chap013 solutions

Chapter 13 - Simple Linear Regression Analysis

CHAPTER 13—Simple Linear Regression Analysis

13.1 The “best” line that can be fitted to the observed data. The slope and the intercept of the least squares line.

LO2

13.2 Because we do not know how y and x are related outside the experimental region.

LO1, LO2

13.3 a. b0 = 15.84, b1 = -0.1279

b. b1 = -179.6475 / 1404.355 = -0.1279, b0 = 10.2125 – (-0.1279*43.98) = 15.84

c. Both the estimate of the mean and the prediction of the point are calculated by putting the temperature of 40 into the equation = 15.84 – 0.1279*40 = 10.724 MMcf.

LO1, LO2

13.4 a. b0 = 14.82 b1 = 5.707

The interpretation of b0 is the starting salary of someone with a GPA of 0.

The interpretation of b1 is for each increase in GPA of 1, salary goes up $5,707

No. The interpretation of b0 does not make practical sense since it indicates that someone with a GPA = 0 would have a starting salary of $14,816, when in fact they would not have graduated with a GPA = 0.

b. y = 14.82 + 5.707(3.25) = 33.36775

That is, $33,367.75

LO1, LO2

13.5 a. b0 = 11.4641 b1 = 24.6022

b0 – 0 copiers, 11.46 minutes of service.

b1 – each additional copier adds 24.6022 minutes of service on average.

No. The interpretation of b0 does not make practical sense since it indicates that 11.46 minutes of service would be required for a customer with no copiers.

b. y = 11.4641 + 24.6022(4) = 109.873, or 109.9 minutes

LO1, LO2

13.6 a. b0 = 7.814 b1 = 2.665

b0 – 0 price difference yields demand of 7.814.

13-1


13-2


b1 – each increase in 1 of price difference increases demand on average by 2.665.

Yes. The interpretation of b0 does make practical sense since it indicates that 781,409 bottles of detergent would be demanded when the price difference with other products is zero.

b. = 7.814 + 2.665 (.10) = 8.0805

LO1, LO2

13.7 a.

xi yi x i2

xi yi

5 71 25 355

62 663 3844 41106

35 381 1225 13335

12 138 144 1656

83 861 6889 71463

14 145 196 2030

46 493 2116 22678

52 548 2704 28496

23 251 529 5773

100 1024 10000 102400

41 435 1681 17835

75 772 5625 57900

13-3


SSxy=∑ x i yi−(∑ x i )(∑ yi)n

¿365 , 027−(548 )(5 , 782 )12

=100 , 982. 33

SSxx=∑ x i2−

(∑ x i )2

n

¿34 , 978−(548 )2

12=9 , 952. 667

b1=SSxy

SSxx

=100 , 982. 339 ,952. 667

=10. 1463

b0= y−b1 x=(578212 )−10 . 1463(548

12 )=18 . 4875

13-4


b. is the estimated increase in mean labor cost (10.1463) for every 1 unit increase in the batch size.

is the estimated mean labor cost (18.4875) when batch size = 0; no.

c.

d. = 18.4880 + 10.1463(60) = 627.266

LO1, LO2

13.8 a. MINITAB output

Regression Analysis: Sale Price versus Size

The regression equation isSale Price = 48.02 + 5.70 Size

b1 = SSxy / SSxx = 1149.18 / 201.6 = 5.70.

b0 = y-bar – b1*x-bar = 155.19 – 5.70*18.8 = 48.03 which is within rounding 48.02.

b. is the estimated increase in mean sales price (5.700) for every hundred square foot increase in home size.

is the estimated mean sales price when square footage = 0. No, the interpretation of makes no practical sense.

c. y=48 . 02+5 .700 x .

d. = 48.02 + 5.700 (20) = 162.02.

That is, $162,020.

LO1, LO2

13.9 (1) Mean of error terms = 0(2) Constant variance(3) Normality(4) Independence

LO3

13-5


13.10 That is, the constant variance and standard deviation of the error term populations.

LO3

13.11

LO3

13.12

LO3

13.13

LO3

13.14s2=2. 8059

28=. 1002 , s=√s2=. 3166

LO3

13.15

s2=746 .762410

=74 .67624 , s=8 .64154

LO3

13.16

LO3

13.17

13-6

s2=SSEn−2

=2 .5688−2

=0 . 428

s=√s2=√0 . 428=0 .654

s2=SSEn−2

=1 .4387−2

=0 .2876

s=√s2=√0 . 2876=0 . 5363

s2=SSEn−2

=191 .701711−2

=21 .3002

s=√s2=√21 .30018=4 . 61521

s2=SSEn−2

=896 .810−2

=112.1

s=√s2=√112.1=10 .58773

s2=SSEn−2

=222 .824210−2

=27 . 8530

s=√s2=√27 .8530=5 .2776


LO3

13.18 Strong ( = .05) or very strong ( = .01) evidence that the regression relationship is significant.

LO4

13.19 Explanations will vary.

LO4

13.20 a. = 15.8379 = -0.1279

b. SSE = 2.5679 s2 = 0.4280 s = 0.6542

c.sb1 = 0.0175 t = -7.3277

t = / sb1 = -0.1279 /0.0175 = -7.3277

d. df = 6 t.025 = 2.447 Reject , Strong evidence of a significant relationship between x and y.

e. t.005 = 3.707 Reject , Very strong evidence of a significant relationship between x and y.

f. p-value =.0003 Reject at all , Extremely strong evidence of a significant relationship between x and y.

g. 95% Cl: [ ± t.025 sb1 ] = -0.1279 ± (2.447)(0.0175) = [-0.1706, -0.0852]

We are 95% confident that the average fuel consumption decreases by between 0.0852 MMcf and 0.1706 MMcf for each 1 degree increase in monthly temperature.

h. 99% Cl: [ ± t.005 sb1 ] = -0.1279 ± (3.707)(0.0175) = [-0.1928, -0.0630]

We are 99% confident that the average fuel consumption decreases by between 0.0630 MMcf and 0.1928 MMcf for each 1 degree increase in monthly temperature.

13-7


i.sb0 = 0.8018 t = 19.7535

t = / sb0 = 15.8379 / 0.8018 = 19.7535

j. p-value < 0.000 Reject at all , Extremely strong evidence that the y-intercept is significant.

k.

LO2, LO3, LO4

13.21 a. = 14.816 = 5.7066

b. SSE = 1.438 s2 = .288 s = .5363

c.sb1 = .3953 t = 14.44

t = / sb1 = 5.7066 /.3953 = 14.44

d. df = 5 t.025 = 2.571 Reject , Strong evidence of a significant relationship between x and y.

e. t.005 = 4.032 Reject , Very strong evidence of a significant relationship between x and y.

f. p-value =.000 Reject at all , Extremely strong evidence of a significant relationship between x and y.

g. 95% Cl: [ ± t.025 sb1 ] = 5.7066 ± (2.571)(.3953) = [4.690, 6.723]

We are 95% confident that the mean starting salary increases by between $4690 and $6723 for each 1.0 increase in GPA.

h. 99% Cl: [ ± t.005 sb1 ] = 5.7066 ± (4.032)(.3953) = [4.113, 7.300]

We are 99% confident that the mean starting salary increases by between $4113 and $7300 for each 1.0 increase in GPA.

i.sb0 = 1.235 t = 12.00

t = / sb0 = 14.816 / 1.235 = 12.00

13-8

sb1=

s

√SSxx

=0 .6542

√1404 .355=0 . 01746

sb0=s√1

n+ x2

SSxx

=. 0 . 6542√18

+(43 .98 )2

1404 .355=0 .8018


j. p-value = .000 Reject at all , Extremely strong evidence that the y-intercept is significant.

k.

LO2, LO3, LO4

13.22 a. = 11.4641 = 24.6022

b. SSE = 191.7017s2 = 21.3002 s = 4.615

c.sb1 = .8045 t = 30.580

t = / sb1 = 24.602 /.8045 = 30.580

d. df = 9 t.025 = 2.262 Reject , strong evidence of a significant relationship between x and y.

e. t.005 = 3.250 Reject , very strong evidence of a significant relationship between x and y.

f. p-value = .000 Reject at all , extremely strong evidence of a significant relationship between x and y.

g. [24.6022 ± 2.262(.8045)] = [22.782, 26.422]

h. [24.6022 ± 3.250(.8045)] = [21.987,27.217]

i.sb0 = 3.4390 t = 3.334

t = / sb0 = 11.464 / 3.439 = 3.334

j. p-value = .0087 Reject at all except .001

k.

LO2, LO3, LO4

13-9

sb1=

s

√SSxx

=.5363

√1.8407=. 3953

sb0=s√1

n+ x2

SSxx

=. 5363√17

+(3 .0814 )2

1 .8407=1. 235

sb1=

s

√SSxx

=4 .61521

√32. 909=. 8045

sb0=s√1

n+ x2

SSxx

=4 . 61521√111

+3. 9092

32. 909=3 . 439


13.23 See the solutions to 13.20 for guidance.

a.

b. SSE=2 .806 , s2=. 100 , s=. 3166

c.sb1

=. 2585 , t=10. 31

d. Reject .

e. Reject .

f. p-value = less than .001; reject at each value of

g. [2.665 ± 2.048(.2585)] = [2.136, 3.194]

h. [2.665 ± 2.763(.2585)] = [1.951, 3.379]

i.sb0

=. 0799 , t=97 . 82

j. p-value = less than .001; reject .

k.

LO2, LO3, LO4


a.

b. SSE = 746.7624, s2=74 . 67624 , s = 8.642

c.sb1

=. 0866 , t=117 .1344

d. Reject .

e. Reject .

f. p-value = .000; reject at each value of

g. [10.1463 ± 2.228(.0866)] = [9.953, 10.339]

h. [10.1463 ± 3.169(.0866)] = [9.872, 10.421]

i.

13-10

sb1=

s

√SSxx

=.31656

√1.49967=. 2585

sb0=s√1

n+ x2

SSxx

=. 31656√130

+. 21332

1 . 49967=. 079883


j. p-value = .003; fail to reject at = .001. Reject at all other values of

k.

LO2, LO3, LO4


a. b0 = 48.02 = 5.7003

b. SSE = 896.8 s2 = 112.1 s = 10.588

c.sb1 = .7457 t = 7.64

t = / sb1 = 5.7003 /.7457 = 7.64

d. df = 8 t.025 = 2.306 Reject

e. t.005 = 3.355 Reject

f. p-value = .000 Reject at all

g. [3.9807,7.4199]

h. [3.198,8.202]

i.sb0 = 14.41 t = 3.33

t = b0 / sb0 = 48.02 / 14.41 = 3.33

j. p-value = .010 Reject at all except .01 and .001

k.

LO2, LO3, LO4

13-11

sb1=

s

√SSxx

=8 .64154

√9952.667=. 086621

sb0=s√1

n+ x2

SSxx

=8. 64154√112

+45. 6672

9952 .667=4 .67658

sb1=

s

√SSxx

=10 .588

√201. 6=.7457

sb0=s√1

10+ x2

SSxx

=10 . 588√110

+18 . 82

201.6=14 . 41


13.26 Find sb1 from Minitab

The regression equation issales = 66.2 + 4.43 ad exp

Predictor Coef SE Coef T PConstant 66.212 5.767 11.48 0.000Ad exp 4.4303 0.5810 7.62 0.000

95% C.I. for β1 [ 4.4303 2.306(.5810) ] = [3.091,5.770]

LO4

13.27 a. b1 = 1.2731. If MeanTaste goes up by 1 MeanPreference will go up by 1.2731.

b. (0.9885, 1.5577). We are 95% confident that this interval contains the true slope.

LO2, LO4

13.28 A confidence interval is for the mean value of y. A prediction interval is for an individual value of y.

LO5

13.29 The distance between xo and x , the average of the previously observed values of x.

LO5

13.30 a. 10.721, [10.130, 11.312]

b. 10.721, [9.015, 12.427]

c. dv = 1/8 + (40-43.98)2 / 1404.355 = 0.1363; dv = (0.241 / 0.6542)2 = 0.1357

d. CI: 15.84 -0.1279 * 40 ± 2.447*0.6542*sqrt(0.1363) = [10.13, 11.31]

PI: 15.84 -0.1279 * 40 ± 2.447*0.6542*sqrt(1.1363) = [9.01, 12.43]

e. Since we are predicting fuel consumption for one day when the average temperature is 40 degrees we must use the prediction interval. Since 9.01 < 9.595 and 12.43 > 11.847 the city cannot be 95% confident it will not pay a fine. For the city to be at least 95% confident the PI would have to be inside the interval [9.595, 11.847].

LO5

13-12


13.31 a. 33.362, [32.813, 33.911]

b. 33.362, [31.878, 34.846]

c.Distance Value=1

7+

(3. 25−3. 0814 )2

1 .8407=. 1583

d. [33.362 ± 2.571(.5363)√ .1583 ] = [32.813, 33.911]

[33.362 ± 2.571(.5363)√1+.1583 ] = [31.878, 34.846]

LO5

13.32 a. 109.873, [106.721, 113.025]

b. 109.873, [98.967, 120.779]

c. We have x = 4, x=3 .90 , SSxx=32 .90 , n=11

distance value = 1

11+(4−3 . 90 )2

32 .90=0 . 090657961

So confidence interval is:

109 . 873±(2. 262)( 4 . 615)√0. 090657961= [106 . 729 ,113 .016 ]this compares (within rounding) to the computer generated output.

For the prediction interval with the same quantities we get

109 . 873±(2. 262)( 4 . 615)√1 . 090657961

= [98.971, 120.775] which also compares within rounding.

d. 113 minutes

LO5

13.33 a. 8.0806; [7.948, 8.213]

b. 8.0806; [7.419, 8.743]

c. dv = (.0648/.316561)2 = .0419

d.s√dist=. 065 , s=. 3166 , dist=(. 065

. 3166 )2

=. 04215

99% C.I.: [8.0806 ± 2.763(.065)] = [7.9016, 8.2596]

13-13


99% P.I.: [8 .0806±2.763( . 3166)√1. 04215 ] = [7 .1877 , 8. 9735 ]

e. (1) 8.4804; [8.360, 8.600](2) 8.4804; [7.821, 9.140]

(3) s√dist=. 059 , s=. 3166 , dist=(. 059

. 3166 )2

=. 03427

99% C.I.: [8.4804 ± 2.763(.059)] = [8.3857, 8.5251]

99% P.I.: [8 .4804±2 .763 (. 3166 )√1 . 03473 ] = [7 .5909 ,9 . 3699 ]

LO5

13.34 a. 627.26, [621.05, 633.47]

b. 627.26, [607.03, 647.49]

c.s√dist=2. 7868 , s=8 . 642, dist=( 2 . 79

8 . 642 )2

=.104000

99% C.I.: [627.26 ± 3.169(2.79)] = [(618.42, 636.10)]

99% P.I.: [627 . 26±3.169(8 .642)√1 .104227 ] = [598 . 48 , 656 .04 ]

LO5

13.35 a. 162.03, [154.04, 170.02]

b. 162.03, [136.34, 187.72]

c. Prediction interval, because it deals with individuals, not an average.

LO5

13.36 Total variation: measures the total amount of variation exhibited by the observed values of y. Unexplained variation: measures the amount of variation in the values of y that is not explained by the model (predictor variable).Explained variation: measures the amount of variation in the values of y that is explained by the predictor variable.

LO6

13.37 Proportion of the total variation in the n observed values of y that is explained by the simple linear regression model.

LO6

13.38 Explained variation = 25.549 – 2.568 = 22.981

r2 = 22.981 / 25.549 = 0.899

r = +sqrt(0.977)*the sign of b1 = -0.948

13-14


89.9% of the variation in fuel consumption can be explained by variation in average temperature.

LO6


r2 = 59.942 / 61.38 = 0.977

r = +sqrt(0.977) = 0.988

97.7% of the variation in starting salary can be explained by variation in GPA.

LO6

13.40 Explained variation = 20,110.5445 – 191.7017 = 19918.8428

r2 = 19918.8428 / 20110.5445 = 0.990

r = +sqrt(0.990) = 0.995

99% of the variation in service time can be explained by variation in number of copiers repaired.

LO6


r2 = 10.653 / 13.459 = 0.792

r = +sqrt(0.792) = 0.890

79.2% of the variation in demand can be explained by variation in price differential.

LO6

13.42 Explained variation = 1,025,339.6667 – 746.7624 = 1,024,592.904

r2 = 1024592.904 / 1025339.6667 == 0.999

r = +sqrt(0.999) = 0.9995

99.9% of the variation in direct labor can be explained by variation in batch size.

LO6


r2 = 6550.7 / 7447.5 = 0.88

r = +sqrt(0.88) = 0.938

88% of the variation in sales price can be explained by variation in square footage.

LO6

13.44 is the actual (unknown) value of the correlation between two variables.

LO7

13.45 Calculate t =

r √n−2

√1−r2

and obtain its associated p-value. If p-value < then you reject.

LO7

13-15


13.46 Reject H0 at all four values of .

LO7

13.47 Reject H0 at all four values of .

LO7

13.48 .

LO8

13.49 t–test on

LO8

13.50 a. F = 22.9808 / (2.5679 / 6)) = 53.6949

b. F.05 = 5.99 df1 = 1, df2 = 6

Since 53.6949 > 5.99, reject H0 with strong evidence of a significant relationship between x and y.

c. F.01 = 13.75 df1 = 1, df2 = 6

Since 53.6949 > 13.75, reject H0 with very strong evidence of a significant relationship between x and y.

d. p-value =0.0003; Reject H0 at all levels of , extremely strong evidence of a significant relationship between x and y.

e. t2 = (-7.33)2 = 53.7289 (approximately equals F = 53.6949)

(t.025)2 = (2.447)2 = 5.99 = F.05

LO8

13.51 a. F = 59.942 / (1.438 / 5) = 208.39

b. F.05 = 6.61 df1 = 1, df2 = 5


c. F.01 = 16.26 df1 = 1, df2 = 5


d. p-value =.000; Reject H0 at all levels of , extremely strong evidence of a significant relationship between x and y.

e. t2 = (14.44)2 = 208.51 (approximately equals F = 208.39)

(t.025)2 = (2.571)2 = 6.61 = F.05

13-16


13-17


LO8

13.52 a. F = 19918.844 / (191.7017 / 9) = 935.149

b. F.05 = 5.12 df1 = 1, df2 = 9


c. F.01 = 10.56 df1 = 1, df2 = 9


d. p-value =less than .001; Reject H0 at all levels of , extremely strong evidence of a significant relationship between x and y.


(t.025)2 = (2.262)2 = 5.12 = F.05

LO8

13.53 a. F = 106.303

b. F.05 =4.20, reject (df1 = 1, df2 = 28). Strong evidence of a significant relationship between x and y.

c. F.01 =7.64, reject (df1 = 1, df2 = 28). Very strong evidence of a significant relationship between x and y.

d. p-value = less than .001, reject . Extremely strong evidence of a significant relationship between x and y.

e. (within rounding error)

(t.025)2 = 4.19 = F.05

LO8

13.54 a. F = 13,720.47

b. Reject .

c. Reject .

d. p-value = .000; reject .

e. (within rounding error)

LO8

13.55 a. F = 6550.7 / (896.8 / 8) = 58.43

13-18


13-19


b. F.05 = 5.32 df1 = 1, df2 = 8

Since 58.43 > 5.32, reject H0.

c. F.01 = 11.26 df1 = 1, df2 = 8

Since 58.43 > 11.3, reject H0.

d. p-value =.000; Reject H0 at all levels of


(t.025)2 = (2.306)2 = 5.32 = F.05

LO8

13.56 They should be plotted against the independent variable and against y . Funneling or curved patterns indicate violations of the regression assumptions.

LO9

13.57 Create a histogram, stem-and-leaf, and normal plot.

LO9

13.58 Transforming the dependent variable.

LO9

13.59 Approximate horizontal band appearance. No violations indicated.

LO9

13.60 Possible violations of the normality and constant variance assumptions.

LO9

13.61 No.

LO9

13.62 a.

b. No

13-20

3( i)−13 n+1

=3( 4 )−133+1

=.3235

. 5000−. 3235=. 1765 ,⇒ z=−. 46

3( i)−13 n+1

=3(10 )−133+1

=. 8529

. 8529−. 5000=. 3529 ,⇒ z=1 . 05


LO9

13.63 The residual plot has somewhat of a cyclical appearance. Since d =.473 is less than dL, 05

= 1.27, we conclude there is positive autocorrelation and since 4 – .473 = 3.527 and this is greater than dU,.05 = 1.45 we conclude that there is not negative autocorrelation.

LO9

13.64 The plot of the residuals shows no pattern to indicate a non-constant variance and are centered around 0.

LO9

13.65 a. ln yt = 2.07012 + 0.25688t

ln y16 = 2.07012 + 025688(16) = 6.1802

b. e6.1802 = 483.09

e5.9945 = 401.22

e6.3659 = 581.67

c. Growth rate = e0.25688 = 1.293

This means the growth rate is expected to be 29.3% per year.

LO9

13.66 a. Yes; see the plot in part c.

b.

c.

13-21

260

240

220

y

x

2.0 2.4 2.8 3.2


d. p–value = .000, reject , significant

e.

LO1, LO2, LO4, LO5

13.67 a. b1 = –6.4424 For every unit increase in width difference, the mean number of accidents are reduced by 6.4 per 100 million vehicles.

b. p-value = .000 Reject H0 at all levels of

c. r2 = .984 98.4% of the variation in accidents is explained by the width difference.

LO2, LO4, LO6

13.68 a. No

b. Possibly not; Don’t take up smoking

LO1, LO2

13-22


13.69 For aggressive stocks a 95% confidence interval for is [ . 0163±t. 025( . 003724 ) ]=[ . 0163±2. 365( . 003724 ) ]=[ . 00749 , .02512 ] where is based on 7 degrees of freedom.We are 95% confident that the effect of a one-month increase in the return length time for an

aggressive stock is to increase the mean value of the average estimate of by between .00749 and .02512.

For defensive stocks a 95% confidence interval for is [–0.00462 ± 2.365(.00084164)] = [–.00661, –.00263].

For neutral stocks a 95% confidence interval for is [.0087255 ± 2.365(.001538)] = [.005088, .01236].

LO2, LO4

13.70 a. Using Figure 13.42, there does seem to be a negative relationship between temperature and o-ring failure.

b. The temperature of 31 was outside the experimental region.

LO1, LO2

13.71 a. There is a relationship since F = 21.13 with a p-value of .0002.

b. b1 = 35.2877, [19.2202,51.3553]

LO4

Internet Exercise -- Answers will vary depending on when data was obtained.

13-23


13.72

The regression equation isGMAT = 184 + 141 GPA

Predictor Coef SE Coef T PConstant 184.27 84.63 2.18 0.034GPA 141.08 25.36 5.56 0.000

S = 21.50 R–Sq = 39.2% R–Sq(adj) = 37.9%

Analysis of Variance

Source DF SS MS F PRegression 1 14316 14316 30.96 0.000Residual Error 48 22197 462Total 49 36513

Predicted Values for New Observations

New Obs Fit SE Fit 95.0% CI 95.0% PI 1 678.06 5.16 ( 667.68, 688.45) ( 633.60, 722.53)

Values of Predictors for New Observations

New Obs GPA1 3.50

LO1, LO2, LO6, LO8

13-24