Part IV: Complete Solutions, Chapter 10
553
Chapter 10: Correlation and Regression
Section 10.1
1.
The explanatory variable is graphed along the horizontal or x axis. The response variable is graphed along
the vertical or y axis.
2.
If two variables are positively correlated, then the response variable will increase as the explanatory
variable increases.
3.
If two variables are negatively correlated, then the response variable will decrease as the explanatory
variable increases.
4.
(a) There is a strong negative linear correlation.
(b) There is a weak or no linear correlation.
(c) There is a strong positive linear correlation.
5.
(a) The points seem close to a straight line, so there is moderate linear correlation.
(b) A straight line does not fit the data well, so there is no linear correlation.
(c) The points seem very close to a straight line, so there is high linear correlation.
6.
(a) The points seem close to a straight line, so there is moderate linear correlation.
(b) The points seem very close to a straight line, so there is high linear correlation.
(c) A straight line does not fit the data well, so there is no linear correlation.
7.
(a) No. The correlation coefficient is a mathematical tool for measuring the strength of a linear
relationship between two variables. As such, it makes no implication about cause or effect. Just
because two variables tend to increase or decrease together does not mean a change in one is causing a
change in the other. A strong correlation between x and y is sometimes due to the presence of lurking
variables.
(b) Increasing population might be a lurking variable causing both variables to increase.
8.
(a) No. The correlation coefficient is a mathematical tool for measuring the strength of a linear
relationship between two variables. As such, it makes no implication about cause or effect. Just
because two variables tend to increase or decrease together does not mean a change in one is causing a
change in the other. A strong correlation between x and y is sometimes due to the presence of lurking
variables.
(b) Rising cost of living might be a lurking variable causing both variables to increase.
9.
(a) No. The correlation coefficient is a mathematical tool for measuring the strength of a linear
relationship between two variables. As such, it makes no implication about cause or effect. Just
because two variables tend to increase or decrease together does not mean a change in one is causing a
change in the other. A strong correlation between x and y is sometimes due to the presence of lurking
variables.
(b) One lurking variable for the increase in average annual income is inflation. Better training might be a
lurking variable responsible for shorter times to run the mile.
10. (a) No. The correlation coefficient is a mathematical tool for measuring the strength of a linear
relationship between two variables. As such, it makes no implication about cause or effect. Just
because two variables tend to increase or decrease together does not mean a change in one is causing a
change in the other. A strong correlation between x and y is sometimes due to the presence of lurking
variables.
(b) A lurking variable might be better health care and treatments.
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Part IV: Complete Solutions, Chapter 10
554
11. (a)
Scatterplot of Kilograms vs Months
200
175
Kilograms
150
125
100
75
50
0
5
10
15
20
25
Months
(b) Since the points lie close to a straight line, the correlation is strong and positive.
n xy    x   y 
5(9,930)  (63)(650)

 0.972
(c) r 
2
2
5(1, 089)  (63) 2 5(95,350)  (650) 2
n x 2    x  n y 2    y 
Since r is positive, y should tend to increase as x increases.
(a)
Scatterplot of Administrative Cost % vs Number of Employees
40
Administrative Cost %
12.
35
30
25
20
15
0
10
20
30
40
50
Number of Employees
60
70
80
(b) Since the points lie fairly close to a straight line, the correlation is moderate and negative.
n xy    x   y 
5(3, 040)  (135)(148)

 0.945
(c) r 
2
2
2 5(4, 674)  (148) 2
2
2
5(7,133)

(135)
n x    x  n y    y 
Since r is negative, y should tend to decrease as x increases.
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Part IV: Complete Solutions, Chapter 10
555
13. (a)
Scatterplot of Wind mph vs Millibars
150
Wind mph
125
100
75
50
930
940
950
960
970
Millibars
980
990
1000
1010
(b) Since the points lie very close to a straight line, the correlation is strong and negative.
n xy    x   y 
6(556,315)  (5,823)(580)

 0.990
(c) r 
2
2
6(5, 655, 779)  (5,823) 2 6(65, 750)  (580) 2
n x 2    x  n y 2    y 
Since r is negative, y should tend to decrease as x increases.
14. (a)
Scatterplot of Depth km vs Magnitude
12
11
10
Depth km
9
8
7
6
5
4
3
2.5
3.0
3.5
Magnitude
4.0
4.5
(b) Since the points are not close to a straight line, the correlation is low and positive.
n xy    x   y 
7(190.18)  (24.1)(53.5)

 0.511
(c) r 
2
2
7(85.75)  (24.1)2 7(458.31)  (53.5)2
n x 2    x  n y 2    y 
Since r is positive, y should tend to increase as x increases.
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Part IV: Complete Solutions, Chapter 10
556
15. (a)
Scatterplot of Home Run Percentage vs Batting Average
8
Home Run Percentage
7
6
5
4
3
2
1
0.250
0.275
0.300
Batting Average
0.325
0.350
(b) Since the points are very close to a straight line, the correlation is strong and positive.
n xy    x   y 
7(8.753)  (1.957)(30.1)

 0.948
(c) r 
2
2
2
2
7(0.553)  (1.957)2 7(150.15)  (30.1) 2
n x    x  n y    y 
Since r is positive, y should tend to increase as x increases.
16. (a)
Scatterplot of Number of Burglaries vs Student Enrollment (1000s)
80
Number of Burglaries
70
60
50
40
30
20
10
10
15
20
Student Enrollment (1000s)
25
30
(b) Since the points are fairly close to a straight line, the correlation is moderate and positive.
n xy    x   y 
8(5, 488.4)  (152.8)(246)

 0.765
(c) r 
2
2
8(3,350.98)  (152.8)2 8(10, 030)  (246)2
n x 2    x  n y 2    y 
Since r is positive, y should tend to increase as x increases.
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Part IV: Complete Solutions, Chapter 10
17. (a)
The axes are scaled equally.
(b)
The y axis is twice the x axis.
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557
Part IV: Complete Solutions, Chapter 10
558
(c)
Scatterplot of y vs x
9
8
7
6
y
5
4
3
2
1
1
2
3
4
5
6
7
x
The x axis is twice the y axis.
(d) Stretching the scale on the y axis makes the line appear steeper. Shrinking the scale on the y axis makes
the line appear flatter. The slope of the line does not change. Only the appearance of the slope changes
as the scale of the y axis changes.
18. (a) We get the same result.
Notice that in the numerator of the formula for r, xy  yx and (x)(y)  (y)(x).
Then notice that the denominator
n x 2    x 
2
n y 2    y   n  y 2    y 
2
2
n x 2    x  .
2
(b) We get the same result.
(c) First set:
3(29)  (8)(9)
r
 0.61859
3(26)  (8) 2 3(41)  (9)2
Second set:
3(29)  (9)(8)
r
 0.61859
3(41)  (9)2 3(26)  (8) 2
19. (a) r  0.972 with n = 5
Since |0.972| = 0.972  0.88 for  = 0.05, r is significant, and we conclude that age and weight of
Shetland ponies are correlated.
(b) r  0.990 with n = 6
Since |0.990| = 0.990  0.92 for  = 0.01, r is significant, and we conclude that the lowest barometric
pressure reading and maximum wind speed for cyclones are correlated.
20. (a) No, becaquse |0.820| = 0.820 < 0.87 for n = 7 and  = 0.01.
Yes, because |0.900| = 0.900  0.80 for n = 9 and  = 0.01.
(b) No, because |0.40| = 0.40 < 0.44 for n = 20 and  = 0.05.
Yes, because |0.40| = 0.40  0.38 for n = 27 and  = 0.05.
(c) No; no; no; a larger sample size means that a smaller |r| might be significant.
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Part IV: Complete Solutions, Chapter 10
559
21. (a)
Scatterplot of Gallons Wasted vs Hours
60
Gallons Wasted
50
40
30
20
10
0
5
10
15
20
Hours
n xy    x   y 
r
n x    x 
2
2
n y    y 
2
2
25

30
35
8(6, 067)  (154)(249)
8(3, 712)  (154)2 8(9,959)  (249)2
 0.991
(b) For variables based on averages, x  19.25 h; sx  10.33 h; y  31.13 gal; s y  17.76 gal.
For variables based on single individuals, x  20.13 h; sx  13.84 h; y  31.87 gal; s y  25.18.
Dividing by larger numbers results in a smaller value.
(c)
Scatterplot of Gallons Wasted vs Hours
70
Gallons Wasted
60
50
40
30
20
10
0
0
r
10
20
Hours
30
8(7, 071)  (161)(255)
8(4,583)  (161) 2 8(12,565)  (255) 2
40
 0.794
(d) 0.991 > 0.791
Yes, by the central limit theorem, the x distribution has a smaller standard deviation than the
corresponding x distribution.
22. (a) No. Interest rate probably affects both investment returns.
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Part IV: Complete Solutions, Chapter 10
560
(b) For w = 0.6x + 0.4y, a = 0.6, b = 0.4
 w  a  x  b y
 w  0.6(7.32)  0.4(13.19)
 w  9.67
 w2  a 2 x2  b 2 2y  2ab x y 
 w2  (0.6)2 (6.59)2  (0.4)2 (18.56)2  2(0.6)(0.4)(6.59)(18.56)(0.424)
 w2  95.64
 w  95.64  9.78
(c) For w = 0.4x + 0.6y, a = 0.4, b = 0.6
 w  0.4(7.32)  0.6(13.19)
 w  10.84
 w2  (0.4)2 (6.59)2  (0.6)2 (18.56)2  2(0.4)(0.6)(6.59)(18.56)(0.424)
 w2  155.85
 w  155.85  12.48
(d) w  0.4x  0.6 y produces higher returns with greater risk as measured by  w .
Section 10.2
Note: In Sections 10.2, 10.3, and 10.4, answers may vary slightly depending on rounding considerations.
1.
The slope is b = –2. When x increases by 1 unit, y decreases by 2 units.
2.
The marginal change is the slope, b = 3 units.
3.
We are extrapolating. Extrapolation is dangerous because the pattern of data may change outside the x
range.
4.
It is negative.
5.
(a) yˆ  318.16  30.878x
(b) There will be about 31 few frost-free days (30.878).
(c) Here, r ≈ –0.981. Note that if the slope is negative, then r is also negative.
6.
(a ) yˆ  0.8565  0.40248x
(b) It increases about 40.25 kcal/24 h.
(c ) Here, r ≈ 0.984.
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Part IV: Complete Solutions, Chapter 10
561
7. (a)
Scatterplot of Entry Level Jobs vs Jobs
9
8
Entry Level Jobs
7
6
5
4
3
2
1
15
20
25
30
35
40
45
50
Jobs
(b) Use a calculator to verify.
 x  202  33.67 jobs
(c) x 
n
6
y
28
   4.67 entry-level jobs
y
n
6
n xy    x   y  6(1, 096)  (202)(28)
b

 0.161
2
6(7, 754)  (202) 2
n x 2    x 
a  y  bx  4.67  (0.161)(33.67)  0.748
ŷ  a  bx or yˆ  0.748  0.161x
(d) See figure of part (a).
(e) r 2  (0.860)2  0.740
This means that 74.0% of the variation in y = number of entry-level jobs can be explained by the
corresponding variation in x = total number of jobs using the least-squares line.
100%  74.0% = 26.0% of the variation is unexplained.
(f) Use x = 40.
yˆ  0.748  0.161(40)
yˆ  5.69 entry-level jobs
8.
(a)
Scatterplot of Weight (kg) vs Age (Weeks)
200
175
Weight (kg)
150
125
100
75
50
0
10
20
Age (Weeks)
(b) Use a calculator to verify.
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30
40
Part IV: Complete Solutions, Chapter 10
562
(c) x 
 x  92  15.33 weeks
n
6
 y  617  102.83 kg
y
b
n
6
n xy    x   y 
n x 2    x 
2

6(13, 642)  (92)(617)
6(2,338)  (92) 2
 4.509
a  y  bx  102.83  (4.509)(15.33)  33.70
yˆ  33.70  4.51x
(d) See figure of part (a).
(e) r 2  (0.998)2  0.996
This means that 99.6% of the variation in y = weight can be explained by the corresponding variation
in x = age using the least-squares line. 100%  99.6% = 0.4% of the variation is unexplained.
(f) Use x = 12.
yˆ  33.70  4.51(12)
yˆ  87.8 kg
(a)
Scatterplot of MPG vs Weight (100s lbs)
30
25
MPG
9.
20
15
10
20
25
30
35
40
Weight (100s lbs)
45
50
55
(b) Use a calculator to verify.
 x 299

 37.375
n
8
 y 167
y

 20.875
n
8
n xy    x   y  8(5,814)  (299)(167)
b

 0.6007
2
8(11,887)  (167) 2
n x 2    x 
a  y  bx  20.875  (0.6007)(37.375)  43.326
yˆ  a  bx or yˆ  43.326  0.6007 x
(d) See figure of part (a).
(c) x 
(e) r 2  (0.946)2  0.895
This means that 89.5% of the variation in y = miles per gallon can be explained by the corresponding
variation in x = weight using the least-squares line. 100%  89.5% = 10.5% of the variation is
unexplained.
(f) Use x = 38.
yˆ  43.326  0.6007(38)  20.5 mpg
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Part IV: Complete Solutions, Chapter 10
563
10. (a)
Scatterplot of Winning Percentage vs Number of Fouls
Winning Percentage
50
45
40
35
30
25
0
1
2
3
Number of Fouls
4
5
6
(b) Use a calculator to verify.
 x 13
(c) x 

 3.25
n
4
 y 154
y

 38.5
n
4
n xy    x   y  4(411)  (13)(154)
b

 3.934
2
4(65)  (13) 2
n x 2    x 
a  y  bx  38.5  (3.934)(3.25)  51.29
yˆ  a  bx or yˆ  51.29  3.934 x
(d) See the figure in part (a).
(e) r 2  (0.988)2  0.976
This means that 97.6% of the variation in y = percent of wins can be explained by the variation in
x = excess fouls using the least-squares line. 100%  97.6% = 2.4% of the variation is unexplained.
(f) Use x = 4.
yˆ  51.29  3.934(4)  35.55%
11. (a)
Scatterplot of Speeding-Related Fatal % vs Age
Speeding-Related Fatal %
40
30
20
10
0
10
20
30
40
50
Age
(b) Use a calculator to verify.
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60
70
80
Part IV: Complete Solutions, Chapter 10
564
 x 329

 47 years
n
7
 y 115
y

 16.43%
n
7
n xy    x   y  7(4015)  (329)(115)
b
,
 0.496
2
7(18, 263)  (329) 2
n x 2    x 
a  y  bx  16.43  (0.496)(47)  39.761
yˆ  a  bx or yˆ  39.761  0.496 x
(d) See the figure in part (a).
(c) x 
(e) r 2  (0.959)2  0.920
This means that 92% of the variation in y = percentage of all fatal accidents due to speeding can be
explained by the corresponding variation in x = age in years of a licensed automobile driver using the
least-squares line. 100%  92% = 8% of the variation is unexplained.
(f) Use x = 25.
yˆ  39.761  0.496(25)  27.36%
12. (a)
Scatterplot of Right-of-Way Fatal % vs Age
Right-of-Way Fatal %
40
30
20
10
0
30
40
50
60
Age
70
80
90
(b) Use a calculator to verify.
 x 372
(c) x 

 62 years
n
6
 y 112
y

 18.67%
n
6
n xy    x   y  6(8, 254)  (372)(112)
b

 0.749
2
6(24,814)  (372)2
n x 2    x 
a  y  bx  18.67  0.749(62)  27.745
yˆ  a  bx or yˆ  27.75  0.75 x
(d) See the figure in part (a).
(e) r 2  (0.943)2  0.889
This means that 88.9% of the variation in y = percentage of fatal accidents due to failure to yield to
right of way can be explained by the corresponding variation in x = age of a licensed driver in years
using the least-squares line. 100%  88.9% = 11.1% of the variation is unexplained.
(f) Use x = 70.
yˆ  27.75  0.75(70)
yˆ  24.75%
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Part IV: Complete Solutions, Chapter 10
565
13. (a)
Scatterplot of Number of Doctors vs Per Capita Income
22
Number of Doctors
20
18
16
14
12
10
8.0
8.5
9.0
Per Capita Income
9.5
10.0
(b) Use a calculator to verify.
 x 53
(c) x 

 $8.83 thousand
n
6
 y 83.7
y

 13.95 physicians per 10,000
n
6
n xy    x   y  6(755.89)  (53)(83.7)
b

 5.756
2
6(471.04)  (53) 2
n x 2    x 
a  y  bx  13.95  5.756(8.83)  36.898
yˆ  a  bx or yˆ  36.898  5.756 x
(d) See the figure in part (a).
(e) r 2  (0.934)2  0.872
This means that 87.2% of the variation in y = number of medical doctors per 10,000 residents can be
explained by the corresponding variation in x = per capita income in thousands of dollars using the
least-squares line. 100%  87.2% = 12.8% of the variation is unexplained.
(f) Use x = 10.
yˆ  36.898  5.756(10)  20.7 physicians per 10,000 residents
14. (a)
Scatterplot of % Change Imprisonment Rate vs % Change Violent Crime
% Change Imprisonment Rate
5.0
2.5
0.0
-2.5
-5.0
-7.5
3
4
5
6
7
8
9
% Change Violent Crime
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10
11
12
Part IV: Complete Solutions, Chapter 10
566
(b) Use a calculator to verify.
 x 44.7
(c) x 

 6.386%
n
7
 y 17.4
y

 2.486%
n
7
n xy    x   y  7( 107.18)  (44.7)( 17.4)
b

 0.129
2
7(315.85)  (44.7) 2
n x 2    x 
a  y  bx  2.486  (0.129)(6.386)  3.311
yˆ  a  bx or yˆ  3.311  0.129 x
(d) See the figure in part (a).
(e) r 2  (0.084)2  0.007
This means that 0.7% of the variation in y = percent change in the rate of imprisonment can be
explained by the corresponding variation in x = percent change in the rate of violent crime using the
least-squares line. 100%  0.7% = 99.3% of the variation is unexplained.
(f) The correlation between the variables is so low that it does not make sense to use the least-squares line
for prediction.
15. (a)
Number of Violent vs % Not Attending
Number of Violent Crimes / 1000
14
12
10
8
6
4
2
0
15
16
17
18
19
20
21
22
% Not Attending / Not Graduated
23
24
(b) Use a calculator to verify.
 x 112.8
(c) x 

 18.8%
n
6
 y 32.4
y

 5.4 crimes per 1,000
n
6
n xy    x   y  6(665.03)  (112.8)(32.4)
b

 1.202
2
6(2,167.14)  (112.8) 2
n x 2    x 
a  y  bx  5.4  1.202(18.8)  17.204
yˆ  a  bx or yˆ  17.204  1.202 x
(d) See the figure in part (a).
(e) r 2  (0.764)2  0.584
This means that 58.4% of the variation in y = reported violent crimes per 1,000 residents can be
explained by the corresponding variation in x = percentage of 16- to 19-year-olds not in school and not
high-school graduates using the least-squares line. 100%  58.4% = 41.6% of the variation is
unexplained.
(f) Use x = 24.
yˆ  17.204  1.202(24)  11.6 crimes per 1,000
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Part IV: Complete Solutions, Chapter 10
567
16. (a)
Scatterplot of Mean Patents vs Research Programs
2.0
1.8
Mean Patents
1.6
1.4
1.2
1.0
0.8
0.6
10
12
14
16
Research Programs
18
20
(b) Use a calculator to verify.
 x 90
(c) x 

 15.0
n
6
 y 8.1
y

 1.35
n
6
n xy    x   y  6(113.8)  (90)(8.1)
b

 0.1100
2
6(1420)  (90) 2
n x 2    x 
a  y  bx  1.35  (0.1100)(15.0)  3.0
yˆ  a  bx or yˆ  3.0  0.11x
(d) See the figure in part (a).
(e) r 2  (0.973)2  0.946
This means that 94.6% of the variation in y = number of patents can be explained by the corresponding
variation in x = number of programs using the least-squares line. 100%  94.6% = 5.4% of the
variation is unexplained.
(f) Use x = 15.
yˆ  3.0  0.11(15)  1.35 patents
17. (a)
Scatterplot of % Unidentified Artifacts vs Elevation
% Unidentified Artifacts
60
50
40
30
20
10
5.0
5.5
6.0
6.5
Elevation
(b) Use a calculator to verify.
Copyright © Houghton Mifflin Company. All rights reserved.
7.0
7.5
Part IV: Complete Solutions, Chapter 10
568
 x 31.25

 6.25
n
5
 y 164
y

 32.8
n
5
n xy    x   y  5(1, 080)  (31.25)(164)
b

 22.0
2
5(197.813)  (31.25) 2
n x 2    x 
a  y  bx  32.8  22.0(6.25)  104.5
yˆ  a  bx or yˆ  104.7  22.0 x
(d) See the figure in part (a).
(c) x 
(e) r 2  (0.913)2  0.833
This means that 83.3% of the variation in y = % unidentified artifacts can be explained by the
corresponding variation in x = elevation. 100%  83.3% = 16.7% of the variation is unexplained.
(f) Use x = 6.5.
yˆ  104.7  22.0(6.5)  38.3 percent
18. (a)
Scatterplot of Temperature vs Chirps / Second
95
Temperature
90
85
80
75
70
14
15
16
17
18
Chirps / Second
19
20
(b) Use a calculator to verify.
 x 249.8
(c) x 

 16.65
n
15
 y 1, 200.6
y

 80.04
n
15
n  xy    x   y  15(20,127.47)  (249.8)(1, 200.6)
b

 3.291
2
15(4, 200.56)  (249.8)2
n x 2    x 
a  y  bx  80.04  3.299(16.65)  25.232
yˆ  a  bx or yˆ  25.232  3.291x
(d) See the figure in part (a).
(e) r 2  (0.835)2  0.697
This means that 69.7% of the variation in y = temperature can be explained by the corresponding
variation in x = chirps per second using the least-squares line. 100%  69.7% = 30.3% of the variation
is unexplained.
(f) Use x = 19.
yˆ  25.232  3.291(19)  87.7F
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Part IV: Complete Solutions, Chapter 10
569
19. (a) Yes. The pattern of residuals appears randomly scattered around the horizontal line at 0.
(b) No. There do not appear to be any outliers.
20. (a)
x
y
y p  43.3263  0.6007 x
Residual = y  y p
27
44
32
47
23
40
34
52
30
19
24
13
29
17
21
14
27.1
16.9
24.1
15.1
29.5
19.3
22.9
12.1
2.9
2.1
0.1
2.1
0.5
2.3
1.9
1.9
Residuals Versus Weight
(response is MPG)
3
2
Residual
1
0
-1
-2
-3
20
25
30
35
40
45
50
55
Weight
(b) The residuals seem to be scattered randomly around the horizontal line at 0. There do not appear to be
any outliers.
21. (a)
(b)
(c)
(d)
Result checks.
Result checks.
Yes.
y  0.143  1.071x
y  0.143  1.071x
y  0.143
x
1.071
1
0.143
y
x
1.071
1.071
or
x  0.9337 y  0.1335
The equation x = 0.9337y  0.1335 does not match part (b) with the symbols x and y exchanged.
(e) In general, switching x and y values produces a different least-squares equation. It is important that
when you perform a linear regression, you know which variable is the explanatory variable and which
is the response variable.
22. (a)
(b)
(c)
(d)
No, a straight line does not fit the data well. The data seem to explode as x increases.
The transformed data fit a straight line better.
Use a calculator to verify.
Use a calculator to verify.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
570
(e)
  100.15  1.413
  100.404  2.535
y  1.413  2.535
x
23. (a)
Scatterplot of Number of Locusts vs Day of Observation
700
Number of Locusts
600
500
400
300
200
100
0
1
2
3
4
5
6
7
Day of Observation
8
9
10
These data are not described well by a straight line because the y values seem to explode as x increases.
(b)
Scatterplot of Log Locusts vs Day of Observation
3.0
2.5
Log Locusts
2.0
1.5
1.0
0.5
0.0
1
2
3
4
5
6
7
Day of Observation
8
9
10
The transformed data are better represented by a straight line.
(c)
yˆ  0.365  0.311x
r  0.998
(d)
  100.365  0.432
  100.311  2.046
y  0.432  2.046 
x
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Part IV: Complete Solutions, Chapter 10
571
24. (a)
Scatterplot of Log Amps vs Log Microseconds
0.65
0.60
Log Amps
0.55
0.50
0.45
0.40
0.35
0.30
0.25
0.3
0.4
0.5
0.6
0.7
Log Microseconds
0.8
0.9
1.0
The transformed data seem to be represented well by a straight line.
(b) Using the transformed data,
yˆ  0.128  0.492 x
r  0.987
(c)
  100.128  1.343
  b  0.492
0.492
yˆ  1.343  x 
25. (a)
Scatterplot of Log Critical Sheer Strength vs Log Boiler Pressure
1.2
Log Critical Sheer Strength
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.6
0.7
0.8
Log Boiler Pressure
0.9
1.0
The transformed data seem to be represented well by a straight line.
(b) Using the transformed data,
yˆ  0.451  1.600 x
r  0.991
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Part IV: Complete Solutions, Chapter 10
572
(c)
  100.451  0.351
  b  1.600
1.600
yˆ  0.354  x 
Section 10.3
1.
The symbol is rho, ρ.
2.
The symbol is beta, β.
3.
As x moves farther from x , the confidence interval for the predicted y becomes wider.
4.
The t value for b equals the t value for r.
5.
(a) The predictor variable is diameter.
(b) Here, a = –0.223, b = 0.8748, and yˆ  0.223  0.7848x.
(c) The P value of b is 0.001. We are testing H0: β = 0 versus H1: β ≠ 0, and we reject H0 and conclude that
the slope is not 0.
(d) Here, r ≈ 0.896. Yes, the correlation coefficient is significant at the α = 0.01 level because the
P value < α.
6.
(a) Se = 4.08.
(b) The standard error for the slope is approximately 0.1471.
(c ) Here, d.f. = n – 2 = 7 and tc = t0.95 = 2.365. E ≈ 2.365 × 0.1471 = 0.348. The interval is
0.7848  0.348   0.437, 1.113 .
7.
(a) Use a calculator to verify.
(b)  = 0.05
H0 :   0
H1:   0
t
r n2

0.784 6  2
 2.526
1 r
1  0.7842
d.f. = n  2 = 6  2 = 4
From Table 6 in Appendix II, 2.526 falls between entries 2.132 and 2.776. Use the one-tailed areas to
find that 0.025 < P value < 0.050.
Since the P-value interval  0.05, we reject H 0 .
At the 5% level of significance, there seems to be a positive correlation between x and y.
(c) Use a calculator to verify.
(d) ŷ  a  bx or yˆ  16.542  0.4117 x
Use x = 70.
yˆ  16.542  0.4117(70)  45.36%
2
(e) d.f. = 4, tc  2.132
E  tc Se 1 
1
n( x  x ) 2
1
6(70  73.167)2

 2.132(2.6964) 1  
 6.31
n n  x 2    x 2
6 6(32.393)  (439)2
The 90% confidence interval is
yˆ  E  y  yˆ  E
45.36  6.31  y  4,536  6.31
39.05  y  51.67
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Part IV: Complete Solutions, Chapter 10
573
(f)  = 0.05
H0:   0
H1:   0
b
1
0.4117
1
2
32,393  (439)2  2.522
 x2    x  
Se
n
2.6964
6
d.f. = n  2 = 6  2 = 4
From Table 6 in Appendix II, 2.522 falls between entries 2.132 and 2.776. Use the one-tailed areas to
find that 0.025 < P value < 0.050.
Since the P-value interval < 0.05, we reject H 0 .
At the 5% level of significance, there seems to be a positive slope between x and y.
(g) d.f. = 4, tc  2.132, b  0.4117
t
E
tc Se
 x2  1n  x 
2

2.132(2.6964)
32,393  16 (439)2
 0.3480
The 90% confidence interval is
bE   bE
0.4117  0.3480    0.4117  0.3480
0.064    0.760
For every percentage increase in successful free throws, the percentage of successful field goals
increases by an amount between 0.06 and 0.76.
8.
(a) Use a calculator to verify.
(b)  = 0.05
H0 :   0
H1:   0
t
r n2
1 r2

0.891 6  2
1  (0.891)2
 3.93
d.f. = n  2 = 6  2 = 4
From Table 6 in Appendix II, 3.93 falls between entries 3.747 and 4.604. Use the two-tailed areas to
find that 0.01 < P value < 0.020. Using a TI-84, P value ≈ 0.0171.
Since the P-value interval  0.05, we reject H 0 .
At the 5% level of significance, there seems to be a correlation between x and y.
(c) Use a calculator to verify.
(d) ŷ  a  bx or yˆ  26.247  65.081x
Use x = 0.300.
yˆ  26.247  65.081(0.300)  6.72
(e) d.f. = 4, tc  1.533, x 
 x  1.842  0.307
n
6
1
n( x  x )2
1
6(0.300  0.307)2
E  tc Se 1  
 1.533(1.6838) 1  
 2.79
n n  x 2    x 2
6 6(0.575838)  (1.842)2
The 80% confidence interval is
yˆ  E  y  yˆ  E
6.72  2.79  y  6.72  2.79
3.93  y  9.51
(f)  = 0.05
H0:   0
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Part IV: Complete Solutions, Chapter 10
574
H1:   0
b
1
65.081
1
2
0.575838  (1.842)2  3.93
 x2    x  
Se
n
1.6838
6
d.f. = n  2 = 6  2 = 4
From Table 6 in Appendix II, 3.93 falls between entries 3.747 and 4.604. Use the two-tailed areas to
find that 0.01 < P value < 0.020. Using a TI-84, P value ≈ 0.0171. Since the P-value interval  0.05,
we reject H 0 .
At the 5% level of significance, the slope seems to be significant.
t
(g) d.f. = 4, tc  2.132, b  65.081
E
tc Se

x2
 x
 1n
2
2.132(1.6838)

0.575838  16 (1.842)2
 35.297
The 90% confidence interval is
bE   bE
65.081  35.297    65.081  35.297
100.38    29.78
For every unit increase in batting average, the percentage strikeouts decrease by between 100% and
29%.
9.
(a) Use a calculator to verify.
(b)  = 0.01
H0 :   0
H1:   0
t
r n2
1 r2

0.976 7  2
1  (0.976)2
 10.02
d.f. = n  2 = 7  2 = 5
From Table 6 in Appendix II, 10.02 falls to the right of entry 6.869. Use the one-tailed areas to find P
value < 0.0005. Using a TI-84, P value ≈ 0.00008.
Since the P-value interval  0.01, we reject H 0 .
At the 1% level of significance, the evidence supports a negative correlation between x and y.
(c) Use a calculator to verify.
(d) ŷ  a  bx or yˆ  3.366  0.0544 x
Use x = 18.
yˆ  3.366  0.0544(18)  2.39 hours
(e) d.f. = 5, tc  1.476, x 
 x  201.4  28.77
n
7
n( x  x )2
1
1
7(18  28.77)2
E  tc Se 1  
 1.476(0.1660) 1  
 0.276
n n  x 2    x 2
7 7(6, 735.46)  (201.4)2
The 80% confidence interval is
yˆ  E  y  yˆ  E
2.39  0.276  y  2.39  0.276
2.11  y  2.67
(f)  = 0.01
H0:   0
H1:   0
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Part IV: Complete Solutions, Chapter 10
575
b
1
0.0544
1
2
6,735.46  (201.4)2  10.02
 x2    x  
Se
n
0.1660
7
From Table 6 in Appendix II, 10.02 falls to the right of entry 6.869. Use the one-tailed areas to find P
value < 0.0005. Using a TI-84, P value ≈ 0.00008.
Since the P-value interval  0.01, we reject H 0 .
At the 1% level of significance, the sample evidence supports a negative slope.
(g) d.f. = 5, tc  2.015, b  0.0544
t
E
tc Se
2
 x2  1n  x 
2.015(0.1660)

6, 735.46  17 (201.4)2
 0.011
The 90% confidence interval is
bE    b E
0.054  0.011    0.054  0.011
0.065    0.043
For a 1-m increase in depth, the optimal time decreases from about 0.04 to 0.07 hour.
10. (a) Use a calculator to verify.
(b)  = 0.01
H0 :   0
H1:   0
t
r n2
1 r2

0.984 5  2
1  (0.984)2
 9.57
d.f. = n  2 = 5  2 = 3
From Table 6 in Appendix II, the sample test statistic falls between entries 5.841 and 12.924. Use the
one-tailed areas to find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0012.
Since the P-value interval  0.01, reject H 0 .
At the 1% level of significance, the sample evidence supports a positive correlation.
(c) Use a calculator to verify.
(d) ŷ  a  bx or yˆ  2.869  6.876 x
Use x = 4.0.
yˆ  2.869  6.876(4.0)  24.63 mm Hg/10.
(e) d.f. = 3, tc  2.353, x 
E  tc Se 1 
 x  21.4  4.28
n
5
n( x  x ) 2
1
1
5(4.0  4.28)2

 2.353(2.5319) 1  
 6.54
n n  x 2    x 2
5 5(103.84)  (21.4)2
The 90% confidence interval is
yˆ  E  y  yˆ  E
24.63  6.54  y  24.63  6.54
18.1  y  31.2
(f)  = 0.01
H0:   0
H1:   0
b
1
6.876
1
2
103.84  (21.4)2  9.504
 x2    x  
Se
n
2.5319
5
d.f. = n  2 = 5  2 = 3
From Table 6 in Appendix II, the sample test statistic falls between entries 5.841 and 12.924. Use the
one-tailed areas to find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0012.
t
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Part IV: Complete Solutions, Chapter 10
576
Since the P-value interval  0.01, reject H 0 .
At the 1% level of significance, the sample evidence supports a positive slope.
(g) d.f. = 3, tc  3.182, b  6.876
E
tc Se
2
 x 2  1n  x 
3.182(2.5319)

103.84 
(21.4)2
5
 2.302
The 95% confidence interval is
bE   bE
6.876  2.302    6.876  2.302
4.57    9.18
For every one-unit increase in oxygen pressure breathing only available air, the oxygen pressure
breathing pure oxygen increases from about 4.57 to 9.18 units.
11. (a) Use a calculator to verify.
(b)  = 0.01
H0 :   0
H1:   0
t
r n2
1 r2

0.956 6  2
1  (0.956) 2
 6.517
d.f. = n  2 = 6  2 = 4
From Table 6 in Appendix II, the sample test statistic falls between entries 4.604 and 8.610. Use onetailed areas to find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0014.
Since the P-value interval  0.01, reject H 0 .
At the 1% level of significance, the sample evidence supports a positive correlation.
(c) Use a calculator to verify.
(d) ŷ  a  bx or yˆ  1.965  0.7577 x
Use x = 14.
yˆ  1.965  0.7577(14)  $12.57 thousand.
(e) d.f. = 4, tc  1.778, x 
 x  79.6  13.267
n
6
n( x  x )2
1
1
6(14  13.267)2
E  tc Se 1  
 1.778(0.1527) 1  
 0.329
n n x 2    x 2
6 6(1, 057.8)  (79.6)2
The 85% confidence interval is
yˆ  E  y  yˆ  E
12.57  0.329  y  12.57  0.329
12.24  y  12.90
(f)  = 0.01
H0:   0
H1:   0
b
1
0.7577
1
2
(1,057.8)  (79.6)2  6.608
 x2    x  
Se
n
0.1527
6
d.f. = n  2 = 6  2 = 4
From Table 6 in Appendix II, the sample test statistic falls between entries 4.604 and 8.610. Use onetailed areas to find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0014.
Since the P-value interval  0.01, reject H 0 .
At the 1% level of significance, the sample evidence supports a positive slope.
t
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Part IV: Complete Solutions, Chapter 10
577
(g) d.f. = 4, tc  2.776, b  0.7577
E
tc Se
 x2  1n  x 
2
2.776(0.1527)

(1, 057.8) 
(79.6)2
6
 0.318
The 95% confidence interval is
bE   bE
0.758  0.318    0.758  0.318
0.44    1.08
For every $1,000 increase in list price, there is an increase in dealer price of between $440 and $1,080.
12. (a) Use a calculator to verify.
(b)  = 0.01
H0 :   0
H1:   0
t
r n2
1 r2

0.977 5  2
 7.936
1  (0.977)2
d.f. = n  2 = 5  2 = 3
From Table 6 in Appendix II, the sample test statistic falls between entries 5.841 and 12.924. Use onetailed areas to find that 0.0005 < P-value < 0.005. Using a TI-84, P-value ≈ 0.0021.
Since the P-value interval  0.01, reject H 0 .
At the 1% level of significance, the sample evidence supports a positive correlation.
(c) Use a calculator to verify.
(d) ŷ  a  bx or yˆ  1.4084  0.8794 x
Use x = 40.
yˆ  1.4084  0.8794(40)  $36.58 thousand
(e) d.f. = 3, tc  3.182, x 
 x  192.5  38.7
n
5
1
n( x  x )2
1
5(40  38.7)2
E  tc Se 1  
 3.182(1.5223) 1  
 5.33
n n  x 2    x 2
5 5(7, 676.71)  (193.5)2
The 95% confidence interval is
yˆ  E  y  yˆ  E
36.58  5.33  y  36.58  5.33
31.25  y  41.91
(f)  = 0.01
H0:   0
H1:   0
b
1
0.8794
1
2
7,676.71  (193.5)2  7.926
 x2    x  
Se
n
1.5223
5
d.f. = n  2 = 5  2 = 3
From Table 6 in Appendix II, the sample test statistic falls between entries 5.841 and 12.924. Use onetailed areas to find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0021. Since the P-value
interval  0.01, reject H 0 .
At the 1% level of significance, the sample evidence supports a positive slope.
t
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Part IV: Complete Solutions, Chapter 10
578
(g) d.f. = 3, tc  2.353, b  0.8794
tc Se
E
1
n

2.353(1.5223)
 0.261
2
(193.5)
   x
7, 676.71 
5
The 90% confidence interval is
bE   bE
0.879  0.261    0.879  0.261
0.618    1.140
For every $1,000 increase in list price, the dealer price is $618 to $1,140 higher.
x2
2
13. (a)  = 0.01
H 0 :  0
H1:  0
r n2
0.90 6  2
t

 4.129
2
1 r
1  (0.90) 2
d. f .  n  2  6  2  4
From Table 6 in Appendix II, 4.129 falls between entries 3.747 and 4.604. Use two-tailed areas to find
0.010 < P value < 0.020. Since the P-value interval > 0.01, we do not reject H 0 . The correlation
coefficient  is not significantly different from 0 at the 0.01 level of significance.
(b)  = 0.01
H0:   0
H1:   0
r n  2 0.90 10  2
t

 5.840
1 r2
1  (0.90) 2
d . f .  n  2  10  2  8
From Table 6 in Appendix II, 5.840 falls to the right of entry 5.041. Use two-tailed areas to find
P value < 0.001. Since the P value  0.01, we reject H 0 . The correlation coefficient  is significantly
different from 0 at the 0.01 level of significance.
(c) From part (a) to part (b), n increased from 6 to 10, and the test statistic t increased from 4.129 to 5.840.
For the same r = 0.90 and the same level of significance   0.01, we rejected H 0 for the larger n but
not for the smaller n.

r n2 
In general, as n increases, the degrees of freedom (n – 2) increase, and the test statistic  t 



1 r2 

increases. This produces a smaller P value.
14. (a) Yes. The t values are equal (differences are due to rounding error).
(b) Yes to all questions.
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Part IV: Complete Solutions, Chapter 10
579
Section 10.4
1.
x1  1.6  3.5x2  7.9 x3  2.0 x4
(a) The response variable is x1. The explanatory variables are x2, x3, and x4.
(b) The constant term is 1.6.
The coefficient 3.5 corresponds to x2 .
The coefficient –7.9 corresponds to x3 .
The coefficient 2.0 corresponds to x4 .
(c) x2  2, x3  1, x4  5
x1  1.6  3.5(2)  7.9(1)  2.0(5)  10.7
The predicted value is 10.7.
(d) In multiple regression, the coefficients of the explanatory variables can be thought of as “slopes” if we
look at one explanatory variable’s coefficient at a time while holding the other explanatory variables as
arbitrary and fixed constants.
x3 and x4 held constant, x2 increased by one unit, the change in x1 would be an increase of 3.5 units.
x3 and x4 held constant, x2 increased by two units, the change in x1 would be an increase of 7 units.
x3 and x4 held constant, x2 decreased by four units, the change in x1 would be a decrease of 14 units.
(e) d . f .  n  k  1  12  3  1  8
A 90% confidence interval for the coefficient of x2 is b2  tS2  2  b2  tS2 .
3.5  1.86(0.419)   2  3.5  1.86(0.419)
2.72   2  4.28
(f)  = 0.05
H 0: 2  0
H1:  2  0
b   2 3.5  0
t 2

 8.35
S2
0.419
d.f. = 8
From Table 6 in Appendix II, 8.35 falls to the right of entry 5.041. Use two-tailed areas to find that Pvalue < 0.001. Since 0.001  0.05, reject H 0 .
We conclude that 2  0 and x2 should be included as an explanatory variable in the least-squares
equation.
2.
x3  16.5  4.0 x1  9.2 x4  1.1x7
(a) The response variable is x3 . The explanatory variables are x1, x4 , and x7 .
(b) The constant term is –16.5.
The coefficient 4.0 corresponds to x1.
The coefficient 9.2 corresponds to x4 .
The coefficient –1.1 corresponds to x7 .
(c) x1  10, x4  1, x7  2
x3  16.5  4.0(10)  9.2(1)  1.1(2)  12.1
The predicted value is 12.1.
(d) In multiple regression, the coefficients of the explanatory variables can be thought of as “slopes” if we
look at one explanatory variable’s coefficient at a time while holding the other explanatory variables as
arbitrary and fixed constants.
x1 and x7 held constant, x4 increased by one unit, the change in x3 is an increase of 9.2 units.
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Part IV: Complete Solutions, Chapter 10
580
x1 and x7 held constant, x4 increased by three units, the change in x3 is an increase of 27.6 units
x1 and x7 held constant, x4 decreased by two units, the change in x3 is a decrease of 18.4 units
(e) d . f .  n  k  1  15  3  1  11
A 90% confidence interval for the coefficient of x4 is b4  tS4   4  b4  tS4 .
9.2  1.796(0.921)   4  9.2  1.796(0.921)
7.55   4  10.85
(f)  = 0.01
H 0 : 4  0
H1:  4  0
b   4 9.2  0
t 4

 9.989
S4
0.921
d.f. = 11
From Table 6 in Appendix II, 9.989 falls to the right of entry 4.437. Use two-tailed areas to find that P
value < 0.001. Since 0.001  0.01, we reject H 0 .
We conclude that 4  0 and x4 should be included as an explanatory variable in the least-squares
equation.
3. (a)
s
 100
x
9.08%
CV 
x
s
x1
150.09
13.63
x2
62.45
9.11
14.59%
x3
195.0
17.31
8.88%
Relative to its mean, x2 has the greatest spread of data values, and x3 has the smallest spread of data
values.
(b) r 2 x1x2  (0.979)2  0.958
r 2 x1x3  (0.971)2  0.943
r 2 x2 x3    0.895
The variable x2 has the greatest influence on x1 (0.958  0.943).
Yes. Both variables x2 and x3 show a strong influence on x1 because 0.958 and 0.943 are close to 1.
95.8% of the variation of x1 can be explained by the corresponding variation in x2 .
94.3% of the variation of x1 can be explained by the corresponding variation in x3 .
(c) R 2  0.977
97.7% of the variation in x1 can be explained by the corresponding variation in x2 and x3 taken
together.
(d) x1  30.99  0.861x2  0.355x3
In multiple regression, the coefficients of the explanatory variables can be thought of as “slopes” if we
look at one explanatory variable’s coefficient at a time while holding the other explanatory variables
as arbitrary and fixed constants.
If age ( x2 ) were held fixed and x3 increased by 10 pounds, the systolic blood pressure would be
expected to increase by 0.335(10) = 3.35.
If weight ( x3 ) were held fixed and x2 increased by 10 years, the systolic blood pressure would be
expected to increase by 0.861(10) = 8.61.
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Part IV: Complete Solutions, Chapter 10
581
(e)  = 0.05
H 0 : i  0
H1: i  0
d . f  n  k  1  11  2  1  8
For  2 , the sample test statistic is t = 3.47 with P value = 0.008.
For  3 , the sample test statistic is t = 2.56 with P value = 0.034.
Since 0.008  0.05 and 0.034  0.05, reject H 0 for each coefficient and conclude that the coefficients of
x2 and x3 are not zero. Explanatory variables xi whose coefficients ( i ) are nonzero, contribute
information in the least-squares equation; i.e., without these xi , the resulting least-squares regression
equation is not as good a fit to the data as is the regression equation that includes these xi .
(f) d . f .  8, t  1.86
A 90% confidence interval for i is
bi  tSi  i  bi  tSi
0.861  1.86(0.2482)   2  0.861  1.86(0.2482)
0.40   2  1.32
0.335  1.86(0.1307)  3  0.335  1.86(0.1307)
0.09  3  0.58
(g) x1  30.99  0.861(68)  0.335(192)  153.9
Michael’s predicted systolic blood pressure is 153.9, and a 90% confidence interval for this new
observation’s value, given these xi (i.e., the prediction interval), is 148.3 to 159.4.
4. (a)
x
s
x1
79.04
12.28
s
100
x
15.53%
x2
79.48
12.50
15.73%
x3
81.48
11.77
14.44%
x4
162.04
24.04
14.83%
CV 
Relative to its mean, each exam had about the same spread of scores. Yes, it seems that all the exams
were about the same level of difficulty.
(b) r 2 x1x2  (0.901) 2  0.812
r 2 x1x3  (0.893) 2  0.797
r 2 x1x4    0.895
r 2 x2 x3  (0.846) 2  0.716
r 2 x2 x4  (0.929) 2  0.863
r 2 x3 x4  (0.972) 2  0.945
Exam 3 had the most influence on the final exam 4 (0.945 > 0.895 > 0.863). Even though exam 3 had
more influence, the other two exams still had influence on the final because 0.895 and 0.863 are close
to 1.
(c) R 2  0.990
99.0% of the variation in x4 can be explained by the corresponding variationin x1,x 2 , and x3 taken
together.
(d) x4  4.34  0.356 x1  0.543x2  1.17 x3
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Part IV: Complete Solutions, Chapter 10
582
In multiple regression, the coefficients of the explanatory variables can be thought of as “slopes” if we
look at one explanatory variable’s coefficient at a time while holding the other explanatory variables
as arbitrary and fixed constants.
If age x1 and x2 are held fixed and x3 is increased by 10 points, the final exam score is expected to
increase by 10(1.17) = 11.7  12 points.
(e)  = 0.05
H 0 : i  0
H1: i  0
d . f  n  k  1  25  3  1  21
For 1, the sample test statistic is t = 2.93 with P value = 0.008.
For  2 , the sample test statistic is t = 5.38 with P value  0.000.
For  3 , the sample test statistic is t = 11.33 with P value  0.000.
Since 0.008  0.05, 0.000  0.05, and 0.000  0.05, reject H 0 for each coefficient and conclude that the
coefficients of x1 , x2 , and x3 are not zero. Explanatory variables xi , whose coefficients ( i ) are
nonzero, contribute information in the least-squares equation; i.e. without these xi , the resulting leastsquares regression equation is not as good a fit to the data as is the regression equation that includes
these xi .
(f) d . f .  21, t  1.721
A 90% confidence interval for i is bi  tSi  i  bi  tSi .
0.356  1.721(0.1214)  1  0.356  1.721(0.1214)
0.147  1  0.565
0.543  1.721(0.1008)   2  0.543  1.721(0.1008)
0.370   2  0.716
1.167  1.721(0.1030)  3  1.167  1.721(0.1030)
0.990  3  1.344
(g) x4  4.34  0.356(68)  0.543(72)  1.17(75)  147
Susan’s predicted score on the final exam is 147, and a 90% confidence interval for this new
observation’s value, given these xi , i.e., the prediction interval, is 142–151, all rounded to whole
numbers.
5. (a)
s
100
x
39.64%
CV 
x
s
x1
85.24
33.79
x2
8.74
3.89
44.51%
x3
4.90
2.48
50.61%
x4
9.92
5.17
52.12%
Relative to its mean, x4 has the largest spread of data values. The larger the CV, the more we expect the
variable to change relative to its average value because a variable with a large CV has a large standard
deviation s relative to x , and s measures “spread,” or variability, in the data. x1 has a small CV
because we divide by a large mean.
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Part IV: Complete Solutions, Chapter 10
583
(b) r 2 x1x2  (0.917) 2  0.841
r 2 x1x3  (0.930) 2  0.865
r 2 x1x4    0.226
r 2 x2 x3  (0.790) 2  0.624
r 2 x2 x4  (0.429) 2  0.184
r 2 x3 x4  (0.299) 2  0.089
The variable x4 has the least influence on box office receipts x1 (0.226 < 0.841 < 0.865).
x2 = production costs, r 2 x1x2  0.841.
84.1% of the variation of box office receipts can be attributed to the corresponding variation in
production costs.
(c) R 2  0.967
96.7% of the variation in x1 can be explained by the corresponding variation in x2 , x3 , and x4 taken
together.
(d) x1  7.68  3.66 x2  7.62 x3  0.83x4
In multiple regression, the coefficients of the explanatory variables can be thought of as “slopes” if we
look at one explanatory variable’s coefficient at a time while holding the other explanatory variables as
arbitrary and fixed constants.
If x2 and x4 were held fixed and x3 were increased by 1 ($1 million), the corresponding change in x1
(box office receipts) would be an increase of $7.62 million.
(e)  = 0.05
H 0 : i  0
H1: i  0
d . f  n  k  1  10  3  1  6
For  2 , the sample test statistic is t = 3.28 with P value = 0.017.
For  3 , the sample test statistic is t = 4.60 with P value = 0.004.
For  4 , the sample test statistic is t = 1.54 with P value = 0.175.
Since 0.017  0.05 and 0.004  0.05, reject H 0 for coefficients  2 and 3 and conclude that the
coefficients of x2 and x3 are not zero.
Since 0.175 > 0.05, do not reject H 0 for the coefficient  4 and conclude that the coefficient of x4 could
be zero. If  4  0, then x4 contributes nothing to the population regression line. We can eliminate the
variable x4 and fit the estimated regression line without it and probably see little, if any, difference
between the predicted values of x1 based on x2 and x3 only and the predicted values of x1 based on
x2 , x3 , and x4 .
(f) d . f .  6, t  1.943
A 90% confidence interval for i is
bi  tSi  i  bi  tSi
3.662  1.943(1.118)   2  3.662  1.943(1.118)
1.49   2  5.83
7.621  1.943(1.657)  3  7.621  1.943(1.657)
4.40  3  10.84
0.8285  1.943(0.5394)   4  0.8285  1.943(0.5394)
0.22   4  1.88
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Part IV: Complete Solutions, Chapter 10
584
(g) x1  7.68  3.66(11.4)  7.62(4.7)  0.83(8.1)  91.94
The prediction is $91.94 million, and a 85% confidence interval for this new observation’s value,
given these xi (i.e., the prediction interval), is $77.6 million to $106.3 million.
(h) x3  0.650  0.102 x1  0.260 x2  0.0899 x4
x3  0.650  0.102(100)  0.260(12)  0.0899(9.2)
x3  5.63
The prediction is $5.63 million, and a 80% confidence interval for this new observation’s value, given
these xi (i.e. the prediction interval), is $4.21 million to $7.04 million.
6. (a)
s
100
x
67.02%
CV 
x
s
x1
286.574
192.062
x2
3.326
2.011
60.46%
x3
387.481
191.168
49.34%
x4
8.100
3.775
46.60%
x5
9.693
5.140
53.03%
x6
7.741
4.896
63.25%
Relative to its mean, x1 has the largest spread of data values, and x4 has a smallest spread of data
values.
(b) r23  0.844
r24  0.749
r25  0.838
r26  0.766
r34  0.906
r35  0.864
r36  0.807
r45  0.795
r46  0.841
r56  0.870
r 2 x1x2  (0.894)2  0.799
r 2 x1x3  (0.946)2  0.895
r 2 x1x4    0.835
r 2 x1x5  (0.954)2  0.910
r 2 x1x6  (0.912)2  0.832
The variable x5 has the greatest influence on annual net sales (0.910 is the largest). The variable x2 has
the least influence on annual net sales (0.799 is the smallest).
(c) R 2  0.993
99.3% of the variation in x1 can be explained by the corresponding variation
in x2 , x3 , x4 , x5 , and x6 taken together.
(d) x1  18.9  16.2 x2  0.175x3  11.5x4  13.6 x5  5.31x6
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Part IV: Complete Solutions, Chapter 10
585
In multiple regression, the coefficients of the explanatory variables can be thought of as “slopes” if we
look at one explanatory variable’s coefficient at a time while holding the other explanatory variables as
arbitrary and fixed constants.
If all explanatory variables except x6 remained fixed and x6 increased by 2, then the annual net sales
are expected to decrease by 2(5.31) = 10.62 or $10,620.
If all explanatory variables except x4 remained fixed and x4 increased by 1 ($1,000), then the annual
net sales are expected to increase by 11.5, or $11,500.
(e)  = 0.05
H 0 : i  0
H1: i  0
d . f  27  5  1  21
For  2 , the sample test statistic is t = 4.57 with P value  0.000.
For  3 , the sample test statistic is t = 3.03 with P value = 0.006.
For  4 , the sample test statistic is t = 4.55 with P-value  0.000.
For  5 , the sample test statistic is t = 7.67 with P value  0.000.
For  6 , the sample test statistic is t = 3.11 with a P value = 0.005.
Since all these P values are  0.05, we reject H 0 for each coefficient and conclude that the coefficients
of x2 , x3 , x4 , x5 , and x6 are not 0.
(f) The predicted annual net sales are 194.41 (or $194.41 thousand), and the 80% confidence interval for
this new observation’s value, given these xi (i.e. the prediction interval), is $160.76 thousand to
$228.06 thousand.
(g) x4  4.14  0.0431x1  0.800 x2  0.00059 x3  0.661x5  0.057 x6
The predicted amount spent on local advertising is $5.571 thousand, and the 80% confidence interval
for this new observation’s value, given these xi (i.e., the prediction interval), is $4.048 thousand to
$7.094 thousand. Advertising costs for this store should be between $4,048 and $7,094.
Chapter 10 Review
1.
We expect r to be close to 0.
2.
A significant r means that we reject the null hypothesis that ρ = 0 and conclude that the population
correlation coefficient is not equal to zero.
3.
Results are more reliable for interpolation.
4.
Here, residual  y  yˆ  8  6  2.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
586
5. (a)
Scatterplot of Mortality Rate vs Age
20
Mortality Rate
19
18
17
16
15
14
1
2
3
Age
4
5
 x 15

3
n
5
 y 86.9
y

 17.38
n
5
n xy    x   y  5(273.4)  (15)(86.9)
b

 1.27
2
5(55)  (15) 2
n x 2    x 
a  y  bx  17.38  1.27(3)  13.57
y  a  bx or y  13.57  1.27 x
(b) x 
(c)
r
n xy    x   y 
n
x2
  x
2
n
y2
  y
2

5(273.4)  (15)(86.9)
5(55)  (15)2 5(1,544.73)  (86.9)2
 0.685
r 2  (0.685) 2  0.469
The correlation coefficient r measures the strength of the linear relationship between a bighorn sheep’s
age and the mortality rate. The coefficient of determination r 2 means that 46.9% of the variation in
mortality rate can be explained by the corresponding variation in age of a bighorn sheep using the
least-squares line.
(d)  = 0.01
H0:   0
H1:   0
r n2
0.685 5  2
t

 1.629
1 r2
1  (0.685) 2
d.f. = n  2 = 5  2 = 3
From Table 6 in Appendix II, 1.629 falls between entries 1.423 and 1.638. Use one-tailed areas to find
that 0.100 < P value < 0.125. Using a TI-84, P value ≈ 0.1011.
Since the P-value interval is > 0.01, we do not reject H 0 .
There does not seem to be a positive correlation between age and mortality rate of bighorn sheep.
(e) No, based on this limited data, predictions from the least-squares line model might be misleading.
There appear to be other lurking variables that affect the mortality rate of sheep in different age groups.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
587
6. (a)
Scatterplot of Annual Salary vs Number of Job Changes
44
42
Annual Salary
40
38
36
34
32
30
0
2
4
6
Number of Job Changes
8
10
 x 60

 6.0
n
10
 y 359
y

 35.9
n
10
n xy    x   y  10(2231)  (60)(359)
b

 0.939024
2
10(442)  (60) 2
n x 2    x 
a  y  bx  35.9  0.939024(6.0)  30.266
yˆ  a  bx or yˆ  30.266  0.939 x
(c)
n xy    x   y 
10(2, 231)  (60)(359)
r

 0.761
2
2
10(442)  (60) 2 10(13, 013)  (359) 2
n x 2    x  n y 2    y 
(b) x 
r 2  (0.761)2  0.579
This means that 57.9% of the variation in salary can be explained by the corresponding variation in
number of job changes using the least-squares line.
(d)  = 0.05
H0:   0
H1:   0
r n  2 0.761 10  2
t

 3.318
1 r2
1  (0.761) 2
d . f .  n  2  10  2  8
From Table 6 in Appendix II, 3.32 falls between entries 2.896 and 3.355. Use one-tailed areas to find
that 0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0053.
Since the P-value interval is  0.05, reject H 0 and conclude that the sample evidence supports a
positive correlation.
(e) Let x = 2.
yˆ  30.266  0.939(2)  32.14
The predicted salary is $32,140.
(f)
Se 
 y 2  a y  b xy  13,013  30.266(359)  0.939024(2, 231)  2.56
n2
Copyright © Houghton Mifflin Company. All rights reserved.
10  2
Part IV: Complete Solutions, Chapter 10
588
(g) E  tc Se 1 
1
n( x  x ) 2

n n  x 2    x 2
 1.860(2.564058) 1 
1
10(2  6) 2

10 10(442)  602
 5.43
A 90% confidence interval for y is
yˆ  E  y  yˆ  E
32.14  5.43  y  32.14  5.43
26.71  y  37.57
(h)  = 0.05
H0:   0
H1:   0
b
1
0.939204
1
2
t
442  (60)2  3.32
 x2    x  
Se
n
2.56
10
d . f .  n  2  10  2  8
From Table 6 in Appendix II, 3.32 falls between entries 2.896 and 3.355. Use one-tailed areas to find
that 0.005 < P value < 0.010. Using a TI-84, P value ≈ 0.0053.
Since the P-value interval  0.05, reject H 0 and conclude that the sample evidence supports a positive
slope.
(i) d.f. = 8, tc  1.860, b  0.939
E
tc Se
x 
2
 x
2

n
1.86(2.56)
2
 0.526
442  60
10
A 90% confidence interval is
bE   bE
0.939  0.526    0.939  0.526
0.413    1.465
At the 90% confidence level, we can say that for each increase of one job change, the annual salary
increases from 0.4 to 1.4 thousand dollars.
7. (a)
Scatterplot of Weight @ 30 years vs Weight @ 1 year
150
Weight @ 30 years
140
130
120
110
15.0
17.5
20.0
22.5
Weight @ 1 year
25.0
27.5
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
589
 x 300

 21.43
n
14
 y 1, 775
y

 126.79
n
14
n xy    x   y  14(38, 220)  (300)(1, 775)
b

 1.285
2
14(6,572)  (300) 2
n x 2    x 
a  y  bx  126.79  (1.285)(21.43)  99.25
yˆ  a  bx or yˆ  99.25  1.285 x
(c)
(b) x 
r
n xy    x   y 
n x 2    x 
2
n y 2    y 
2

14(38, 220)  (300)(1, 775)
14(6,572)  (300) 2 14(226,125)  (1, 775) 2
 0.468
r 2  (0.468) 2  0.219
The coefficient of determination r 2 means that 21.9% of the variation in weight of a 30-year-old
female can be explained by the corresponding variation in the weight of a 1-year-old baby girl using
the least-squares line.
(d)  = 0.01
H0:   0
H1:   0
r n  2 0.468 14  2
t

 1.834
1 r2
1  (0.468) 2
d . f .  n  2  14  2  12
From Table 6 in Appendix II, the sample test statistic falls between entries 1.782 and 2.179. Use onetailed areas to find that 0.025 < P value < 0.050. Using a TI-84, P value ≈ 0.0457. Since the P-value
interval > 0.01, do not reject H 0 . At the 1% level of significance, there does not seem to be a positive
correlation between weight of baby and weight of adult.
(e) Let x = 20.
yˆ  99.25  1.285(20)  124.95
The predicted weight is 124.95 pounds, but since r is not significant, the prediction might not be very
reliable.
(f)
Se 
 y 2  a y  b xy 
(g) E  tc Se 1 
n2
226,125  99.25(1, 775)  1.285(38, 220)
 8.38
14  2
1
n( x  x ) 2

n n  x 2    x 2
 2.179(8.38) 1 
1
14(20  21.43) 2

14 14(6,572)  (300) 2
 19.03
A 95% confidence interval for y is
yˆ  E  y  yˆ  E
124.95  19.03  y  124.95  19.03
105.92  y  143.98
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
590
(h)  = 0.01
H0:   0
H1:   0
b
1
1.285
1
2
6,572  (300)2  1.84
 x2    x  
Se
n
8.38
14
d . f .  n  2  14  2  12
From Table 6 in Appendix II, the sample test statistic falls between entries 1.782 and 2.179. Use onetailed areas to find that 0.025 < P value < 0.050. Using a TI-84, P value ≈ 0.0457. Since the P-value
interval > 0.01 for , do not reject H 0 . At the 1% level of significance, there does not seem to be a
positive slope between weight of baby and weight of adult.
t
(i) d.f. = 12, tc  1.356, b = 1.285
tc Se
E
 x2 
 x
2
n

1.356(8.38)
(300)2
6,572  14
 0.949
An 80% confidence interval is
bE   bE
1.285  0.949    1.285  0.949
0.34    2.22
At the 80% confidence level, we can say that for each additional pound a female infant weighs at 1
year, the adult weight increases by 0.34–2.22 pounds.
8. (a)
Scatterplot of Number of Buyers vs Number of Visits
11
10
Number of Buyers
9
8
7
6
5
4
3
2
5
10
15
20
Number of Visits
25
30
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Part IV: Complete Solutions, Chapter 10
591
 x 248

 16.53  16.53
n
15
 y 97
y

 6.46  6.47
n
15
n xy    x   y  15(1,825)  (248)(97)
b

 0.292784
2
15(4,856)  (248) 2
n x 2    x 
(b) x 
a  y  bx  6.46  0.292784(16.53)  1.626
yˆ  a  bx or yˆ  1.626  0.293x
r
(c)
n xy    x   y 
n
x2
  x
2
n
y2
  y
2

15(1,825)  (248)(97)
15(4,856)  (248) 2 15(731)  (97) 2
 0.790
 0.624
This means that 62.4% of the variation in number of people who bought insurance that week can be
explained by the corresponding variation in number of visits made each week using the least-squares
line.
(d)  = 0.01
H0:   0
H1:   0
r n  2 0.790 15  2
t

 4.65
1 r2
1  (0.790) 2
r2
 (0.790) 2
d . f .  n  2  15  2  13
From Table 6 in Appendix II, 4.65 falls to the right of entry 4.221. Use one-tailed areas to find that P
value < 0.005. Using a TI-84, P value ≈ 0.0002. Since the P value  0.01, reject H 0 . At the 1% level of
significance, there is sufficient evidence to show a positive correlation between number of visits and
number of sales.
(e) Let x = 18.
yˆ  1.626  0.293(18)  6.9
The predicted number of sales is 6.9, or 7.
(f)
Se 
 y 2  a y  b xy  731  1.626(97)  0.292784(1,825)  1.7309
(g) E  tc Se 1 
n2
15  2
n( x  x ) 2
1

n n  x 2    x 2
 2.160(1.730940) 1 
1
15(18  16.53) 2

15 15(4,856)  (248) 2
4
A 95% confidence interval for y is
yˆ  E  y  yˆ  E
74 y  74
3  y  11
(h)  = 0.01
H0:   0
H1:   0
b
1
0.292784
1
2
4,856  (248)2  4.65
 x2    x  
Se
n
1.7309
15
d . f .  n  2  15  2  13
t
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
592
From Table 6 in Appendix II, 4.65 falls to the right of entry 4.221. Use one-tailed areas to find that P
value < 0.0005. Using a TI-84, P value ≈ 0.0002. Since the P value  0.01, reject H 0 . At the 1% level
of significance, there is sufficient evidence to show a positive slope between number of visits and
number of sales.
(i) d.f. = 13, tc  1.350, b = 0.292784
tc Se
E
x
2

 x
2

1.350(1.731)
(248)2
 0.085
4856  15
n
An 80% confidence interval is
bE   bE
0.292784  0.085    0.292784  0.085
0.208    0.378
At the 80% confidence level, we can say that for each additional visit made, between 0.21 and 0.38
more sales are made.
9. (a)
Scatterplot of Number of Employees vs Weight of Mail
17.5
Number of Employees
15.0
12.5
10.0
7.5
5.0
5
10
15
Weight of Mail
20
25
 x 131

 16.375  
n
8
 y 81
y

 10.125  10.13
n
8
n xy    x   y  8(1, 487)  (131)(81)
b

 0.554118  0.554
2
8(2, 435)  (131) 2
n x 2    x 
a  y  bx  10.125  0.554118(16.375)  1.051
yˆ  a  bx or yˆ  1.051  0.554 x
(b) x 
(c)
r
n xy    x   y 
n x 2    x 
2
n y 2    y 
2

8(1, 487)  (131)(81)
8(2, 435)  (131)2 8(927)  (81) 2
 0.913
r 2  (0.913) 2  0.833
The coefficient of determination r 2 means that 83.3% of the variation in number of employees can be
explained by the corresponding variation in weight of incoming mail using the least-squares line.
(d)  = 0.01
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Part IV: Complete Solutions, Chapter 10
593
H0:   0
H1:   0
r n2
0.913 8  2
t

 5.48
2
1 r
1  (0.913) 2
d. f .  n  2  8  2  6
From Table 6 in Appendix II, the sample test statistic falls between entries 3.707 and 5.959. Use onetailed areas to find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0008. Since the P-value
interval  0.01, reject H 0 . At the 1% level of significance, there is sufficient evidence to show a
positive correlation between pounds of mail and number of employees required to process the mail.
(e) Use x = 15.
yˆ  1.051  0.554(15)  9.36
About nine employees should be assigned mail duty.
(f)
Se 
 y 2  a y  b xy  927  1.051(81)  0.554118(1, 487)  1.73
n2
(g) E  tc Se 1 
82
n( x  x ) 2
1

n n  x 2    x 2
1 8(15  16.375) 2
 2.447(1.73) 1  
8 8(2, 435)  (131) 2
 4.5
A 95% confidence interval for y is
yˆ  E  y  yˆ  E
9.36  4.5  y  9.36  4.5
4.86  y  13.86
(h)  = 0.01
H0:   0
H1:   0
b
1
0.554118
1
2
t
2435  (131)2  5.45
 x2    x  
Se
n
1.73
8
d. f .  n  2  8  2  6
From Table 6 in Appendix II, the sample test statistic falls between entries 3.707 and 5.959. Use onetailed areas to find that 0.0005 < P value < 0.005. Using a TI-84, P value ≈ 0.0008. Since the P-value
interval  0.01, reject H 0 . At the 1% level of significance, there is sufficient evidence to show a
positive slope between pounds of mail and number of employees required to process the mail.
(i) d . f .  6, tc  1.440, b  0.554
tc Se
E

x2

 x
n
2

1.440(1.73)
2, 435 
(131)2
8
 0.146
An 80% confidence interval is
bE   bE
0.554  0.146    0.554  0.146
0.41    0.70
At the 80% confidence level, we can say that for each additional pound of mail, between 0.4 and 0.7
additional employees are needed.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
594
10.
(a)
Scatterplot of Crime Rate vs Population % Change
200
175
Crime Rate
150
125
100
75
50
0
5
10
15
20
Population % Change
25
30
 x 72

 12.0
n
6
 y 589
y

 98.16  98.17
n
6
n xy    x   y  6(9, 499)  (72)(589)
b

 5.1071  5.11
2
6(1,340)  (72)2
n x 2    x 
(b) x 
a  y  bx  98.16  5.1071(12.0)  36.881  36.9
yˆ  a  bx or yˆ  36.9  5.11x
(c)
r
n xy    x   y 
n
x2
  x
2
n
y2
  y
2

6(9, 499)  (72)(589)
6(1,340)  (72)2 6(72, 277)  (589) 2
 0.927
r 2  (0.927) 2  0.859
This means that 85.9% of the variation in crime rate can be explained by the corresponding variation in
percent population change.
(d)  = 0.01
H0:   0
H1:   0
r n2
0.927 6  2
t

 4.94
2
1 r
1  (0.927) 2
d. f .  n  2  6  2  4
From Table 6 in Appendix II, the sample test statistic falls between entries 4.604 and 8.610. Use twotailed areas to find that 0.001 < P value < 0.010. Using a TI-84, P value ≈ 0.0079. Since the P-value
interval < 0.01, reject H 0 . At the 1% level of significance, there is sufficient evidence to show a
nonzero correlation between population change and crime rate.
Copyright © Houghton Mifflin Company. All rights reserved.
Part IV: Complete Solutions, Chapter 10
595
(e) Let x = 12
yˆ  36.9  5.11(12)  98.2
The predicted crime rate is 98.2 crimes per thousand.
(f)
Se 
 y 2  a y  b xy 
72, 277  36.881(589)  5.1071(9, 499)
 22.59
62
n2
(g) E  tc Se 1 
1
n( x  x ) 2

n n  x 2    x 2
1
6(12  12)2
 1.533(22.59) 1  
 37.41
6 6(1,340)  (72)2
A 80% confidence interval for y is
yˆ  E  y  yˆ  E
98.2  37.41  y  98.2  37.41
60.8  y  135.6, or about 61 to 136 crimes per thousand.
(h)  = 0.01
H0:   0
H1:   0
b
1
5.1071
1
2
1,340  (72)2  4.93
 x2    x  
Se
n
22.59
6
d. f .  n  2  6  2  4
From Table 6 in Appendix II, the sample test statistic falls between entries 4.604 and 8.610. Use twotailed areas to find that 0.001 < P value < 0.010. Using a TI-84, P value ≈ 0.0079. Since the P-value
interval < 0.01, reject H 0 . At the 1% level of significance, there is sufficient evidence to show a
nonzero slope between population change and crime rate.
t
(i)
d . f .  4, tc  1.533, b  5.11
E
tc Se
 x
2

1.533(22.59)
(72)2
 1.59
1,340  6
  n
An 80% confidence interval is
x2
bE   bE
5.11  1.59    5.11  1.59
3.52    6.70
At the 80% confidence level, we can say that for every percentage point increase in population, expect
the crime rate per 1,000 to increase from between 3.52 and 6.70 crimes per thousand.
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