship between x = price and y = score.

c.

x

xi 2638

 263.8
n
10

y

( xi  x )( yi  y )  14,601.40
b1 

yi 672

 67.2
n
10

( xi  x ) 2  258,695.60

( xi  x )( yi  y ) 14,601.40

 .05644
( xi  x )2
258,695.60

b0  y  b1 x  67.2  (.05644)(263.8)  52.311
yˆ  52.311  .05644 x
d.

The slope is .05644. For a $100 higher price, the score can be expected to increase
= 5.644, or about 6 points.

e.

yˆ  52.311  .05644(225)  65

14 ­ 10

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100(.05644)


Simple Linear Regression
10. a.

b.

There appears to be a positive linear relationship between x = age and y = salary.

c.

x

xi 885

 59
n

15

y

yi 30,939

 2062.6
n
15

( xi  x )( yi  y )  175, 265
b1 

( xi  x ) 2  1174

( xi  x )( yi  y ) 175, 265

 149.2888
( xi  x ) 2
1174

b0  y  b1 x  2062.6  (149.2888)(59)  6745.44
yˆ  6745.44  149.29 x
d.

yˆ  6745.44  149.29 x  6745.44  149.29(72)  4003 or $4,003,000

11. a.
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Chapter 14

b.

There appears to be a positive linear relationship between x = price and y = road-test score.

c.

x

xi 339.6

 28.3
n
12

y

( xi  x )( yi  y )  309.90
b1 

yi 930

 77.5
n
12

( xi  x ) 2  346.38

( xi  x )( yi  y ) 309.90

 .8947
( xi  x ) 2
346.38

b0  y  b1 x  77.5  (.8947)(28.3)  52.18
yˆ  52.18  .8947 x
d. The slope is .8947. A sporty car that has a ten thousand dollar higher price can be expected to
have a 10(.8947) = 8.947, or approximately a 9 point higher road-test score.
e.

yˆ  52.18  .8947(36.7)  85

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Simple Linear Regression
12. a.

b.

The scatter diagram indicates a positive linear relationship between x = weight and y = price.
Thus, it appears that PWC’s that weigh more have a higher price.

c.

x  xi / n  7730 /10  773

y  yi / n  92, 200 /10  9220

( xi  x )( yi  y )  332, 400

( xi  x )2  14,810

b1 

( xi  x )( yi  y ) 332, 400

 22.4443
( xi  x ) 2
14,810

b0  y  b1 x  9220  (22.4443)(773)  8129.4439
yˆ  8129.4439  22.4443x
d.

yˆ  8129.4439  22.4443 x  8129.4439  22.4443(750)  8703.78
Thus, the estimate of the price of Jet Ski with a weight of 750 pounds is approximately $8704.

e.

No. The relationship between weight and price is not deterministic.

f.

The weight of the Kawasaki SX-R 800 is so far below the lowest weight for the data used to
develop the estimated regression equation that we would not recommend using the estimated
regression equation to predict the price for this model.

13. a.
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Reasonable Amount of Itemized
Deductions ($1000s)

Chapter 14

30.0
25.0
20.0
15.0
10.0
5.0
0.0
0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

Adjusted Gross Income ($1000s)
b.

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

xi 399

 57
n
7

y

( xi  x )( yi  y )  1233.7
b1 

yi 97.1


 13.8714
n
7
( xi  x ) 2  7648

( xi  x )( yi  y ) 1233.7

 0.1613
( xi  x ) 2
7648

b0  y  b1 x  13.8714  (0.1613)(57)  4.6773
y 4.68  016
. x
c.

y 4.68  016
. x 4.68  016
. (52.5) 13.08 or approximately $13,080. 
The agent's request for an audit appears to be justified.

14 ­ 14

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Simple Linear Regression
14. a.

b.

There appears to be a positive linear relationship between x = features rating and y = PCW World
Rating.

c.

x

xi 784

 78.4
n
10

y

( xi  x )( yi  y )  147.20
b1 

yi 777

 77.7
n
10

( xi  x )2  284.40

( xi  x )( yi  y ) 147.20


 .51758
( xi  x ) 2
284.40

b0  y  b1 x  77.7  (.51758)(78.4)  37.1217
yˆ  37.1217  .51758 x
d.
15. a.

yˆ  37.1217  .51758(70)  73.35 or 73
The estimated regression equation and the mean for the dependent variable are:
yi 0.2  2.6 xi

y 8

The sum of squares due to error and the total sum of squares are
SSE  ( yi  yi ) 2 12.40

SST  ( yi  y ) 2 80

Thus,  SSR = SST ­ SSE = 80 ­ 12.4 = 67.6
b.

r2 = SSR/SST = 67.6/80 = .845
The least squares line provided a very good fit; 84.5% of the variability in y has been explained by 
the least squares line.

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Chapter 14

c.    rxy  .845  .9192
16. a.

The estimated regression equation and the mean for the dependent variable are:
yˆi  68  3 x

y  35

The sum of squares due to error and the total sum of squares are
SSE  �( yi  yˆi ) 2  230

SST  �( yi  y ) 2  1850

Thus, SSR = SST ­ SSE = 1850 ­ 230 = 1620
b.

r2 = SSR/SST = 1620/1850 = .876
The least squares line provided an excellent fit; 87.6% of the variability in y has been explained by 
the estimated regression equation.

c.

rxy  .876  .936
Note: the sign for r is negative because the slope of the estimated regression equation is negative.
(b1 = ­3)

17.

The estimated regression equation and the mean for the dependent variable are:
yˆi  7.6  .9 x

y  16.6

The sum of squares due to error and the total sum of squares are
SSE  �( yi  yˆi ) 2  127.3

SST  �( yi  y ) 2  281.2

Thus, SSR = SST ­ SSE = 281.2 – 127.3 = 153.9
r2 = SSR/SST = 153.9/281.2 = .547
We see that 54.7% of the variability in y has been explained by the least squares line.
rxy  .547  .740
18. a.

The estimated regression equation and the mean for the dependent variable are:
yˆ 1790.5  581.1x

y 3650

The sum of squares due to error and the total sum of squares are
2
SSE ( yi  y�
i ) 85,135.14

SST  ( yi  y ) 2 335,000

Thus,  SSR = SST ­ SSE = 335,000 ­ 85,135.14 = 249,864.86
b.

r2 = SSR/SST = 249,864.86/335,000 = .746
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Simple Linear Regression

c.
19. a.

We see that 74.6% of the variability in y has been explained by the least squares line.
rxy  .746  .8637
The estimated regression equation and the mean for the dependent variable are:
yˆ = 80 + 4x

y = 108

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

SSE  �( yi  yˆ i ) 2  170

SST  �( yi  y ) 2  2442

Thus,  SSR = SST ­ SSE = 2442 ­ 170 = 2272

b.

r2 = SSR/SST = 2272/2442 = .93
We see that 93% of the variability in y has been explained by the least squares line.

c.

rxy  .93  .96

20. a.

The scatter diagram indicates a positive linear relationship between x = price and y = score.
x  xi / n  29, 600 /10  2960
( xi  x )( yi  y )  34,840
b1 

y  yi / n  496 /10  49.6

( xi  x ) 2  2, 744, 000

( xi  x )( yi  y )
34,840

 .012697
2
( xi  x )
2, 744,000

b0  y  b1 x  49.6  (.012697)(2960)  12.0169
14 ­ 17

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 14

b.

yˆ  12.0169  .0127 x
The sum of squares due to error and the total sum of squares are
SSE  �( yi  yˆi ) 2  540.0446

SST  �( yi  y ) 2  982.4

Thus, SSR = SST ­ SSE = 982.4 – 540.0446 = 442.3554
r2 = SSR/SST = 442.3554/982.4 = .4503
The fit provided by the estimated regression equation is not that good; only 45.03% of the 
variability in y has been explained by the least squares line. 
c.

yˆ  12.0169  .0127 x  12.0169  .0127(3200)  52.66
The estimate of the overall score for a 42-inch plasma television is approximately 53.

21. a.

x

xi 3450


 575
n
6

y

( xi  x )( yi  y )  712,500
b1 

yi 33, 700

 5616.67
n
6
( xi  x ) 2  93, 750

( xi  x )( yi  y ) 712,500

7.6
( xi  x ) 2
93, 750

b0  y  b1 x 5616.67  (7.6)(575) 1246.67
y 1246.67  7.6 x
b.

$7.60

c.

The sum of squares due to error and the total sum of squares are:
2
SSE  ( yi  y�
i )  233,333.33

SST  ( yi  y ) 2 5, 648,333.33

Thus,  SSR = SST ­ SSE = 5,648,333.33 ­ 233,333.33 = 5,415,000
r2 = SSR/SST = 5,415,000/5,648,333.33 = .9587
We see that 95.87% of the variability in y has been explained by the estimated regression equation.
d.
22. a.

y 1246.67  7.6 x 1246.67  7.6(500) $5046.67
y = 74

SSE = 173.88

The total sum of squares is

SST  �( yi  y ) 2  756

14 ­ 18

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Simple Linear Regression
Thus, SSR = SST ­ SSE = 756 – 173.88 = 582.12

r2 = SSR/SST = 582.12/756 = .77
b.

The estimated regression equation provided a good fit because 77% of the variability in y has been 
explained by the least squares line.

c.

rxy  .77  .88
This reflects a strong positive linear relationship between price and rating.

23. a.

s2  =  MSE  =  SSE / (n ­ 2)  =  12.4 / 3  =  4.133

b.

s  MSE  4.133 2.033

c.

( xi  x ) 2 10
sb1 

d.

t

s
( xi  x )

2



2.033
10

0.643

b1
2.6

 4.044
sb1 .643

Using t table (3 degrees of freedom), area in tail is between .01 and .025
p­value is between .02 and .05
Using Excel or Minitab, the p­value corresponding to t = 4.04 is .0272.
Because p­value  , we reject H0: 1 = 0
e.

MSR  = SSR / 1  =  67.6
F  = MSR / MSE  = 67.6 / 4.133  = 16.36
Using F table (1 degree of freedom numerator and 3 denominator), p­value is between .025 and .05
Using Excel or Minitab, the p­value corresponding to F = 16.36 is .0272.
Because p­value  , we reject H0: 1 = 0
Source
of Variation
Regression

Error
Total

24. a.
b.

Sum
of Squares
67.6
12.4
80.0

Degrees
of Freedom
1
3
4

Mean
Square
67.6
4.133

F
16.36

p­value
.0272

s2 = MSE = SSE/(n ­ 2) = 230/3 = 76.6667

s  MSE  76.6667  8.7560
14 ­ 19

© 2010 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 14

c.

( xi  x ) 2  180
sb1 

d.

t

s
( xi  x )

2



8.7560
180

 0.6526

b1
3

 4.59
sb1 .653

Using t table (3 degrees of freedom), area in tail is less than .01; p­value is less than .02
Using Excel or Minitab, the p­value corresponding to t = ­4.59 is .0193.
Because p­value  , we reject H0: 1 = 0
e.

MSR = SSR/1 = 1620
F  = MSR/MSE = 1620/76.6667 = 21.13
Using F table (1 degree of freedom numerator and 3 denominator), p­value is less than .025
Using Excel or Minitab, the p­value corresponding to F = 21.13 is .0193.
Because p­value  , we reject H0: 1 = 0
Source
of Variation
Regression
Error
Total

25. a.

Sum
of Squares
 230
1620
1850

Degrees
of Freedom
1
3
4

Mean
Square
230
76.6667

F
21.13

p­value
.0193

s2 = MSE = SSE/(n ­ 2) = 127.3/3 = 42.4333
s  MSE  42.4333  6.5141

b.

( xi  x ) 2  190
sb1 

t

s
( xi  x )

2



6.5141
190

 0.4726

b1
.9

 1.90
sb1 .4726

Using t table (3 degrees of freedom), area in tail is between .05 and .10
 p­value is between .10 and .20
Using Excel or Minitab, the p­value corresponding to t = 1.90 is .1530.
14 ­ 20

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Simple Linear Regression
Because p­value >  , we cannot reject H0: 1 = 0; x and y do not appear to be related.
c.

MSR = SSR/1 = 153.9 /1 = 153.9
F = MSR/MSE = 153.9/42.4333 = 3.63

Using F table (1 degree of freedom numerator and 3 denominator),  p­value is greater than .10
Using Excel or Minitab, the p­value corresponding to F = 3.63 is .1530.
Because p­value >  , we cannot reject H0: 1 = 0; x and y do not appear to be related.

26. a.

In solving exercise 18, we found SSE = 85,135.14 
s2 = MSE = SSE/(n ­ 2) = 85,135.14/4 = 21,283.79
s  MSE  21,283.79 14589
.
( xi  x ) 2 0.74
sb1 

t

s
( xi  x )

2



145.89
0.74

169.59

b1
581.1


 3.43
sb1 169.59

Using t table (4 degrees of freedom), area in tail is between .01 and .025
 p­value is between .02 and .05
Using Excel or Minitab, the p­value corresponding to t = 3.43 is .0266.
Because p­value  , we reject H0: 1 = 0; there is a significant relationship between grade point 
average and monthly salary
b.

MSR = SSR/1 = 249,864.86/1 = 249,864.86
F  = MSR/MSE = 249,864.86/21,283.79 = 11.74
Using F table (1 degree of freedom numerator and 4 denominator),  p­value is between .025 and .05
Using Excel or Minitab, the p­value corresponding to F = 11.74 is .0266.
Because p­value  , we reject H0: 1 = 0

c.
Source
of Variation
Regression
Error

Sum
of Squares
249864.86
85135.14

Degrees
of Freedom
1

4
14 ­ 21

Mean
Square
 249864.86
21283.79

F
11.74

p­value
.0266

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 14
Total
27. a.

x

xi 37

 3.7
n
10

335000
y

( xi  x )( yi  y )  315.2
b1 

5

yi 1654

 165.4
n
10
( xi  x ) 2  10.1

( xi  x )( yi  y ) 315.2

31.2079
( xi  x ) 2
10.1

b0  y  b1 x  165.4  (31.2079)(3.7)  49.9308
yˆ  49.9308  31.2079 x
b.

2
2
SSE =  ( yi  y�
i ) 2487.66    SST =  ( yi  y ) = 12,324.4

Thus, SSR = SST ­ SSE = 12,324.4 ­ 2487.66 = 9836.74
MSR = SSR/1 = 9836.74
MSE  = SSE/(n ­ 2) = 2487.66/8 = 310.96
F  = MSR / MSE  = 9836.74/310.96 = 31.63
Using F table (1 degree of freedom numerator and 8 denominator), p­value is less than .01
Using Excel or Minitab, the p­value corresponding to F = 31.63 is .001.
Because p­value  , we reject H0: 1 = 0
Upper support and price are related.
c.

r2 = SSR/SST = 9,836.74/12,324.4 = .80
The estimated regression equation provided a good fit; we should feel comfortable using the 
estimated regression equation to estimate the price given the upper support rating.

d.
28.

y�= 49.93 + 31.21(4) = 174.77
The sum of squares due to error and the total sum of squares are
SSE  �( yi  yˆi ) 2  12,953.09

SST  �( yi  y ) 2  66, 200

Thus, SSR = SST - SSE = 66,200 – 12,953.09 = 53,246.91
s2 = MSE = SSE / (n - 2) = 12,953.09 / 9 = 1439.2322
s  MSE  1439.2322  37.9372
We can use either the t test or F test to determine whether temperature rating and price are
related.
14 ­ 22

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Simple Linear Regression
We will first illustrate the use of the t test.
Note: from the solution to exercise 10 ( xi  x ) 2  1912
s

sb1 

t

( xi  x )

b1
sb

1



5.2772
.8676

2



37.9372

1912

 .8676

 6.0825

Using t table (9 degrees of freedom), area in tail is less than .005; p-value is less than .01
Using Excel or Minitab, the p-value corresponding to t = -6.0825 is .000.
Because p-value  , we reject H0: 1 = 0
Because we can reject H0: 1 = 0 we conclude that temperature rating and price are related.
Next we illustrate the use of the F test.
MSR = SSR / 1 = 53,246.91
F = MSR / MSE = 53,246.91 / 1439.2322 = 37.00
Using F table (1 degree of freedom numerator and 9 denominator), p-value is less than .01
Using Excel or Minitab, the p-value corresponding to F = 37.00 is .000.
Because p-value  , we reject H0: 1 = 0
Because we can reject H0: 1 = 0 we conclude that temperature rating and price are related.
The ANOVA table is shown below.
Source
of Variation
Regression
Error
Total
29.

Sum
of Squares
53,246.91
12,953.09
66,200

Degrees
of Freedom
1
9
10

Mean
Square
53,246.91
1439.2322

F
37.00

p­value
.000

F
92.83

p­value
.0006

SSE = ( yi  yˆi )2  233,333.33     SST = ( yi  y ) 2 = 5,648,333.33
Thus, SSR = SST – SSE= 5,648,333.33 –233,333.33 = 5,415,000
MSE = SSE/(n ­ 2) = 233,333.33/(6 ­ 2) = 58,333.33
MSR = SSR/1 = 5,415,000
F  = MSR / MSE  = 5,415,000 / 58,333.25 = 92.83 
Source of 

Variation
Regression

Sum
of Squares
5,415,000.00

Degrees of
Freedom
1
14 ­ 23

Mean
Square
5,415,000

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 14
Error
Total

   233,333.33
5,648,333.33

4
5

58,333.33

Using F table (1 degree of freedom numerator and 4 denominator),  p­value is less than .01
Using Excel or Minitab, the p­value corresponding to F = 92.83 is .0006.
Because p­value  , we reject H0: 1 = 0.  Production volume and total cost are related.
30.

SSE = ( yi  yˆi )2  173.88    SST = ( yi  y ) 2 = 756
Thus, SSR = SST – SSE = 756 – 173.88 = 582.12
s2 = MSE = SSE/(n­2) = 173.88/6 = 28.98

s  28.98  5.3833
 ( xi  x ) 2  = 8,155,000
sb1 

t

s
( xi  x )

2



5.3833
 .001885
8,155, 000

b1 .008449


 4.48
sb1 .001885

Using t table (1 degree of freedom numerator and 8 denominator), area in tail is less than .005
 p­value is less than .01
Using Excel or Minitab, the p­value corresponding to t = 4.48 is .0042.
Because p­value  , we reject H0: 1 = 0
There is a significant relationship between price and rating.
31.

SSE = 540.04 and SST = 982.40
Thus, SSR = SST - SSE = 982.40 – 540.04 = 442.36
s2 = MSE = SSE / (n - 2) = 540.04 / 8 = 67.5050
s  MSE  67.5050  8.216
MSR = SSR / 1 = 442.36
F = MSR / MSE = 442.36 / 67.5050 = 6.55
Using F table (1 degree of freedom numerator and 8 denominator), p-value is between .025 and .

05
Using Excel or Minitab, the p-value corresponding to F = 6.55 is .034.
14 ­ 24

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Simple Linear Regression
Because p-value  , we reject H0: 1 = 0
Conclusion: price and overall score are related

32. a.

b.

s  =  2.033
x 3

( xi  x ) 2 10

s y�p s

2
1 ( xp  x )
1 (4  3) 2

2.033 
1.11
2
n ( xi  x )
5
10

y 0.2  2.6 x 0.2  2.6(4) 10.6
y p t / 2 s y p
10.6    3.182 (1.11)  =  10.6    3.53
or 7.07 to 14.13

c.

sind s 1 

d.

y p t / 2 sind

2
1 ( xp  x )
1 (4  3) 2


2.033
1


2.32
n ( xi  x ) 2
5
10

10.6    3.182 (2.32)  =  10.6    7.38
or 3.22 to 17.98
33. a.     s  =  8.7560
b.

  x  11

s yˆp  s

( xi  x ) 2  180
2

1 ( xp  x )
1 (8  11) 2


8.7560

 4.3780
n ( xi  x ) 2
5
180

yˆ  68  3 x  68  3(8)  44
y p t / 2 s y p

44  3.182 (4.3780) = 44  13.93
or 30.07 to 57.93

14 ­ 25

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.