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 leastsquares 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 Learn how to use a residual plot to make a judgement as to the validity of the regression assumptions recognize outliers and identify influential observations
7 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
8 Be able to compute the sample correlation coefficient from the regression analysis output
9 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
standardized residual plot
outlier
influential observation
leverage
Solutions
1 a
b There appears to be a 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 Summations needed to compute the slope and yintercept are
e
2 a
b There appears to be a 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 Summations needed to compute the slope and yintercept are
e
3 a
b Summations needed to compute the slope and yintercept are
c
4 a
b There appears to be a 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 Summations needed to compute the slope and yintercept are
e pounds
5 a
b There appears to be a 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
Summations needed to compute the slope and yintercept are
d A one million dollar increase in media expenditures will increase case sales by approximately 1442 million
e
6 a
b There appears to be a linear relationship between x and y
c Summations needed to compute the slope and yintercept are
d A one percent increase in the percentage of flights arriving on time will decrease the number of complaints per 100000 passengers by 007
e
7 a
b Let x DJIA and y S&P Summations needed to compute the slope and yintercept are
c or approximately 1500
8 a Summations needed to compute the slope and yintercept are
b Increasing the number of times an ad is aired by one will increase the number of household exposures by approximately 307 million
c
9 a
b Summations needed to compute the slope and yintercept are
c
10 a
b Let x performance score and y overall rating Summations needed to compute the slope and yintercept are
c or approximately 84
11 a
b There appears to be a linear relationship between the variables
c The summations needed to compute the slope and the yintercept are
d
12 a
b There appears to be a positive linear relationship between the number of employees and the revenue
c Let x number of employees and y revenue Summations needed to compute the slope and yintercept are
d
13 a
b The summations needed to compute the slope and the yintercept are
c or approximately 13080
The agent＇s request for an audit appears to be justified
14 a
b The summations needed to compute the slope and the yintercept are
c
15 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 80 124 676
b r2 SSRSST 67680 845
The least squares line provided a very good fit 845 of the variability in y has been explained by the least squares line
c
16 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 11480 633 10847
b r2 SSRSST 1084711480 945
The least squares line provided an excellent fit 945 of the variability in y has been explained by the estimated regression equation
c
Note the sign for r is negative because the slope of the estimated regression equation is negative
(b1 188)
17 The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 112 53 59
r2 SSRSST 59112 527
We see that 527 of the variability in y has been explained by the least squares line
18 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 335000 8513514 24986486
b r2 SSRSST 24986486335000 746
We see that 746 of the variability in y has been explained by the least squares line
c
19 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 4758210 754714 4003496
b r2 SSRSST 40034964758210 84
We see that 84 of the variability in y has been explained by the least squares line
c
20 a Let x income and y home price Summations needed to compute the slope and yintercept are
b The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 1137309 – 201737 935572
r2 SSRSST 9355721137309 82
We see that 82 of the variability in y has been explained by the least squares line
c or approximately 173500
21 a The summations needed in this problem are
b 760
c The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 564833333 23333333 5415000
r2 SSRSST 5415000564833333 9587
We see that 9587 of the variability in y has been explained by the estimated regression equation
d
22 a The summations needed in this problem are
b The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 1998 12724495 7255505
r2 SSRSST 72555051998 03631
Approximately 37 of the variability in change in executive compensation is explained by the twoyear change in the return on equity
c
It reflects a linear relationship that is between weak and strong
23 a s2 MSE SSE (n 2) 124 3 4133
b
c
d
t025 3182 (3 degrees of freedom)
Since t 404 > t05 3182 we reject H0 b1 0
e MSR SSR 1 676
F MSR MSE 676 4133 1636
F05 1013 (1 degree of freedom numerator and 3 denominator)
Since F 1636 > F05 1013 we reject H0 b1 0
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
676
1
676
1636
Error
124
3
4133
Total
800
4
24 a s2 MSE SSE (n 2) 633 3 211
b
c
d
t025 3182 (3 degrees of freedom)
Since t 718 < t025 3182 we reject H0 b1 0
e MSR SSR 1 847
F MSR MSE 10847 211 5141
F05 1013 (1 degree of freedom numerator and 3 denominator)
Since F 5141 > F05 1013 we reject H0 b1 0
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
10847
1
10847
5141
Error
633
3
211
Total
11480
4
25 a s2 MSE SSE (n 2) 530 3 177
b
t025 3182 (3 degrees of freedom)
Since t 182 < t025 3182 we cannot reject H0 b1 0 x and y do not appear to be related
c MSR SSR1 590 1 590
F MSRMSE 590177 333
F05 1013 (1 degree of freedom numerator and 3 denominator)
Since F 333 < F05 1013 we cannot reject H0 b1 0 x and y do not appear to be related
26 a s2 MSE SSE (n 2) 8513514 4 2128379
t025 2776 (4 degrees of freedom)
Since t 343 > t025 2776 we reject H0 b1 0
b MSR SSR 1 24986486 1 24986486
F MSR MSE 24986486 2128379 1174
F05 771 (1 degree of freedom numerator and 4 denominator)
Since F 1174 > F05 771 we reject H0 b1 0
c
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
24986486
1
24986486
1174
Error
8513514
4
2128379
Total
335000
5
27 The sum of squares due to error and the total sum of squares are
SSE SST 2442
Thus SSR SST SSE 2442 170 2272
MSR SSR 1 2272
SSE SST SSR 2442 2272 170
MSE SSE (n 2) 170 8 2125
F MSR MSE 2272 2125 10692
F05 532 (1 degree of freedom numerator and 8 denominator)
Since F 10692 > F05 532 we reject H0 b1 0
Years of experience and sales are related
28 SST 41173 SSE 16137 SSR 25036
MSR SSR 1 25036
MSE SSE (n 2) 16137 13 12413
F MSR MSE 25036 12413 2017
F05 467 (1 degree of freedom numerator and 13 denominator)
Since F 2017 > F05 467 we reject H0 b1 0
29 SSE 23333333 SST 564833333 SSR 5415000
MSE SSE(n 2) 23333333(6 2) 5833333
MSR SSR1 5415000
F MSR MSE 5415000 5833325 9283
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
541500000
1
5415000
9283
Error
23333333
4
5833333
Total
564833333
5
F05 771 (1 degree of freedom numerator and 4 denominator)
Since F 9283 > 771 we reject H0 b1 0 Production volume and total cost are related
30 Using the computations from Exercise 22
SSE 12724495 SST 1998 SSR 7255505
458339286
t025 2571
Since t 169 < 2571 we cannot reject H0 b1 0
There is no evidence of a significant relationship between x and y
31 SST 1137309 SSE 201737 SSR 935572
MSR SSR 1 935572
MSE SSE (n 2) 201737 16 1260856
F MSR MSE 935572 1260856 7420
F01 853 (1 degree of freedom numerator and 16 denominator)
Since F 7420 > F01 853 we reject H0 b1 0
32 a s 2033
b
106 ± 3182 (111) 106 ± 353
or 707 to 1413
c
d
106 ± 3182 (232) 106 ± 738
or 322 to 1798
33 a s 1453
b
2469 ± 3182 (68) 2469 ± 216
or 2253 to 2685
c
d
2469 ± 3182 (161) 2469 ± 512
or 1957 to 2981
34 s 133
228 ± 3182 (85) 228 ± 270
or 40 to 498
228 ± 3182 (158) 228 ± 503
or 227 to 731
35 a s 14589
203378 ± 2776 (6854) 203378 ± 19027
or 184351 to 222405
b
203378 ± 2776 (16119) 203378 ± 44746
or 158632 to 248124
36 a
b s 35232
80859 ± 2160 (1055) 80859 ± 2279
or 7858 to 8314
c
80859 ± 2160 (3678) 80859 ± 7944
or 7292 to 8880
37 a
s2 188 s 137
1308 ± 2571 (52) 1308 ± 134
or 1174 to 1442 or 11740 to 14420
b sind 147
1308 ± 2571 (147) 1308 ± 378
or 930 to 1686 or 9300 to 16860
c Yes 20400 is much larger than anticipated
d Any deductions exceeding the 16860 upper limit could suggest an audit
38 a
b
s2 MSE 5833333 s 24152
504667 ± 4604 (26750) 504667 ± 123157
or 381510 to 627824
c Based on one month 6000 is not out of line since 381510 to 627824 is the prediction interval However a sequence of five to seven months with consistently high costs should cause concern
39 a Summations needed to compute the slope and yintercept are
b SST 3906514 SSE 4145141 SSR 34920000
r2 SSRSST 3492000039065141 0894
The estimated regression equation explained 894 of the variability in y a very good fit
c s2 MSE 41451418 518143
27063 ± 2262 (886) 27063 ± 2004
or 25059 to 29067
d
27063 ± 2262 (2442) 27063 ± 5524
or 21539 to 32587
40 a 9
b 200 + 721x
c 13626
d SSE SST SSR 519841 415873 103968
MSE 1039687 14853
F MSR MSE 415873 14853 2800
F05 559 (1 degree of freedom numerator and 7 denominator)
Since F 28 > F05 559 we reject H0 B1 0
e 200 + 721(50) 3805 or 380500
41 a 61092 + 8951x
b
t025 2306 (1 degree of freedom numerator and 8 denominator)
Since t 601 > t025 2306 we reject H0 B1 0
c 61092 + 8951(25) 2849 or 2849 per month
42 a 800 + 500x
b 30
c F MSR MSE 68286821 8317
F05 420 (1 degree of freedom numerator and 28 denominator)
Since F 8317 > F05 420 we reject H0 B1 0
Branch office sales are related to the salespersons
d 80 + 50 (12) 680 or 680000
43 a The Minitab output is shown below
The regression equation is
Price 118 + 218 Income
Predictor Coef SE Coef T P
Constant 1180 1284 092 0380
Income 21843 02780 786 0000
S 6634 RSq 861 RSq(adj) 847
Analysis of Variance
Source DF SS MS F P
Regression 1 27179 27179 6175 0000
Residual Error 10 4401 440
Total 11 31580
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 7579 247 ( 7029 8128) ( 6002 9156)
b r2 861 The least squares line provided a very good fit
c The 95 confidence interval is 7029 to 8128 or 70290 to 81280
d The 95 prediction interval is 6002 to 9156 or 60020 to 91560
44 ab The scatter diagram shows a linear relationship between the two variables
c The Minitab output is shown below
The regression equation is
Rental 371 0779 Vacancy
Predictor Coef SE Coef T P
Constant 37066 3530 1050 0000
Vacancy 07791 02226 350 0003
S 4889 RSq 434 RSq(adj) 398
Analysis of Variance
Source DF SS MS F P
Regression 1 29289 29289 1226 0003
Residual Error 16 38237 2390
Total 17 67526
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 1759 251 ( 1227 2290) ( 594 2923)
2 2826 142 ( 2526 3126) ( 1747 3905)
Values of Predictors for New Observations
New Obs Vacancy
1 250
2 113
d Since the pvalue 0003 is less than a 05 the relationship is significant
e r2 434 The least squares line does not provide a very good fit
f The 95 confidence interval is 1227 to 2290 or 1227 to 2290
g The 95 prediction interval is 1747 to 3905 or 1747 to 3905
45 a
b The residuals are 348 247 483 16 and 522
c
With only 5 observations it is difficult to determine if the assumptions are satisfied However the plot does suggest curvature in the residuals that would indicate that the error term assumptions are not satisfied The scatter diagram for these data also indicates that the underlying relationship between x and y may be curvilinear
d
The standardized residuals are 132 59 111 40 149
e The standardized residual plot has the same shape as the original residual plot The curvature observed indicates that the assumptions regarding the error term may not be satisfied
46 a
b
The assumption that the variance is the same for all values of x is questionable The variance appears to increase for larger values of x
47 a Let x advertising expenditures and y revenue
b SST 1002 SSE 31028 SSR 69172
MSR SSR 1 69172
MSE SSE (n 2) 31028 5 620554
F MSR MSE 69172 620554 1115
F05 661 (1 degree of freedom numerator and 5 denominator)
Since F 1115 > F05 661 we conclude that the two variables are related
c
d The residual plot leads us to question the assumption of a linear relationship between x and y Even though the relationship is significant at the 05 level of significance it would be extremely dangerous to extrapolate beyond the range of the data
48 a
b The assumptions concerning the error term appear reasonable
49 a Let x return on investment (ROE) and y priceearnings (PE) ratio
b
c There is an unusual trend in the residuals The assumptions concerning the error term appear questionable
50 a The MINITAB output is shown below
The regression equation is
Y 661 + 0402 X
Predictor Coef Stdev tratio p
Constant 6610 3206 206 0094
X 04023 02276 177 0137
s 1262 Rsq 385 Rsq(adj) 261
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 4972 4972 312 0137
Error 5 7957 1591
Total 6 12929
Unusual Observations
Obs X Y Fit StdevFit Residual StResid
1 135 14500 12042 487 2458 211R
R denotes an obs with a large st resid
The standardized residuals are 211 108 14 38 78 04 41
The first observation appears to be an outlier since it has a large standardized residual
b
24+
*
STDRESID
12+
*
00+ *
* *
*
12+ *
++++++YHAT
1100 1150 1200 1250 1300 1350
The standardized residual plot indicates that the observation x 135y 145 may be an outlier note that this observation has a standardized residual of 211
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 observation x 135y 145 may be an outlier the implication is that for simple linear regression an outlier can be identified by looking at the scatter diagram
51 a The Minitab output is shown below
The regression equation is
Y 130 + 0425 X
Predictor Coef Stdev tratio p
Constant 13002 2396 543 0002
X 04248 02116 201 0091
s 3181 Rsq 402 Rsq(adj) 302
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 4078 4078 403 0091
Error 6 6072 1012
Total 7 10150
Unusual Observations
Obs X Y Fit StdevFit Residual StResid
7 120 2400 1810 120 590 200R
8 220 1900 2235 278 335 216RX
R denotes an obs with a large st resid
X denotes an obs whose X value gives it large influence
The standardized residuals are 100 41 01 48 25 65 200 216
The last two observations in the data set appear to be outliers since the standardized residuals for these observations are 200 and 216 respectively
b Using MINITAB we obtained the following leverage values
28 24 16 14 13 14 14 76
MINITAB identifies an observation as having high leverage if hi > 6n for these data 6n 68 75 Since the leverage for the observation x 22 y 19 is 76 MINITAB would identify observation 8 as a high leverage point Thus we conclude that observation 8 is an influential observation
c
240+ *
Y
200+ *
*
*
160+ *
*
*
120+ *
++++++X
00 50 100 150 200 250
The scatter diagram indicates that the observation x 22 y 19 is an influential observation
52 a The Minitab output is shown below
The regression equation is
Amount 409 + 0196 MediaExp
Predictor Coef SE Coef T P
Constant 4089 2168 189 0096
MediaExp 019552 003635 538 0001
S 5044 RSq 783 RSq(adj) 756
Analysis of Variance
Source DF SS MS F P
Regression 1 73584 73584 2893 0001
Residual Error 8 20351 2544
Total 9 93935
Unusual Observations
Obs MediaExp Amount Fit SE Fit Residual St Resid
1 120 3630 2755 330 875 230R
R denotes an observation with a large standardized residual
b Minitab identifies observation 1 as having a large standardized residual thus we would consider observation 1 to be an outlier
53 a The Minitab output is shown below
The regression equation is
Exposure 86 + 771 Aired
Predictor Coef SE Coef T P
Constant 855 2165 039 0703
Aired 77149 05119 1507 0000
S 3488 RSq 966 RSq(adj) 962
Analysis of Variance
Source DF SS MS F P
Regression 1 276434 276434 22717 0000
Residual Error 8 9735 1217
Total 9 286169
Unusual Observations
Obs Aired Exposure Fit SE Fit Residual St Resid
1 950 7588 7244 320 344 246RX
R denotes an observation with a large standardized residual
X denotes an observation whose X value gives it large influence
b Minitab identifies observation 1 as having a large standardized residual thus we would consider observation 1 to be an outlier Minitab also identifies observation 1 as an influential observation
54 a The Minitab output is shown below
The regression equation is
Salary 707 + 000482 MktCap
Predictor Coef SE Coef T P
Constant 7070 1180 599 0000
MktCap 00048154 00008076 596 0000
S 3798 RSq 664 RSq(adj) 645
Analysis of Variance
Source DF SS MS F P
Regression 1 5129071 5129071 3555 0000
Residual Error 18 2596647 144258
Total 19 7725718
Unusual Observations
Obs MktCap Salary Fit SE Fit Residual St Resid
6 507217 33250 31495 3386 1755 102 X
17 120967 1162 12895 864 11733 317R
R denotes an observation with a large standardized residual
X denotes an observation whose X value gives it large influence
b Minitab identifies observation 6 as having a large standardized residual and observation 17 as an observation whose x value gives it large influence A standardized residual plot against the predicted values is shown below
55 No Regression or correlation analysis can never prove that two variables are casually related
56 The estimate of a mean value is an estimate of the average of all y values associated with the same x The estimate of an individual y value is an estimate of only one of the y values associated with a particular x
57 To determine whether or not there is a significant relationship between x and y However if we reject B1 0 it does not imply a good fit
58 a The Minitab output is shown below
The regression equation is
Price 926 + 0711 Shares
Predictor Coef SE Coef T P
Constant 9265 1099 843 0000
Shares 07105 01474 482 0001
S 1419 RSq 744 RSq(adj) 712
Analysis of Variance
Source DF SS MS F P
Regression 1 46784 46784 2322 0001
Residual Error 8 16116 2015
Total 9 62900
b Since the pvalue corresponding to F 2322 001 < a 05 the relationship is significant
c 744 a good fit The least squares line explained 744 of the variability in Price
d
59 a The Minitab output is shown below
The regression equation is
Options 383 + 0296 Common
Predictor Coef SE Coef T P
Constant 3834 5903 065 0529
Common 029567 002648 1117 0000
S 1104 RSq 919 RSq(adj) 912
Analysis of Variance
Source DF SS MS F P
Regression 1 15208 15208 12472 0000
Residual Error 11 1341 122
Total 12 16550
b approximately 406 million shares of options grants outstanding
c 919 a very good fit The least squares line explained 919 of the variability in Options
60 a The Minitab output is shown below
The regression equation is
IBM 0275 + 0950 S&P 500
Predictor Coef StDev T P
Constant 02747 09004 031 0768
S&P 500 09498 03569 266 0029
S 2664 RSq 470 RSq(adj) 403
Analysis of Variance
Source DF SS MS F P
Regression 1 50255 50255 708 0029
Error 8 56781 7098
Total 9 107036
b Since the pvalue 0029 is less than a 05 the relationship is significant
c r2 470 The least squares line does not provide a very good fit
d Woolworth has higher risk with a market beta of 125
61 a
b It appears that there is a positive linear relationship between the two variables
c The Minitab output is shown below
The regression equation is
High 239 + 0898 Low
Predictor Coef SE Coef T P
Constant 23899 6481 369 0002
Low 08980 01121 801 0000
S 5285 RSq 781 RSq(adj) 769
Analysis of Variance
Source DF SS MS F P
Regression 1 17923 17923 6418 0000
Residual Error 18 5027 279
Total 19 22949
d Since the pvalue corresponding to F 6418 000 < a 05 the relationship is significant
e 781 a good fit The least squares line explained 781 of the variability in high temperature
f
62 The MINITAB output is shown below
The regression equation is
Y 105 + 0953 X
Predictor Coef Stdev tratio p
Constant 10528 3745 281 0023
X 09534 01382 690 0000
s 4250 Rsq 856 Rsq(adj) 838
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 86005 86005 4762 0000
Error 8 14447 1806
Total 9 100453
Fit StdevFit 95 CI 95 PI
3913 149 ( 3569 4257) ( 2874 4952)
a 105 + 953 x
b Since the pvalue corresponding to F 4762 000 < a 05 we reject H0 b1 0
c The 95 prediction interval is 2874 to 4952 or 2874 to 4952
d Yes since the expected expense is 3913
63 a The Minitab output is shown below
The regression equation is
Defects 222 0148 Speed
Predictor Coef SE Coef T P
Constant 22174 1653 1342 0000
Speed 014783 004391 337 0028
S 1489 RSq 739 RSq(adj) 674
Analysis of Variance
Source DF SS MS F P
Regression 1 25130 25130 1133 0028
Residual Error 4 8870 2217
Total 5 34000
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 14783 0896 ( 12294 17271) ( 9957 19608)
b Since the pvalue corresponding to F 1133 028 < a 05 the relationship is significant
c 739 a good fit The least squares line explained 739 of the variability in the number of defects
d Using the Minitab output in part (a) the 95 confidence interval is 12294 to 17271
64 a There appears to be a negative linear relationship between distance to work and number of days absent
b The MINITAB output is shown below
The regression equation is
Y 810 0344 X
Predictor Coef Stdev tratio p
Constant 80978 08088 1001 0000
X 034420 007761 443 0002
s 1289 Rsq 711 Rsq(adj) 675
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 32699 32699 1967 0002
Error 8 13301 1663
Total 9 46000
Fit StdevFit 95 CI 95 PI
6377 0512 ( 5195 7559) ( 3176 9577)
c Since the pvalue corresponding to F 41967 is 002 < a 05 We reject H0 b1 0
d r2 711 The estimated regression equation explained 711 of the variability in y this is a reasonably good fit
e The 95 confidence interval is 5195 to 7559 or approximately 52 to 76 days
65 a Let X the age of a bus and Y the annual maintenance cost
The MINITAB output is shown below
The regression equation is
Y 220 + 132 X
Predictor Coef Stdev tratio p
Constant 22000 5848 376 0006
X 13167 1780 740 0000
s 7550 Rsq 873 Rsq(adj) 857
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 312050 312050 5475 0000
Error 8 45600 5700
Total 9 357650
Fit StdevFit 95 CI 95 PI
7467 298 ( 6780 8154) ( 5595 9339)
b Since the pvalue corresponding to F 5475 is 000 < a 05 we reject H0 b1 0
c r2 873 The least squares line provided a very good fit
d The 95 prediction interval is 5595 to 9339 or 55950 to 93390
66 a Let X hours spent studying and Y total points earned
The MINITAB output is shown below
The regression equation is
Y 585 + 0830 X
Predictor Coef Stdev tratio p
Constant 5847 7972 073 0484
X 08295 01095 758 0000
s 7523 Rsq 878 Rsq(adj) 862
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 32497 32497 5742 0000
Error 8 4528 566
Total 9 37025
Fit StdevFit 95 CI 95 PI
8465 367 ( 7619 9311) ( 6535 10396)
b Since the pvalue corresponding to F 5742 is 000 < a 05 we reject H0 b1 0
c 8465 points
d The 95 prediction interval is 6535 to 10396
67 a The Minitab output is shown below
The regression equation is
Audit 0471 +0000039 Income
Predictor Coef SE Coef T P
Constant 04710 05842 081 0431
Income 000003868 000001731 223 0038
S 02088 RSq 217 RSq(adj) 174
Analysis of Variance
Source DF SS MS F P
Regression 1 021749 021749 499 0038
Residual Error 18 078451 004358
Total 19 100200
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 08828 00523 ( 07729 09927) ( 04306 13349)
b Since the pvalue 0038 is less than a 05 the relationship is significant
c r2 217 The least squares line does not provide a very good fit
d The 95 confidence interval is 7729 to 9927
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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 leastsquares 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 Learn how to use a residual plot to make a judgement as to the validity of the regression assumptions recognize outliers and identify influential observations
7 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
8 Be able to compute the sample correlation coefficient from the regression analysis output
9 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
standardized residual plot
outlier
influential observation
leverage
Solutions
1 a
b There appears to be a 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 Summations needed to compute the slope and yintercept are
e
2 a
b There appears to be a 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 Summations needed to compute the slope and yintercept are
e
3 a
b Summations needed to compute the slope and yintercept are
c
4 a
b There appears to be a 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 Summations needed to compute the slope and yintercept are
e pounds
5 a
b There appears to be a 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
Summations needed to compute the slope and yintercept are
d A one million dollar increase in media expenditures will increase case sales by approximately 1442 million
e
6 a
b There appears to be a linear relationship between x and y
c Summations needed to compute the slope and yintercept are
d A one percent increase in the percentage of flights arriving on time will decrease the number of complaints per 100000 passengers by 007
e
7 a
b Let x DJIA and y S&P Summations needed to compute the slope and yintercept are
c or approximately 1500
8 a Summations needed to compute the slope and yintercept are
b Increasing the number of times an ad is aired by one will increase the number of household exposures by approximately 307 million
c
9 a
b Summations needed to compute the slope and yintercept are
c
10 a
b Let x performance score and y overall rating Summations needed to compute the slope and yintercept are
c or approximately 84
11 a
b There appears to be a linear relationship between the variables
c The summations needed to compute the slope and the yintercept are
d
12 a
b There appears to be a positive linear relationship between the number of employees and the revenue
c Let x number of employees and y revenue Summations needed to compute the slope and yintercept are
d
13 a
b The summations needed to compute the slope and the yintercept are
c or approximately 13080
The agent＇s request for an audit appears to be justified
14 a
b The summations needed to compute the slope and the yintercept are
c
15 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 80 124 676
b r2 SSRSST 67680 845
The least squares line provided a very good fit 845 of the variability in y has been explained by the least squares line
c
16 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 11480 633 10847
b r2 SSRSST 1084711480 945
The least squares line provided an excellent fit 945 of the variability in y has been explained by the estimated regression equation
c
Note the sign for r is negative because the slope of the estimated regression equation is negative
(b1 188)
17 The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 112 53 59
r2 SSRSST 59112 527
We see that 527 of the variability in y has been explained by the least squares line
18 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 335000 8513514 24986486
b r2 SSRSST 24986486335000 746
We see that 746 of the variability in y has been explained by the least squares line
c
19 a The estimated regression equation and the mean for the dependent variable are
The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 4758210 754714 4003496
b r2 SSRSST 40034964758210 84
We see that 84 of the variability in y has been explained by the least squares line
c
20 a Let x income and y home price Summations needed to compute the slope and yintercept are
b The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 1137309 – 201737 935572
r2 SSRSST 9355721137309 82
We see that 82 of the variability in y has been explained by the least squares line
c or approximately 173500
21 a The summations needed in this problem are
b 760
c The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 564833333 23333333 5415000
r2 SSRSST 5415000564833333 9587
We see that 9587 of the variability in y has been explained by the estimated regression equation
d
22 a The summations needed in this problem are
b The sum of squares due to error and the total sum of squares are
Thus SSR SST SSE 1998 12724495 7255505
r2 SSRSST 72555051998 03631
Approximately 37 of the variability in change in executive compensation is explained by the twoyear change in the return on equity
c
It reflects a linear relationship that is between weak and strong
23 a s2 MSE SSE (n 2) 124 3 4133
b
c
d
t025 3182 (3 degrees of freedom)
Since t 404 > t05 3182 we reject H0 b1 0
e MSR SSR 1 676
F MSR MSE 676 4133 1636
F05 1013 (1 degree of freedom numerator and 3 denominator)
Since F 1636 > F05 1013 we reject H0 b1 0
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
676
1
676
1636
Error
124
3
4133
Total
800
4
24 a s2 MSE SSE (n 2) 633 3 211
b
c
d
t025 3182 (3 degrees of freedom)
Since t 718 < t025 3182 we reject H0 b1 0
e MSR SSR 1 847
F MSR MSE 10847 211 5141
F05 1013 (1 degree of freedom numerator and 3 denominator)
Since F 5141 > F05 1013 we reject H0 b1 0
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
10847
1
10847
5141
Error
633
3
211
Total
11480
4
25 a s2 MSE SSE (n 2) 530 3 177
b
t025 3182 (3 degrees of freedom)
Since t 182 < t025 3182 we cannot reject H0 b1 0 x and y do not appear to be related
c MSR SSR1 590 1 590
F MSRMSE 590177 333
F05 1013 (1 degree of freedom numerator and 3 denominator)
Since F 333 < F05 1013 we cannot reject H0 b1 0 x and y do not appear to be related
26 a s2 MSE SSE (n 2) 8513514 4 2128379
t025 2776 (4 degrees of freedom)
Since t 343 > t025 2776 we reject H0 b1 0
b MSR SSR 1 24986486 1 24986486
F MSR MSE 24986486 2128379 1174
F05 771 (1 degree of freedom numerator and 4 denominator)
Since F 1174 > F05 771 we reject H0 b1 0
c
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
24986486
1
24986486
1174
Error
8513514
4
2128379
Total
335000
5
27 The sum of squares due to error and the total sum of squares are
SSE SST 2442
Thus SSR SST SSE 2442 170 2272
MSR SSR 1 2272
SSE SST SSR 2442 2272 170
MSE SSE (n 2) 170 8 2125
F MSR MSE 2272 2125 10692
F05 532 (1 degree of freedom numerator and 8 denominator)
Since F 10692 > F05 532 we reject H0 b1 0
Years of experience and sales are related
28 SST 41173 SSE 16137 SSR 25036
MSR SSR 1 25036
MSE SSE (n 2) 16137 13 12413
F MSR MSE 25036 12413 2017
F05 467 (1 degree of freedom numerator and 13 denominator)
Since F 2017 > F05 467 we reject H0 b1 0
29 SSE 23333333 SST 564833333 SSR 5415000
MSE SSE(n 2) 23333333(6 2) 5833333
MSR SSR1 5415000
F MSR MSE 5415000 5833325 9283
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Regression
541500000
1
5415000
9283
Error
23333333
4
5833333
Total
564833333
5
F05 771 (1 degree of freedom numerator and 4 denominator)
Since F 9283 > 771 we reject H0 b1 0 Production volume and total cost are related
30 Using the computations from Exercise 22
SSE 12724495 SST 1998 SSR 7255505
458339286
t025 2571
Since t 169 < 2571 we cannot reject H0 b1 0
There is no evidence of a significant relationship between x and y
31 SST 1137309 SSE 201737 SSR 935572
MSR SSR 1 935572
MSE SSE (n 2) 201737 16 1260856
F MSR MSE 935572 1260856 7420
F01 853 (1 degree of freedom numerator and 16 denominator)
Since F 7420 > F01 853 we reject H0 b1 0
32 a s 2033
b
106 ± 3182 (111) 106 ± 353
or 707 to 1413
c
d
106 ± 3182 (232) 106 ± 738
or 322 to 1798
33 a s 1453
b
2469 ± 3182 (68) 2469 ± 216
or 2253 to 2685
c
d
2469 ± 3182 (161) 2469 ± 512
or 1957 to 2981
34 s 133
228 ± 3182 (85) 228 ± 270
or 40 to 498
228 ± 3182 (158) 228 ± 503
or 227 to 731
35 a s 14589
203378 ± 2776 (6854) 203378 ± 19027
or 184351 to 222405
b
203378 ± 2776 (16119) 203378 ± 44746
or 158632 to 248124
36 a
b s 35232
80859 ± 2160 (1055) 80859 ± 2279
or 7858 to 8314
c
80859 ± 2160 (3678) 80859 ± 7944
or 7292 to 8880
37 a
s2 188 s 137
1308 ± 2571 (52) 1308 ± 134
or 1174 to 1442 or 11740 to 14420
b sind 147
1308 ± 2571 (147) 1308 ± 378
or 930 to 1686 or 9300 to 16860
c Yes 20400 is much larger than anticipated
d Any deductions exceeding the 16860 upper limit could suggest an audit
38 a
b
s2 MSE 5833333 s 24152
504667 ± 4604 (26750) 504667 ± 123157
or 381510 to 627824
c Based on one month 6000 is not out of line since 381510 to 627824 is the prediction interval However a sequence of five to seven months with consistently high costs should cause concern
39 a Summations needed to compute the slope and yintercept are
b SST 3906514 SSE 4145141 SSR 34920000
r2 SSRSST 3492000039065141 0894
The estimated regression equation explained 894 of the variability in y a very good fit
c s2 MSE 41451418 518143
27063 ± 2262 (886) 27063 ± 2004
or 25059 to 29067
d
27063 ± 2262 (2442) 27063 ± 5524
or 21539 to 32587
40 a 9
b 200 + 721x
c 13626
d SSE SST SSR 519841 415873 103968
MSE 1039687 14853
F MSR MSE 415873 14853 2800
F05 559 (1 degree of freedom numerator and 7 denominator)
Since F 28 > F05 559 we reject H0 B1 0
e 200 + 721(50) 3805 or 380500
41 a 61092 + 8951x
b
t025 2306 (1 degree of freedom numerator and 8 denominator)
Since t 601 > t025 2306 we reject H0 B1 0
c 61092 + 8951(25) 2849 or 2849 per month
42 a 800 + 500x
b 30
c F MSR MSE 68286821 8317
F05 420 (1 degree of freedom numerator and 28 denominator)
Since F 8317 > F05 420 we reject H0 B1 0
Branch office sales are related to the salespersons
d 80 + 50 (12) 680 or 680000
43 a The Minitab output is shown below
The regression equation is
Price 118 + 218 Income
Predictor Coef SE Coef T P
Constant 1180 1284 092 0380
Income 21843 02780 786 0000
S 6634 RSq 861 RSq(adj) 847
Analysis of Variance
Source DF SS MS F P
Regression 1 27179 27179 6175 0000
Residual Error 10 4401 440
Total 11 31580
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 7579 247 ( 7029 8128) ( 6002 9156)
b r2 861 The least squares line provided a very good fit
c The 95 confidence interval is 7029 to 8128 or 70290 to 81280
d The 95 prediction interval is 6002 to 9156 or 60020 to 91560
44 ab The scatter diagram shows a linear relationship between the two variables
c The Minitab output is shown below
The regression equation is
Rental 371 0779 Vacancy
Predictor Coef SE Coef T P
Constant 37066 3530 1050 0000
Vacancy 07791 02226 350 0003
S 4889 RSq 434 RSq(adj) 398
Analysis of Variance
Source DF SS MS F P
Regression 1 29289 29289 1226 0003
Residual Error 16 38237 2390
Total 17 67526
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 1759 251 ( 1227 2290) ( 594 2923)
2 2826 142 ( 2526 3126) ( 1747 3905)
Values of Predictors for New Observations
New Obs Vacancy
1 250
2 113
d Since the pvalue 0003 is less than a 05 the relationship is significant
e r2 434 The least squares line does not provide a very good fit
f The 95 confidence interval is 1227 to 2290 or 1227 to 2290
g The 95 prediction interval is 1747 to 3905 or 1747 to 3905
45 a
b The residuals are 348 247 483 16 and 522
c
With only 5 observations it is difficult to determine if the assumptions are satisfied However the plot does suggest curvature in the residuals that would indicate that the error term assumptions are not satisfied The scatter diagram for these data also indicates that the underlying relationship between x and y may be curvilinear
d
The standardized residuals are 132 59 111 40 149
e The standardized residual plot has the same shape as the original residual plot The curvature observed indicates that the assumptions regarding the error term may not be satisfied
46 a
b
The assumption that the variance is the same for all values of x is questionable The variance appears to increase for larger values of x
47 a Let x advertising expenditures and y revenue
b SST 1002 SSE 31028 SSR 69172
MSR SSR 1 69172
MSE SSE (n 2) 31028 5 620554
F MSR MSE 69172 620554 1115
F05 661 (1 degree of freedom numerator and 5 denominator)
Since F 1115 > F05 661 we conclude that the two variables are related
c
d The residual plot leads us to question the assumption of a linear relationship between x and y Even though the relationship is significant at the 05 level of significance it would be extremely dangerous to extrapolate beyond the range of the data
48 a
b The assumptions concerning the error term appear reasonable
49 a Let x return on investment (ROE) and y priceearnings (PE) ratio
b
c There is an unusual trend in the residuals The assumptions concerning the error term appear questionable
50 a The MINITAB output is shown below
The regression equation is
Y 661 + 0402 X
Predictor Coef Stdev tratio p
Constant 6610 3206 206 0094
X 04023 02276 177 0137
s 1262 Rsq 385 Rsq(adj) 261
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 4972 4972 312 0137
Error 5 7957 1591
Total 6 12929
Unusual Observations
Obs X Y Fit StdevFit Residual StResid
1 135 14500 12042 487 2458 211R
R denotes an obs with a large st resid
The standardized residuals are 211 108 14 38 78 04 41
The first observation appears to be an outlier since it has a large standardized residual
b
24+
*
STDRESID
12+
*
00+ *
* *
*
12+ *
++++++YHAT
1100 1150 1200 1250 1300 1350
The standardized residual plot indicates that the observation x 135y 145 may be an outlier note that this observation has a standardized residual of 211
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 observation x 135y 145 may be an outlier the implication is that for simple linear regression an outlier can be identified by looking at the scatter diagram
51 a The Minitab output is shown below
The regression equation is
Y 130 + 0425 X
Predictor Coef Stdev tratio p
Constant 13002 2396 543 0002
X 04248 02116 201 0091
s 3181 Rsq 402 Rsq(adj) 302
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 4078 4078 403 0091
Error 6 6072 1012
Total 7 10150
Unusual Observations
Obs X Y Fit StdevFit Residual StResid
7 120 2400 1810 120 590 200R
8 220 1900 2235 278 335 216RX
R denotes an obs with a large st resid
X denotes an obs whose X value gives it large influence
The standardized residuals are 100 41 01 48 25 65 200 216
The last two observations in the data set appear to be outliers since the standardized residuals for these observations are 200 and 216 respectively
b Using MINITAB we obtained the following leverage values
28 24 16 14 13 14 14 76
MINITAB identifies an observation as having high leverage if hi > 6n for these data 6n 68 75 Since the leverage for the observation x 22 y 19 is 76 MINITAB would identify observation 8 as a high leverage point Thus we conclude that observation 8 is an influential observation
c
240+ *
Y
200+ *
*
*
160+ *
*
*
120+ *
++++++X
00 50 100 150 200 250
The scatter diagram indicates that the observation x 22 y 19 is an influential observation
52 a The Minitab output is shown below
The regression equation is
Amount 409 + 0196 MediaExp
Predictor Coef SE Coef T P
Constant 4089 2168 189 0096
MediaExp 019552 003635 538 0001
S 5044 RSq 783 RSq(adj) 756
Analysis of Variance
Source DF SS MS F P
Regression 1 73584 73584 2893 0001
Residual Error 8 20351 2544
Total 9 93935
Unusual Observations
Obs MediaExp Amount Fit SE Fit Residual St Resid
1 120 3630 2755 330 875 230R
R denotes an observation with a large standardized residual
b Minitab identifies observation 1 as having a large standardized residual thus we would consider observation 1 to be an outlier
53 a The Minitab output is shown below
The regression equation is
Exposure 86 + 771 Aired
Predictor Coef SE Coef T P
Constant 855 2165 039 0703
Aired 77149 05119 1507 0000
S 3488 RSq 966 RSq(adj) 962
Analysis of Variance
Source DF SS MS F P
Regression 1 276434 276434 22717 0000
Residual Error 8 9735 1217
Total 9 286169
Unusual Observations
Obs Aired Exposure Fit SE Fit Residual St Resid
1 950 7588 7244 320 344 246RX
R denotes an observation with a large standardized residual
X denotes an observation whose X value gives it large influence
b Minitab identifies observation 1 as having a large standardized residual thus we would consider observation 1 to be an outlier Minitab also identifies observation 1 as an influential observation
54 a The Minitab output is shown below
The regression equation is
Salary 707 + 000482 MktCap
Predictor Coef SE Coef T P
Constant 7070 1180 599 0000
MktCap 00048154 00008076 596 0000
S 3798 RSq 664 RSq(adj) 645
Analysis of Variance
Source DF SS MS F P
Regression 1 5129071 5129071 3555 0000
Residual Error 18 2596647 144258
Total 19 7725718
Unusual Observations
Obs MktCap Salary Fit SE Fit Residual St Resid
6 507217 33250 31495 3386 1755 102 X
17 120967 1162 12895 864 11733 317R
R denotes an observation with a large standardized residual
X denotes an observation whose X value gives it large influence
b Minitab identifies observation 6 as having a large standardized residual and observation 17 as an observation whose x value gives it large influence A standardized residual plot against the predicted values is shown below
55 No Regression or correlation analysis can never prove that two variables are casually related
56 The estimate of a mean value is an estimate of the average of all y values associated with the same x The estimate of an individual y value is an estimate of only one of the y values associated with a particular x
57 To determine whether or not there is a significant relationship between x and y However if we reject B1 0 it does not imply a good fit
58 a The Minitab output is shown below
The regression equation is
Price 926 + 0711 Shares
Predictor Coef SE Coef T P
Constant 9265 1099 843 0000
Shares 07105 01474 482 0001
S 1419 RSq 744 RSq(adj) 712
Analysis of Variance
Source DF SS MS F P
Regression 1 46784 46784 2322 0001
Residual Error 8 16116 2015
Total 9 62900
b Since the pvalue corresponding to F 2322 001 < a 05 the relationship is significant
c 744 a good fit The least squares line explained 744 of the variability in Price
d
59 a The Minitab output is shown below
The regression equation is
Options 383 + 0296 Common
Predictor Coef SE Coef T P
Constant 3834 5903 065 0529
Common 029567 002648 1117 0000
S 1104 RSq 919 RSq(adj) 912
Analysis of Variance
Source DF SS MS F P
Regression 1 15208 15208 12472 0000
Residual Error 11 1341 122
Total 12 16550
b approximately 406 million shares of options grants outstanding
c 919 a very good fit The least squares line explained 919 of the variability in Options
60 a The Minitab output is shown below
The regression equation is
IBM 0275 + 0950 S&P 500
Predictor Coef StDev T P
Constant 02747 09004 031 0768
S&P 500 09498 03569 266 0029
S 2664 RSq 470 RSq(adj) 403
Analysis of Variance
Source DF SS MS F P
Regression 1 50255 50255 708 0029
Error 8 56781 7098
Total 9 107036
b Since the pvalue 0029 is less than a 05 the relationship is significant
c r2 470 The least squares line does not provide a very good fit
d Woolworth has higher risk with a market beta of 125
61 a
b It appears that there is a positive linear relationship between the two variables
c The Minitab output is shown below
The regression equation is
High 239 + 0898 Low
Predictor Coef SE Coef T P
Constant 23899 6481 369 0002
Low 08980 01121 801 0000
S 5285 RSq 781 RSq(adj) 769
Analysis of Variance
Source DF SS MS F P
Regression 1 17923 17923 6418 0000
Residual Error 18 5027 279
Total 19 22949
d Since the pvalue corresponding to F 6418 000 < a 05 the relationship is significant
e 781 a good fit The least squares line explained 781 of the variability in high temperature
f
62 The MINITAB output is shown below
The regression equation is
Y 105 + 0953 X
Predictor Coef Stdev tratio p
Constant 10528 3745 281 0023
X 09534 01382 690 0000
s 4250 Rsq 856 Rsq(adj) 838
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 86005 86005 4762 0000
Error 8 14447 1806
Total 9 100453
Fit StdevFit 95 CI 95 PI
3913 149 ( 3569 4257) ( 2874 4952)
a 105 + 953 x
b Since the pvalue corresponding to F 4762 000 < a 05 we reject H0 b1 0
c The 95 prediction interval is 2874 to 4952 or 2874 to 4952
d Yes since the expected expense is 3913
63 a The Minitab output is shown below
The regression equation is
Defects 222 0148 Speed
Predictor Coef SE Coef T P
Constant 22174 1653 1342 0000
Speed 014783 004391 337 0028
S 1489 RSq 739 RSq(adj) 674
Analysis of Variance
Source DF SS MS F P
Regression 1 25130 25130 1133 0028
Residual Error 4 8870 2217
Total 5 34000
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 14783 0896 ( 12294 17271) ( 9957 19608)
b Since the pvalue corresponding to F 1133 028 < a 05 the relationship is significant
c 739 a good fit The least squares line explained 739 of the variability in the number of defects
d Using the Minitab output in part (a) the 95 confidence interval is 12294 to 17271
64 a There appears to be a negative linear relationship between distance to work and number of days absent
b The MINITAB output is shown below
The regression equation is
Y 810 0344 X
Predictor Coef Stdev tratio p
Constant 80978 08088 1001 0000
X 034420 007761 443 0002
s 1289 Rsq 711 Rsq(adj) 675
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 32699 32699 1967 0002
Error 8 13301 1663
Total 9 46000
Fit StdevFit 95 CI 95 PI
6377 0512 ( 5195 7559) ( 3176 9577)
c Since the pvalue corresponding to F 41967 is 002 < a 05 We reject H0 b1 0
d r2 711 The estimated regression equation explained 711 of the variability in y this is a reasonably good fit
e The 95 confidence interval is 5195 to 7559 or approximately 52 to 76 days
65 a Let X the age of a bus and Y the annual maintenance cost
The MINITAB output is shown below
The regression equation is
Y 220 + 132 X
Predictor Coef Stdev tratio p
Constant 22000 5848 376 0006
X 13167 1780 740 0000
s 7550 Rsq 873 Rsq(adj) 857
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 312050 312050 5475 0000
Error 8 45600 5700
Total 9 357650
Fit StdevFit 95 CI 95 PI
7467 298 ( 6780 8154) ( 5595 9339)
b Since the pvalue corresponding to F 5475 is 000 < a 05 we reject H0 b1 0
c r2 873 The least squares line provided a very good fit
d The 95 prediction interval is 5595 to 9339 or 55950 to 93390
66 a Let X hours spent studying and Y total points earned
The MINITAB output is shown below
The regression equation is
Y 585 + 0830 X
Predictor Coef Stdev tratio p
Constant 5847 7972 073 0484
X 08295 01095 758 0000
s 7523 Rsq 878 Rsq(adj) 862
Analysis of Variance
SOURCE DF SS MS F p
Regression 1 32497 32497 5742 0000
Error 8 4528 566
Total 9 37025
Fit StdevFit 95 CI 95 PI
8465 367 ( 7619 9311) ( 6535 10396)
b Since the pvalue corresponding to F 5742 is 000 < a 05 we reject H0 b1 0
c 8465 points
d The 95 prediction interval is 6535 to 10396
67 a The Minitab output is shown below
The regression equation is
Audit 0471 +0000039 Income
Predictor Coef SE Coef T P
Constant 04710 05842 081 0431
Income 000003868 000001731 223 0038
S 02088 RSq 217 RSq(adj) 174
Analysis of Variance
Source DF SS MS F P
Regression 1 021749 021749 499 0038
Residual Error 18 078451 004358
Total 19 100200
Predicted Values for New Observations
New Obs Fit SE Fit 950 CI 950 PI
1 08828 00523 ( 07729 09927) ( 04306 13349)
b Since the pvalue 0038 is less than a 05 the relationship is significant
c r2 217 The least squares line does not provide a very good fit
d The 95 confidence interval is 7729 to 9927
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