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Chapter 14

Simple Linear Regression

Learning Objectives

1. Understand how regression analysis can be used to develop an equation that estimatesmathematically how two variables are related.

2. Understand the differences between the regression model, the regression equation, and theestimated 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 computethe sample correlation coefficient from the regression analysis output.

5. Understand the assumptions necessary for statistical inference and be able to test for a significantrelationship.

6. Know how to develop confidence interval estimates ofygiven a specific value ofxin both the caseof a mean value ofyand an individual value ofy.

7. Learn how to use a residual plot to make a judgement as to the validity of the regressionassumptions, recognize outliers, and identify influential observations.

8. Know the definition of the following terms:

independent and dependent variablesimple linear regressionregression modelregression equation and estimated regression equationscatter diagramcoefficient of determinationstandard error of the estimateconfidence intervalprediction intervalresidual plotstandardized residual plotoutlierinfluential observation

leverage

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Chapter 14

Solutions:

1 a.

b. There appears to be a linear relationship betweenx andy.

c. Many different straight lines can be drawn to provide a linear approximation of therelationship betweenx and y; in part d we will determine the equation of a straight linethat best represents the relationship according to the least squares criterion.

d. Summations needed to compute the slope and y-intercept are:

215 40 ( )( ) 26 ( ) 10i i i i ix y x x y y x x = = = =

1 2

( )( ) 262.6

10( )

i i

i

x x y yb

x x

= = =

b y b x0 1 8 2 6 3 0 2= = =( . )( ) .

0.2 2.6y x= +

e. 0.2 2.6(4) 10.6y= + =

14 - 2

0

2

4

6

8

10

1214

16

0 1 2 3 4 5 6

x

y

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2. a.

b. There appears to be a linear relationship betweenx andy.

c. Many different straight lines can be drawn to provide a linear approximation of therelationship betweenx and y; in part d we will determine the equation of a straight line thatbest represents the relationship according to the least squares criterion.

d. Summations needed to compute the slope andy-intercept are:

219 116 ( )( ) 57.8 ( ) 30.8i i i i ix y x x y y x x = = = =

1 2

( )( ) 57.81.8766

30.8( )

i i

i

x x y yb

x x

= = =

b y b x0 1 232 18766 38 30 3311= = =. ( . )( . ) .

. .y x= 30 33 188

e. . . ( ) .y= =30 33 188 6 19 05

14 - 3

0

5

10

15

20

25

30

35

0 1 2 3 4 5 6 7 8 9

x

y

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Chapter 14

3. a.

b. Summations needed to compute the slope andy-intercept are:

226 17 ( )( ) 11.6 ( ) 22.8i i i i ix y x x y y x x = = = =

1 2

( )( ) 11.60.5088

22.8( )

i i

i

x x y yb

x x

= = =

b y b x0 1 34 05088 2 0 7542= = =. ( . )(5. ) .

. .y x= +0 75 0 51

c. . . ( ) .y= + =0 75 0 51 4 2 79

14 - 4

0

1

2

3

45

6

7

0 1 2 3 4 5 6 7 8 9

x

y

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100

105

110

115

120

125

130

135

61 62 63 64 65 66 67 68 69

x

y


4. a.

b.There appears to be a linear relationship betweenx andy.

c. Many different straight lines can be drawn to provide a linear approximation of therelationship betweenx and y; in part d we will determine the equation of a straight line thatbest represents the relationship according to the least squares criterion.


2325 585 ( )( ) 110 ( ) 20i i i i ix y x x y y x x = = = =

1 2

( )( ) 1105.5

20( )

i i

i

x x y yb

x x

= = =

b y b x0 1 117 5 65 2405= = = (5. )( ) .

. .y x= +2405 55

e. . . . . ( )y x= + = + =2405 55 2405 55 63 106 pounds

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Chapter 14

5. a.

0

500

1000

1500

2000

2500

1 1.5 2 2.5 3 3.5 4 4.5

Baggage Capacity

Price($)

b. Letx= baggage capacity andy= price ($).

There appears to be a linear relationship betweenx andy.

c. Many different straight lines can be drawn to provide a linear approximation of therelationship betweenx andy; in part (d) we will determine the equation of a straight linethat best represents the relationship according to the least squares criterion.


229.5 11,110 ( )( ) 2909.8889 ( ) 4.5556i i i i ix y x x y y x x = = = =

1 2

( )( ) 2909.9889638.7561

4.5556( )

i i

i

x x y yb

x x

= = =

0 1 1234.4444 (638.7561)(3.2778) 859.26b y b x= = =

859.26 638.76y x= +

e. A one point increase in the baggage capacity rating will increase the price by approximately $639.

f. 859.26 638.76 859.26 638.76(3) $1057y x= + = + =

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6. a.

0

5

10

15

20

25

30

50.00 70.00 90.00 110.00 130.00 150.00 170.00 190.00

Salary

Bonus

b. Letx= salary andy= bonus.

There appears to be a linear relationship betweenx andy.

c. Summations needed to compute the slope andy-intercept are:

21420 160 ( )( ) 1114 ( ) 6046i i i i ix y x x y y x x = = = =

1 2

( )( ) 1114.18425

6046( )

i i

i

x x y yb

x x

= = =

0 1 16 (.18425)(142) 10.16b y b x= = =

10.16 .18y x

= +

d. $1000 increase in salary will increase the bonus by $180.

e 10.16 .18 10.16 .18(120) 11.95y x= + = + = or approximately $12,000

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Chapter 14

7. a.

b. Letx= DJIA andy= S&. S!""at#n% needed t "'!te te %'e and y*#nte+e't a+e

2104,850 14,233 ( )( ) 268,921 ( ) 1,806,384i i i i ix y x x y y x x = = = =

1 2

( )( ) 268,9210.14887

1,806,384( )

i i

i

x x y yb

x x

= = =

0 1 1423.3 (.14887)(10,485) 137.629b y b x= = =

137.63 0.1489y x= +

. 137.63 0.1489(11,000) 1500.27y= + = + a''+-#"ate 1500

8. a.

14 - 8

1300

1350

1400

1450

1500

1550

9600 9800 10000 10200 10400 10600 10800 11000 11200

DJIA

S&

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0

200

400

600

800

1000

1200

1400

1600

1800

0 1 2 3 4 5 6

Sidetrack Capability

Price

b. /e+e a''ea+% t be a #nea+ +eat#n%#' beteenx = %#det+a a'ab##t andy = '+#e, #t #e+'+#ed "de% a#n a #e+ ee and#n.

. S!""at#n% needed t "'!te te %'e and *#nte+e't a+e

228 10,621 ( )( ) 4003.2 ( ) 19.6i i i i ix y x x y y x x = = = =

1 2

( )( ) 4003.2204.2449

19.6( )

i i

i

x x y yb

x x

= = =

0 1 1062.1 (204.2449)(2.8) 490.21b y b x= = =

490.21 204.24y x= +

d. 490.21 204.24 490.21 204.24(4) 1307y x= + = + =

9. a.

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Chapter 14

b. Summations needed to compute the slope andy-intercept are:

270 1080 ( )( ) 568 ( ) 142i i i i ix y x x y y x x = = = =

1 2

( )( ) 5684

142( )

i i

i

x x y yb

x x

= = =

b y b x0 1 108 4 7 80= = =( )( )

y x= +80 4

c. ( )y x= + = + =80 4 80 4 9 116

14 - 10

50

60

70

80

90

100

110

120

130

140

150

0 2 4 6 8 10 12 14x

y

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10. a.

b. Letx= 'e++"ane %+e andy= e+a +at#n. S!""at#n% needed t "'!te te %'e and y*#nte+e't a+e

22752 1177 ( )( ) 1723.73 ( ) 11,867.73i i i i ix y x x y y x x = = = =

1 2

( )( ) 1723.730.1452

11,867.73( )

i i

i

x x y yb

x x

= = =

0 1 78.4667 (.1452)(183.4667) 51.82b y b x= = =

51.82 0.145y x= +

. 51.82 0.145(225) 84.4y= + = + a''+-#"ate 84

14 - 11

60

65

70

75

80

85

90

95

100 120 140 160 180 200 220 240 260

e+$+"an&e S&$+e

53e+a))6at#n2

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Chapter 14

11. a.

0

5

10

15

20

25

30

35

10 15 20 25 30 35

Arrivals (% )

Departures(%

)

b. There appears to be a linear relationship between the variables.

c. Letx= percentage of late arrivals andy= percentage of late departures.

The summations needed to compute the slope and they-intercept are:

2273 265 ( )( ) 101 ( ) 114.4i i i i ix y x x y y x x = = = =

1 2

( )( ) 101.8829

114.4( )

i i

i

x x y yb

x x

= = =

0 1 21 (.8829)(21.4) 2.106b y b x= = =

2.106 .883y x= +

d. A one percent increase in the percentage of late arrivals will increase the percentage of late arrivalsby .88 or slightly less than one percent.

e. 2.106 .883 2.106 .883(22) 21.53y x= + = + =

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12. a.

0

500,000

1,000,000

1,500,000

2,000,000

2,500,000

3,000,000

3,500,000

4,000,000

4,500,000

5,000,000

0 200 400 600 800 1,000 1,200 1,400 1,600 1,800

!"be+ $ "'$ee%

+$d!t#$n

b. /e+e a''ea+% t be a '%#t#e #nea+ +eat#n%#' beteen te n!"be+ e"'ee% and te'+d!t#n !t'!t.

. Letx= n!"be+ e"'ee% andy= '+d!t#n !t'!t. S!""at#n% needed t "'!te te %'eandy*#nte+e't a+e

258,990,16640,6 == ii yx

27272,894,161.)(499.81828,166,725,))((2== xxyyxx iii

7935.821,22727.161,894,2

8182.499,725,166,8

)(

))((

21 ==

=

xx

yyxx

bi

ii

479,384.180*(301.8182)2,821.7935*45772,284.4510 === xbyb

xy 2,821.793504*79,384.18 +=

d. y -79,834.1804 + 2,821.7935 (400) = 1,048,883

13. a.

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Chapter 14

b. The summations needed to compute the slope and they-intercept are:

2399 97.1 ( )( ) 1233.7 ( ) 7648i i i i ix y x x y y x x = = = =

1 2

( )( ) 1233.70.16131

7648( )

i i

i

x x y yb

x x

= = =

b y b x0 1 1387143 016131 4 67675= = =. ( . )(57) .

. .y x= +4 68 016

c. . . . . (52. ) .y x= + = + =4 68 016 4 68 016 5 1308 or approximately $13,080.

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

14 - 14

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0.0 20.0 40.0 60.0 80.0 100.0 120.0 140.0

x

y

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14. a.

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0 100 200 300 400 500 600 700 800 900

Quantity of the house

VacancyRate(%)

b. Letx= number of houses andy= vacancy rate

The summations needed to compute the slope and they-intercept are:

62,592)(98.862,5))((312697,42==== xxyyxxyx iiiii

0099.620,592

98.862,5

)(

))((21 =

=

=

xx

yyxxb

i

ii

11.21)294.276(0099.376.1810 =+== xbyb

xy 0099.11.21 =

c. 13.19)200(0099.11.21 ==y

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

. .y x yi i= + =0 2 2 6 8

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

SS SS/= = = =( ) . ( )y y y yi i i2 212 40 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 inyhas been explained bythe least squares line.

c. r= = +. .845 9192

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Chapter 14


30.33 1.88 23.2iy x y= =


2 2SS ( ) 6.33 SS/ ( ) 114.80i i iy y y y= = = =

Thus, SSR = SST - SSE = 114.80 - 6.33 = 108.47

b. r2= SSR/SST = 108.47/114.80 = .945

The least squares line provided an excellent fit; 94.5% of the variability inyhas been explained bythe estimated regression equation.

c. r= = . .945 9721

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

(b1= -1.88)

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

.75 .51 3.4iy x y= + =


2 2SS ( ) 5.3 SS/ ( ) 11.2i i iy y y y= = = =

Thus, SSR = SST - SSE = 11.2 - 5.3 = 5.9

r2= SSR/SST = 5.9/11.2 = .527

We see that 52.7% of the variability inyhas been explained by the least squares line.

r= = +. .527 7259


1790.5 581.1 3650y x y= + =


2 2SS ( ) 85,135.14 SS/ ( ) 335,000i i iy y y y= = = =

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

b. r2= SSR/SST = 249,864.86/335,000 = .746


c. r= = +. .746 8637


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137.63 .1489 1423.3y x y= + =


2 2SS ( ) 7547.14 SS/ ( ) 47,582.10i i iy y y y= = = =

Thus, SSR = SST - SSE = 47,582.10 - 7547.14 = 40,034.96

b. r2= SSR/SST = 40,034.96/47,582.10 = .84

We see that 84% of the variability inyhas been explained by the least squares line.

c. .84 .92r= = +

20. a. Letx= #n"e andy= "e '+#e. S!""at#n% needed t "'!te te %'e andy*#nte+e't a+e

074,163)(242,41))((752,2706,142==== xxyyxxyx iiiii

253.074,163

242,41

)(

))((21 ==

=

xx

yyxxb

i

ii

6.80)225,1)(2529.0(3333.22910 === xbyb

xy 253.6.80 +=

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

==== 6667.466,18)(2960.036,8)(22

yySSTyySSE iii

Thus, SSR = SST - SSE = 18,466.6667 8,036.2960 = 10,430.3707

r2= SSR/SST = 10,430.3707/18,466.6667 = .5648


7515.5648. ==r

. 75.159)950(253.6.80 =+=y

21. a. The summations needed in this problem are:

23450 33,700 ( )( ) 712,500 ( ) 93,750i i i i ix y x x y y x x = = = =

1 2

( )( ) 712,5007.6

93,750( )

i i

i

x x y yb

x x

= = =

0 1 5616.67 (7.6)(575) 1246.67b y b x= = =

. .y x= +1246 67 7 6

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Chapter 14

b. $7.60

c. The sum of squares due to error and the total sum of squares are:

2 2SS ( ) 233,333.33 SS/ ( ) 5,648,333.33i i iy y y y= = = =

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 inyhas been explained by the estimated regression equation.

d. . . . . (500) $5046.y x= + = + =1246 67 7 6 1246 67 7 6 67

22. a. Letx= speed (ppm) andy= price ($)

The summations needed in this problem are:

2

149.9 10,221 ( )( ) 34,359.81 ( ) 291.389i i i i ix y x x y y x x = = = =

1 2

( )( ) 34,359.81117.9173

291.389( )

i i

i

x x y yb

x x

= = =

0 1 1022.1 (117.9173)(14.99) 745.80b y b x= = =

745.80 117.917y x= +

b. The sum of squares due to error and the total sum of squares are:

2 2SS ( ) 1,678, 294 SS/ ( ) 5,729,911i i iy y y y= = = =

Thus, SSR = SST - SSE = 5,729,911 - 1,678,294 = 4,051,617

r2= SSR/SST = 4,051,617/5,729,911 = 0.7071

Approximately 71% of the variability in price is explained by the speed.

c. .7071 .84r= = +

It reflects a linear relationship that is between weak and strong.

23. a. s2 = MSE = SSE / (n- 2) = 12.4 / 3 = 4.133

b. s= = =:S 4133 2 033. .

c. 2( ) 10ix x =

1 2

2.0330.643

10( )b

i

ss

x x= = =

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d. t b

sb= = =1

1

2 6

6434 04

.

..

Using ttable (3 degrees of freedom), area in tail is between .01 and .025

p-value is between .02 and .05

Actualp-value = .0272

Becausep-value , we rejectH0: 1= 0

e. MSR = SSR / 1 = 67.6

F = MSR / MSE = 67.6 / 4.133 = 16.36

Using Ftable (1 degree of freedom numerator and 3 denominator),p-value is between .025 and .05



Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Regression 67.6 1 67.6 16.36

Error 12.4 3 4.133

Total 80.0 4

24. a. s2 = MSE = SSE / (n- 2) = 6.33 / 3 = 2.11

b. s= = =:S 211 1453. .

c. 2( ) 30.8ix x =

1 2

1.4530.262

30.8( )b

i

ss

x x= = =

d. t b

sb= =

= 1

1

188

262718

.

..

Using ttable (3 degrees of freedom), area in tail is less than .005;p-value is less than .01


Becausep-value

, we rejectH0: 1= 0

e. MSR = SSR / 1 = 8.47

F = MSR / MSE = 108.47 / 2.11 = 51.41

Using Ftable (1 degree of freedom numerator and 3 denominator),p-value is less than .01


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Chapter 14



Regression 108.47 1 108.47 51.41

Error 6.33 3 2.11

Total 114.80 4

25. a. s2 = MSE = SSE / (n- 2) = 5.30 / 3 = 1.77

s= = =:S 177 133. .

b. 2( ) 22.8ix x =

1 2

1.330.28

22.8( )b

i

ss

x x= = =

t

b

sb= = =

1

1

51

28 182

.

. .


p-value is between .10 and .20


Becausep-value > , we cannot rejectH0: 1= 0;xandydo not appear to be related.

c. MSR = SSR/1 = 5.90 /1 = 5.90

F= MSR/MSE = 5.90/1.77 = 3.33

Using Ftable (1 degree of freedom numerator and 3 denominator), p-value is greater than .10


Becausep-value > , we cannot rejectH0: 1= 0;xandydo not appear to be related.

26. a. s2 = MSE = SSE / (n- 2) = 85,135.14 / 4 = 21,283.79

s= = =:S 21 28379 14589, . .

2( ) 0.74ix x =

1 2

145.89169.59

0.74( )b

i

ss

x x= = =

t b

sb= = =1

1

58108

16959343

.

..


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p-value is between .02 and .05Actualp-value = .0266


b. MSR = SSR / 1 = 249,864.86 / 1 = 249.864.86

F = MSR / MSE = 249,864.86 / 21,283.79 = 11.74

Using Ftable (1 degree of freedom numerator and 4 denominator), p-value is between .025 and .05



c.


Regression 249864.86 1 249864.86 11.74

Error 85135.14 4 21283.79

Total 335000 5

27. a. Summations needed to compute the slope andy-intercept are:

237 1654 ( )( ) 315.2 ( ) 10.1i i i i ix y x x y y x x = = = =

1 2

( )( ) 315.231.2079

10.1( )

i i

i

x x y yb

x x

= = =

0 1 165.4 (31.2079)(3.7) 19.93b y b x= = =

19.93 31.21y x= +

b. SSE =2( ) 2487.66i iy y = SST =

2( )iy 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

F.05 = 5.32 (1 degree of freedom numerator and 8 denominator)

Since F= 31.63 > F.05= 5.32 we rejectH0: 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 theestimated regression equation to estimate the price given the upper support rating.

d. y = 19.93 + 31.21(4) = 144.77

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Chapter 14

28. SST = 411.73 SSE = 161.37 SSR = 250.36

MSR = SSR / 1 = 250.36

MSE = SSE / (n- 2) = 161.37 / 13 = 12.413

F= MSR / MSE = 250.36 / 12.413= 20.17

Using Ftable (1 degree of freedom numerator and 13 denominator), p-value is less than .01


Becausep-value , we rejectH0: 1= 0.

29. SSE = 233,333.33 SST = 5,648,333.33 SSR = 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


Regression 5,415,000.00 1 5,415,000 92.83

Error 233,333.33 4 58,333.33

Total 5,648,333.33 5



Becausep-value , we rejectH0: 1= 0. Production volume and total cost are related.

30. Using the computations from Exercise 22,

SSE = 1,678,294 SST = 5,729,911 SSR = 4,051,617

s2= MSE = SSE/(n-2) = 1,678,294/8 = 209,786.8

209,786.8 458.02s= =

2( )ix x = 291.389

1 2

458.0226.832

291.389( )b

i

ss

x x= = =

1

1 117.9173 4.39526.832b

bt

s= = =

Using ttable (1 degree of freedom numerator and 8 denominator), area in tail is less than .005

p-value is less than .01


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There is a significant relationship between xandy.

31. SST = 11,373.09 SSE = 2017.37 SSR = 9355.72

MSR = SSR / 1 = 9355.72

MSE = SSE / (n- 2) = 2017.37/ 16 = 126.0856

F= MSR / MSE = 9355.72/ 126.0856 = 74.20



Becausep-value , we rejectH0: 1= 0.

32. a. s = 2.033

23 ( ) 10ix x x= =

'

2 2'

2

( )1 1 (4 3)2.033 1.11

5 10( )y

i

x xs s

n x x

= + = + =

b. . . . . ( ) .y x= + = + =0 2 2 6 0 2 2 6 4 10 6

; y t sy' ' 2

10.6 3.182 (1.11) = 10.6 3.53

or 7.07 to 14.13

c.

2 2'

#nd 2

( )1 1 (4 3)1 2.033 1 2.32

5 10( )i

x xs s

n x x

= + + = + + =

d. ;y t s' #nd 2

10.6 3.182 (2.32) = 10.6 7.38

or 3.22 to 17.98

33. a. s = 1.453

b. 23.8 ( ) 30.8ix x x= =

'

2 2'

2

( )1 1 (3 3.8)1.453 .068

5 30.8( )y

i

x xs s

n x x

= + = + =

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Chapter 14

. . . . ( ) .y x= = =30 33 188 30 33 188 3 24 69

; y t sy' ' 2

24.69 3.182 (.68) = 24.69 2.16

or 22.53 to 26.85

c.

2 2'

#nd 2

( )1 1 (3 3.8)1 1.453 1 1.61

5 30.8( )i

x xs s

n x x

= + + = + + =

d. ;y t s' #nd 2

24.69 3.182 (1.61) = 24.69 5.12

or 19.57 to 29.81

34. s = 1.33

25.2 ( ) 22.8ix x x= =

'

2 2'

2

( )1 1 (3 5.2)1.33 0.85

5 22.8( )y

i

x xs s

n x x

= + = + =

. . . . ( ) .y x= + = + =0 75 0 51 0 75 051 3 2 28

; y t sy' ' 2

2.28 3.182 (.85) = 2.28 2.70

or -.40 to 4.98

2 2'

#nd 2

( )1 1 (3 5.2)1 1.33 1 1.58

5 22.8( )i

x xs s

n x x

= + + = + + =

;y t s' #nd 2

2.28

3.182 (1.58) = 2.28

5.03

or -2.27 to 7.31

35. a. s = 145.89

23.2 ( ) 0.74ix x x= =

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'

2 2'

2

( )1 1 (3 3.2)145.89 68.54

6 0.74( )y

i

x xs s

n x x

= + = + =

1790.5 581.1 1790.5 581.1(3) 3533.8y x= + = + =

; y t sy' ' 2

3533.8 2.776 (68.54) = 3533.8 190.27

or $3343.53 to $3724.07

b.

2 2'

#nd 2

( )1 1 (3 3.2)1 145.89 1 161.19

6 0.74( )i

x xs s

n x x

= + + = + + =

;y t s' #nd 2

3533.8

2.776 (161.19) = 3533.8

447.46

or $3086.34 to $3981.26

36. a. 51.82 .1452 51.82 .1452(200) 80.86y x= + = + =

b. s = 3.5232

2183.4667 ( ) 11,867.73ix x x= =

'

2 2'

2

( )1 1 (200 183.4667)3.5232 1.055

15 11,867.73( )y

i

x xs s

n x x

= + = + =

; y t sy' ' 2

80.86 2.160 (1.055) = 80.86 2.279

or 78.58 to 83.14

c.

2 2'

#nd 2

( )1 1 (200 183.4667)1 3.5232 1 3.678

15 11,867.73( )i

x xs s

n x x

= + + = + + =

;y t s' #nd 2

80.86 2.160 (3.678) = 80.86 7.944

or 72.92 to 88.80

37. a. 257 ( ) 7648ix x x= =

s2= 1.88 s= 1.37

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Chapter 14

'

2 2'

2

( )1 1 (52.5 57)1.37 0.52

7 7648( )y

i

x xs s

n x x

= + = + =

; y t sy' ' 2

13.08 2.571 (.52) = 13.08 1.34

or 11.74 to 14.42 or $11,740 to $14,420

b. sind= 1.47

13.08 2.571 (1.47) = 13.08 3.78

or 9.30 to 16.86 or $9,300 to $16,860

c. Yes, $20,400 is much larger than anticipated.

d. Any deductions exceeding the $16,860 upper limit could suggest an audit.

38. a. . . (500) $5046.y= + =1246 67 7 6 67

b. 2575 ( ) 93,750ix x x= =

s2= MSE = 58,333.33 s= 241.52

2 2'

#nd 2

( )1 1 (500 575)1 241.52 1 267.50

6 93,750( )i

x xs s

n x x

= + + = + + =

;y t s' #nd 2

5046.67 4.604 (267.50) = 5046.67 1231.57

or $3815.10 to $6278.24

c. Based on one month, $6000 is not out of line since $3815.10 to $6278.24 is the prediction interval.However, a sequence of five to seven months with consistently high costs should cause concern.

39. a. Letx= miles of track andy= weekday ridership in thousands.

Summations needed to compute the slope andy-intercept are:

2203 309 ( )( ) 1471 ( ) 838i i i i ix y x x y y x x = = = =

1 2

( )( ) 14711.7554

838( )

i i

i

x x y yb

x x

= = =

0 1 44.1429 (1.7554)(29) 6.76b y b x= = =

6.76 1.755y x= +

b. SST =3620.9 SSE = 1038.7 SSR = 2582.1

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r2= SSR/SST = 2582.1/3620.9 = .713

The estimated regression equation explained 71.3% of the variability iny; a good fit.

c. s2= MSE = 1038.7/5 = 207.7

207.7 14.41s= =

'

2 2'

2

( )1 1 (30 29)14.41 5.47

7 838( )y

i

x xs s

n x x

= + = + =

6.76 1.755 6.76 1.755(30) 45.9y x= + = + =

; y t sy' ' 2

45.9 2.571(5.47) = 45.9 14.1

or 31.8 to 60

d.

2 2'

#nd 2

( )1 1 (30 29)1 14.41 1 15.41

7 838( )i

x xs s

n x x

= + + = + + =

;y t s' #nd 2

45.9 2.571(15.41) = 45.9 39.6

or 6.3 to 85.5

The prediction interval is so wide that it would not be of much value in the planning process. Alarger data set would be beneficial.

40. a. 9

b. y = 20.0 + 7.21x

c. 1.3626

d. SSE = SST - SSR = 51,984.1 - 41,587.3 = 10,396.8

MSE = 10,396.8/7 = 1,485.3

F= MSR / MSE = 41,587.3 /1,485.3 = 28.00



Becausep-value , we reject H0:B1= 0.

e. y = 20.0 + 7.21(50) = 380.5 or $380,500

41. a. y = 6.1092 + .8951x

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Chapter 14

b.1

1 1 .8951 0 6.01.149b

b Bt

s

= = =

Using ttable (1 degree of freedom numerator and 8 denominator), area in tail is less than .005

p-value is less than .01


Becausep-value , we reject H0:B1= 0

c. y = 6.1092 + .8951(25) = 28.49 or $28.49 per month

42 a. y = 80.0 + 50.0x

b. 30

c. F= MSR / MSE = 6828.6/82.1 = 83.17



Becausep-value

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e. The 95% prediction interval is 24,285,484 to 40,667,127 .

44. a/b. The scatter diagram shows a linear relationship between the two variables.

c. The Minitab output is shown below:

The regression equation isouse = 41.& 1.33 acancy(

Predictor Coe !" Coe T P

Constant 41.490 $.333 $.&& 0.001

acancy( -1.3301 0.3&22 -3.7# 0.013

! = 3.0#923 '-!q = 74.0( '-!q)ad* = $#.#(

,naysis o ariance

!ource / !! ! P

'egression 1 13$.07 13$.07 14.2$ 0.013

'esidua "rror & 47.72 9.&4

Tota $ 1#3.79

d. Since thep-value = 0.013 is less than = .05, the relationship is significant.

e. 2r = .688. The least squares line does not provide a very good fit.

f. The 95% confidence interval is 14.53 t 20.56or NT$14,530 to NT$20,560.

g. The 95% prediction interval is 4.55 t 22.57or NT$4,550 to NT$22,570.

45. a. 270 76 ( )( ) 200 ( ) 126i i i i ix y x x y y x x = = = =

1 2( )( ) 200 1.5873

126( )i i

i

x x y ybx x

= = =

0 1 15.2 (1.5873)(14) 7.0222b y b x= = =

7.02 1.59y x= +

b. The residuals are 3.48, -2.47, -4.83, -1.6, and 5.22

c.

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Chapter 14

*6

*4

*2

0

2

4

6

0 5 10 15 20 25

x

6e%#d

!a)%

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29.4 1.55y x= +

b. SS/ = 1002 SS = 310.28 SS = 691.72

:S = SS ; 1 = 691.72

:S = SS ; (n* 2) = 310.28; 5 = 62.0554

F= :S ; :S = 691.72; 62.0554= 11.15

Using Ftable (1 degree of freedom numerator and 5 denominator), p-value is between .01 and .025


Becausep-value = .05, e n!de tat te t a+#abe% a+e +eated.

.

*15

*10

*5

0

5

10

25 30 35 40 45 50 55 60 65

+ed#&ted @a)!e%

6e%#d!a)%

d. /e +e%#d!a 't ead% !% t ?!e%t#n te a%%!"'t#n a #nea+ +eat#n%#' beteenxandy. ent! te +eat#n%#' #% %#n##ant at te .05 ee %#n##ane, #t !d be e-t+e"edane+!% t e-t+a'ate bend te +ane te data.

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Chapter 14

48. a. 80 4y x= +

*8

*6

*4

*2

0

2

4

6

8

0 2 4 6 8 10 12 14

x

6e%#d!a)%

b. /e a%%!"'t#n% ne+n#n te e+++ te+" a''ea+ +ea%nabe.

49. a. Letx= +et!+n n #ne%t"ent () andy= '+#e;ea+n#n% (;) +at#.

32.13 3.22y x= +

b.

*1.5

*1

*0.5

0

0.5

1

1.5

2

0 10 20 30 40 50 60

x

Standa+d#>ed

6e%#d!

a)%

. /e+e #% an !n!%!a t+end #n te +e%#d!a%. /e a%%!"'t#n% ne+n#n te e+++ te+" a''ea+?!e%t#nabe.

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50. a. The MINITAB output is shown below:

The regression equation is

Y = 66.1 + 0.402 X

Predictor Coef SE Coef T p

Constant 66.10 32.06 2.06 0.094

X 0.4023 0.2276 1.77 0.137

S = 12.62 R-sq = 38.5% R-sq(adj) = 26.1%

Analysis of Variance

SOURCE DF SS MS F p

Regression 1 497.2 497.2 3.12 0.137

Residual Error 5 795.7 159.1

Total 6 1292.9

Unusual Observations

Obs. X Y Fit SEFit Residual St.Resid

1 135 145.00 120.42 4.87 24.58 2.11R

R denotes an observation with a large standardized residual.

/e %tanda+d#>ed +e%#d!a% a+e 2.11, *1.08, .14, *.38, *.78, *.04, *.41

/e #+%t b%e+at#n a''ea+% t be an !t#e+ %#ne #t a% a a+e %tanda+d#>ed +e%#d!a.

b.

2.4+

- *

STDRESID-

-

-

1.2+ -

-

-

- *

0.0+ *

-

- * *

- *

-

-1.2+ *

-

--+---------+---------+---------+---------+---------+----YHAT

110.0 115.0 120.0 125.0 130.0 135.0

The standardized residual plot indicates that the observationx = 135,y = 145 may be an outlier;note that this observation has a standardized residual of 2.11.

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Chapter 14

c. The scatter diagram is shown below -

Y - *

-

-

135+

-

- * * -

-

120+ * *

-

-

- *

-

105+

-

- *

----+---------+---------+---------+---------+---------+--X

105 120 135 150 165 180

The scatter diagram also indicates that the observationx = 135,y = 145 may be an outlier; theimplication is that for simple linear regression an outlier can be identified by looking at the scatterdiagram.

51. a. The Minitab output is shown below:


Y = 13.0 + 0.425 X

Predictor Coef SE Coef T p

Constant 13.002 2.396 5.43 0.002

X 0.4248 0.2116 2.01 0.091

S = 3.181 R-sq = 40.2% R-sq(adj) = 30.2%


SOURCE DF SS MS F p

Regression 1 40.78 40.78 4.03 0.091


Total 7 101.50

Unusual Observations

Obs. X Y Fit Stdev.Fit Residual St.Resid

7 12.0 24.00 18.10 1.20 5.90 2.00R

8 22.0 19.00 22.35 2.78 -3.35 -2.16RX

R denotes an observation with a large standardized residual.

X denotes an observation whose X value gives it large influence.

The standardized residuals are: -1.00, -.41, .01, -.48, .25, .65, -2.00, -2.16

The last two observations in the data set appear to be outliers since the standardized residuals forthese observations are 2.00 and -2.16, respectively.

b. Using MINITAB, we obtained the following leverage values:

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.28, .24, .16, .14, .13, .14, .14, .76

MINITAB identifies an observation as having high leverage if hi> 6/n; for these data, 6/n=6/8 = .75. Since the leverage for the observationx= 22,y= 19 is .76, MINITAB wouldidentify observation 8 as a high leverage point. Thus, we conclude that observation 8 is aninfluential observation.

c. 24.0+ *

-

Y -

-

-

20.0+ *

- *

- *

-

-

16.0+ *

- *

-

- *

-

12.0+ *

-

+---------+---------+---------+---------+---------+------X

0.0 5.0 10.0 15.0 20.0 25.0

The scatter diagram indicates that the observation x = 22,y = 19 is an influential observation.


The regression equation is,5ount = 4.09 + 0.19$ edia"6


Constant 4.0#9 2.1$# 1.#9 0.09$

edia"6 0.19&&2 0.03$3& &.3# 0.001

! = &.044 '-!q = 7#.3( '-!q)ad* = 7&.$(

,naysis o ariance

!ource / !! ! P

'egression 1 73&.#4 73&.#4 2#.93 0.001

'esidua "rror # 203.&1 2&.44

Tota 9 939.3&

8nusua :ser;ations

:s edia"6 ,5ount it !" it 'esidua !t 'esid

1 120 3$.30 27.&& 3.30 #.7& 2.30'

' denotes an o:ser;ation

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Chapter 14

observation 1 to be an outlier.

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Price = 1&.2 + 0.01#1 ou5e


Constant 1&.193 2.004 7. 0.000

ou5e 0.01#0#$ 0.002079 #.70 0.000

! = &.297$0 '-!q = #1.7( '-!q)ad* = #0.$(

,naysis o ariance

!ource / !! ! P

'egression 1 2123.# 2123.# 7&.$# 0.000

'esidua "rror 17 477.1 2#.1

Tota 1# 2$00.9

b. /e #n +e%#d!a 't e+##e% te%e b%e+at#n%.

Fitt ed Value

Standardized

Residual

5550454035302520

2

1

0

-1

-2

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Chapter 14

54. a. /e :#n#tab !t'!t #% %n be


!aary = 707 + 0.004#2 >tCa


Constant 707.0 11#.0 &.99 0.000

>tCa 0.004#1&4 0.000#07$ &.9$ 0.000

! = 379.# '-!q = $$.4( '-!q)ad* = $4.&(

,naysis o ariance

!ource / !! ! P

'egression 1 &129071 &129071 3&.&& 0.000

'esidua "rror 1# 2&9$$47 1442

Tota 19 772&71#

8nusua :ser;ations

:s >tCa !aary it !" it 'esidua !t 'esid

$ &07217 332&.0 3149.& 33#.$ 17&.& 1.02 ?

17 1209$7 11$.2 12#9.& #$.4 -1173.3 -3.17'

' denotes an o:ser;ation ed +e%#d!a 't aa#n%t te'+ed#ted a!e% #% %n be

*4

*3

*2

*1

0

1

2

3

500 1000 1500 2000 2500 3000 3500

+ed#&ted @a!e%

Standa+d#>ed

6e%#d!a)%

55. No. Regression or correlation analysis can never prove that two variables are causally related.

56. The estimate of a mean value is an estimate of the average of allyvalues associated with the samex. The estimate of an individualyvalue is an estimate of only one of theyvalues associated with aparticularx.

57. The purpose of testing whether 1 0 = is to determine whether or not there is a significant

relationship betweenxandy. However, rejecting 1 0 = does not necessarily imply a good fit. For

example, if 1 0 = is rejected and r2is low, there is a statistically significant relationship betweenx

andybut the fit is not very good.

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58. a. /e :#n#tab !t'!t #% %n be


Price = 9.2$ + 0.711 !hares


Constant 9.2$& 1.099 #.43 0.000

!hares 0.710& 0.1474 4.#2 0.001

! = 1.419 '-!q = 74.4( '-!q)ad* = 71.2(

,naysis o ariance

!ource / !! ! P

'egression 1 4$.7#4 4$.7#4 23.22 0.001

'esidua "rror # 1$.11$ 2.01&

Tota 9 $2.900

b. S#ne tep*a!e ++e%'nd#n t = 23.22 = .001 B = .05, te +eat#n%#' #% %#n##ant.

. 2r = .744C a d #t. /e ea%t %?!a+e% #ne e-'a#ned 74.4 te a+#ab##t #n +#e.

d. 9.26 .711(6) 13.53y= + =

59. a. /e :#n#tab !t'!t #% %n be


tions = - 3.#3 + 0.29$ Co55on


Constant -3.#34 &.903 -0.$& 0.&29

Co55on 0.29&$7 0.02$4# 11.17 0.000

! = 11.04 '-!q = 91.9( '-!q)ad* = 91.2(

,naysis o ariance

!ource / !! ! P

'egression 1 1&20# 1&20# 124.72 0.000

'esidua "rror 11 1341 122

Tota 12 1$&&0

b. 3.83 .296(150) 40.57y= + = C a''+-#"ate 40.6 "##n %a+e% 't#n% +ant% !t%tand#n.

. 2r = .919C a e+ d #t. /e ea%t %?!a+e% #ne e-'a#ned 91.9 te a+#ab##t #n 't#n%.



Horizon = 0.275 + 0.950 S&P 500

Predictor Coef SE Coef T P

Constant 0.2747 0.9004 0.31 0.768

S&P 500 0.9498 0.3569 2.66 0.029

S = 2.664 R-Sq = 47.0% R-Sq(adj) = 40.3%


Source DF SS MS F P

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Chapter 14

Regression 1 50.255 50.255 7.08 0.029


Total 9 107.036

b. Since thep-value = 0.029 is less than = .05, the relationship is significant.

c. r2= .470. The least squares line does not provide a very good fit.

d. Texas Instruments has higher risk with a market beta of 1.25.

61. a.

40

50

60

70

80

90

100

35 45 55 65 75 85

L$ /e"'e+at! +e

=#2(/e"'e+at!+e

b. It a''ea+% tat te+e #% a '%#t#e #nea+ +eat#n%#' beteen te t a+#abe%.

. /e :#n#tab !t'!t #% %n be


igh = 23.9 + 0.#9# @o

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