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5.
The heights and weights for the men (dots) are on
6. (a) Let xrepresent velocity and let yrepresent acceleration.
(b)
0.5 1 1.5 2 2.5 3 3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
Velocity
Acceleration
(c) No, the point (2.92, 5.00) is an outlier.
(d) No effect. Converting units from meters to centimeters and from seconds to minutes involves multi-
plying by a constant, which does not change the correlation coefficient.
7. (a) Let xrepresent temperature, yrepresent stirring rate, and zrepresent yield.
Page 348
SECTION 7.1 349
(b) No, the result might be due to confounding, since the correlation between temperature and stirring
8. (a) Let xrepresent temperature, yrepresent stirring rate, and zrepresent yield.
Then x= 126.5, y= 45, z= 75.59, Pn
i=1(xi−x)2= 2420, Pn
i=1(yi−y)2= 2000,
Page 349
350 CHAPTER 7
9. (a) x= 201.3, y= 202.2, Pn
i=1(xi−x)2= 2164.1, Pn
i=1(yi−y)2= 1571.6,
(b) The null and alternate hypotheses are H0:ρ≥0.9 versus H1:ρ < 0.9.
(c) r= 0.930698, n= 10, U=r√n−2/√1−r2= 7.1965.
10. (a) r= 0.15, n= 300, U=r√n−2/√1−r2= 2.61903.
Page 350
SECTION 7.1 351
11. r=−0.9515, W=1
2ln 1 + r
1−r=−1.847395, σW=p1/(30000 −3) = 0.00577379.
12. r= 0.25, n= 134, U=r√n−2/√1−r2= 2.96648.
13. r=−0.509, n= 23, U=r√n−2/√1−r2=−2.7098.
14. x= 0 and Pn
i=1(xi−x)2= 10.
y= [−2 + (−1) + 0 + 1 + y]/5 = (y−2)/5, so y= 5y+ 2.
Express Pn
i=1(xi−x)(yi−y) , Pn
i=1(yi−y)2, and rin terms of y:
Pn
352 CHAPTER 7
(a) Substitute r= 1 to obtain −y2= 0, so y= 0, and y= 5(0) + 2 = 2.
Section 7.2
1. (a) 245.82 + 1.13(65) = 319.27 pounds
2. (a) −196.32 + 2.42(102.7) = 52.21 ksi.
3. r2= 1 −Pn
i=1(yi−byi)2
Pn
i=1(yi−y)2= 1 −1450
9615 = 0.8492.
Page 352
SECTION 7.2 353
i=1(yi−byi)2
i=1(yi−y)2= 1 −33.9
5. (a) −0.2967 + 0.2738(70) = 18.869 in.
6. n= 40, Pn
i=1(xi−x)2= 98,775, Pn
i=1(yi−y)2= 19.10, x= 26.36, y= 0.5188,
Pn
i=1(xi−x)(yi−y) = 826.94.
i=1(xi−x)(yi−y)
7. b
β1=rsy/sx= (0.85)(1.9)/1.2 = 1.3458. b
β0=y−b
β1x= 30.4−1.3458(8.1) = 19.499.
354 CHAPTER 7
8. (a)
4 4.5 5 5.5 6
50
60
70
80
90
100
110
Diameter (mm)
Strength (kN/mm)
The linear model is appropriate.
(b) x= 5.1, y= 79.9, Pn
i=1(xi−x)2= 3.3, Pn
i=1(yi−y)2= 2994.9,
Pn
(c) The fitted values are found by computing, for each i, the quantity byi=b
β0+b
β1xi. The residuals are
(d) Increase by 28.939394(0.3) = 8.682 kN/mm.
Page 354
9. (a)
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0
1
2
3
4
Damping Ratio
Frequency (Hz)
(b) x= 0.527778, y= 1.752778, Pn
i=1(xi−x)2= 0.476111, Pn
i=1(yi−y)2= 13.672761,
Pn
i=1(xi−x)(yi−y) = −2.335389.
10. (a)
5 10 15 20 25 30
0
0.5
1
1.5
2
Temperature (°C)
Corrosion (mm/yr)
(b) x= 17.75556, y= 1.02444, Pn
i=1(xi−x)2= 485.5022, Pn
i=1(yi−y)2= 0.88302,
Page 355
356 CHAPTER 7
(c) 0.035509(10) = 0.35509 mm/yr
(f) The residuals are the values ei=yi−byifor each i. They are shown in the following table.
Fitted Value Residual
x y by=b
β0+b
β1x e =y−by
(g) r=Pn
i=1(xi−x)(yi−y)
pPn
i=1(xi−x)2pPn
i=1(yi−y)2= 0.832627.
Page 356
SECTION 7.2 357
(h) The total sum of squares is Pn
i=1(yi−y)2= 0.8830.
11. (a)
4 4.5 5 5.5 6 6.5
1000
1500
2000
2500
3000
pH
Yield (pounds per acre)
The linear model is appropriate.
(b) x= 5.342857, y= 1847.285714, Pn
i=1(xi−x)2= 1.577143, Pn
i=1(yi−y)2= 1398177.429,
Pn
i=1(xi−x)(yi−y) = 1162.514286.
Page 357
358 CHAPTER 7
12. b
β1=rsy/sx= 0.7(100/2) = 35. b
β0=y−b
β1x= 1350 −35(5) = 1175.
14. (a)
0 2 4 6 8 10
0
2
4
6
8
10
12
x
y
(b) No, because xand yare not linearly related.
Page 358
SECTION 7.2 359
(c)
0 2 4 6 8 10
−2
−1
0
1
2
3
x
z
(d) x= 5.5, z= 0.41, Pn
i=1(xi−x)2= 82.5, Pn
i=1(zi−z)2= 18.2494, Pn
i=1(xi−x)(zi−z) = 38.55.
15. (iii) equal to $34,900. Since 70 inches is equal to x, the predicted yvalue, bywill be equal to y= 34,900.
16. (a) Let xrepresent time and y=T1represent temperature.
Page 359
360 CHAPTER 7
(c) Let xrepresent time and y=T3represent temperature.
x= 10, y= 63.857143, Pn
i=1(xi−x)2= 770, Pn
i=1(yi−y)2= 18480.57143,
(d) b
β0= (17.714286 + 17.320346 + 15.675325)/3 = 16.9033.
Section 7.3
1. (a) x= 65.0, y= 29.05, Pn
i=1(xi−x)2= 6032.0, Pn
i=1(yi−y)2= 835.42, Pn
i=1(xi−x)(yi−y) =
1988.4, n= 12.
i=1(xi−x)2= 0.329642 and b
Page 360
SECTION 7.3 361
i=1(xi−x)(yi−y)]2
i=1(xi−x)2Pn
i=1(yi−y)2= 0.784587. s2=(1 −r2)Pn
i=1(yi−y)2
(c) s=√17.996003 = 4.242170. sb
β0=ss1
n+x2
Pn
i=1(xi−x)2= 3.755613.
(d) b
β1= 0.329642, sb
β1= 0.0546207, n= 12. There are 12 −2 = 10 degrees of freedom.
(e) x= 40, by= 7.623276 + 0.329642(40) = 20.808952.
(f) x= 40, by= 7.623276 + 0.329642(40) = 20.808952.
2. (a) n−2 = 7 −2 = 5
Page 361
362 CHAPTER 7
3. (a) The slope is −0.7524; the intercept is 88.761.
4. (a) The slope is −0.13468.
Page 362
(b) b
βAand b
βBare independent and normally distributed with means βAand βB, respectively, and esti-
6. (a) x= 1.237222, y= 1.274074, Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.336755. s=r(1 −r2)Pn
(c) s= 0.169191. sb
β1=s
pPn
i=1(xi−x)2= 0.0775689.
364 CHAPTER 7
(f) by=b
β0+b
β1(1.2) = 1.259238, spred =ss1 + 1
n+(x−x)2
Pn
i=1(xi−x)2= 0.170774.
7. (a) x= 1.547286, y= 0.728571, Pn
i=1(xi−x)2= 0.141471, Pn
i=1(yi−y)2= 0.0246857,
Pn
Page 364
SECTION 7.3 365
(f) by= 0.709806, spred =ss1 + 1
n+(x−x)2
Pn
i=1(xi−x)2= 0.0236098.
8. (a) x= 0.840370, y= 0.826667 Pn
(b) The null and alternate hypotheses are H0:β0= 0 versus H1:β06= 0.
i=1(yi−y)2
(c) The null and alternate hypotheses are H0:β1= 1 versus H1:β16= 1.
i=1(xi−x)(yi−y)]2
i=1(yi−y)2
Page 365
366 CHAPTER 7
(e) by= 0.787257, sby=ss1
n+(x−x)2
Pn
i=1(xi−x)2= 0.0132258.
9. (a) x= 21.5075 y= 4.48, Pn
i=1(xi−x)2= 1072.52775, Pn
i=1(yi−y)2= 112.624,
Pn
i=1(xi−x)(yi−y) = 239.656, n= 40.
(b) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.475485, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 1.246816,
Page 366
SECTION 7.3 367
(c) The prediction is b
β0+b
β1(20) = −0.325844 + 0.223450(20) = 4.143150.
10. (a) If the standard deviation sis held constant, the width of a confidence interval is proportional to
1/pPn
i=1(xi−x)2.
(c) If the standard deviation sis held constant, the width of a confidence interval is proportional to
s1
n+(x−x)2
Pn
i=1(xi−x)2.
Page 367
