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SUPPLEMENTARY EXERCISES FOR CHAPTER 7 381
(b)
0 10 20 30 40 50 60
−30
−20
−10
0
10
20
30
Order of Observations
Residual
No violations of the standard assumptions are in-
dicated.
5. (a) x= 50, y= 47.909091, Pn
i=1(xi−x)2= 0.941818 and b
(b) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.998805, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 1.138846.
(c) The null and alternate hypotheses are H0:β1= 1 versus H1:β16= 1.
Page 381
382 CHAPTER 7
(d) Yes, since we can conclude that β16= 1, we can conclude that the machine is out of calibration.
(e) x= 20, by=b
β0+b
β1(20) = 19.65455.
(f) x= 80, by=b
β0+b
β1(80) = 76.163636.
6. (a) x= 4.036667, y= 51.3333, Pn
i=1(xi−x)2= 21.150533, Pn
i=1(yi−y)2= 3411.333333,
7. (a) x= 92.0, y= 0.907407, Pn
i=1(xi−x)2= 514.0, Pn
i=1(yi−y)2= 4.738519,
i=1(xi−x)2= 0.0391051 and b
Page 382
SUPPLEMENTARY EXERCISES FOR CHAPTER 7 383
(b) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.165877, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 0.397618.
(c) x= 93, by=b
β0+b
β1(93) = 0.94651.
(e) x= 93, by=b
β0+b
β1(93) = 0.94651.
8. (a) x= 4.43, y= 3.451875, Pn
i=1(xi−x)2= 11.2828, Pn
i=1(yi−y)2= 37.7994875,
Page 383
384 CHAPTER 7
(b) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.480861, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 0.808768.
(c) By b
β1(1.5) = 1.269242(1.5) = 1.904.
(e) Let y5.0=β0+β1(5.0) denote the mean lipase production when the cell mass is 5.0 g/L.
9. (a) ln y=β0+β1ln x, where β0= ln kand β1=r.
(b) Let ui= ln xiand let vi= ln yi.
i=1(ui−u)2= 0.650328 and b
Page 384
SUPPLEMENTARY EXERCISES FOR CHAPTER 7 385
(c) The null and alternate hypotheses are H0:r= 0.5 versus H1:r6= 0.5.
10. (a) x= 28.078261, y= 29.221739, Pn
i=1(xi−x)2= 287.999130, Pn
i=1(yi−y)2= 1692.179130,
(b) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.247383, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 7.787545.
(c) By 1.205626(2) = 2.411 days.
Page 385
386 CHAPTER 7
(e) Let y30 =β0+β1(30) denote the mean flight period for butterflies whose wingspan is 30 mm.
(f) x= 28.5, by=b
β0+b
β1(28.5) = 28.713280.
11. (a) x= 41.428157, y= 33.428571, Pn
i=1(xi−x)2= 6135.428571, Pn
i=1(yi−y)2= 4899.428157,
(b) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.199916, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 18.073814,
Page 386
SUPPLEMENTARY EXERCISES FOR CHAPTER 7 387
(c) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.199916, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 18.073814,
12. (iv). Since Newton’s law of cooling is a physical law, the plot of ln temperature versus time should
13. (a) x= 225, y= 86.48333, Pn
i=1(xi−x)2= 37500, Pn
i=1(yi−y)2= 513.116667,
(b) The null and alternate hypotheses are H0:β0= 0 versus H1:β06= 0.
Page 387
388 CHAPTER 7
(c) The null and alternate hypotheses are H0:β1= 0 versus H1:β16= 0.
(d)
80 85 90 95
−1
−0.5
0
0.5
1
Fitted Value
Residual
The linear model appears to be appropriate.
(e) b
β1= 0.116533, sb
(f) x= 225, by=b
β0+b
β1(225) = 86.483333.
(g) x= 225, by=b
β0+b
β1(225) = 86.483333.
Page 388
SUPPLEMENTARY EXERCISES FOR CHAPTER 7 389
14. (a) by=−0.23429 + 0.72868(2.5) = 1.5874.
15. (ii). The standard deviation sbyis not given in the output. To compute sby, the quantity Pn
i=1(xi−x)2
16. The width of the confidence interval is proportional to sby=ss1
n+(x−x)2
Pn
i=1(xi−x)2.
17. (a) If f= 1/2 then 1/f = 2. The estimate is b
t= 145.736 −0.05180(2) = 145.63.
Page 389
390 CHAPTER 7
18. (a) If yis unbiased, then y= true value + ε. If the calculated value xis equal to the true value, then
y=x+ε.
(d) r2=[Pn
i=1(xi−x)(yi−y)]2
Pn
i=1(xi−x)2Pn
i=1(yi−y)2= 0.946673, s=r(1 −r2)Pn
i=1(yi−y)2
n−2= 0.0655887.
(e) It is plausible that the calculated value is accurate. Neither of the hypotheses β0= 0 nor β1= 1 can
be rejected.
19. (a) We need to minimize the sum of squares S=P(yi−b
βxi)2.
Page 390
SUPPLEMENTARY EXERCISES FOR CHAPTER 7 391
β=Pc2
20. First note that Pn
i=1(xi−x) = Pn
i=1 xi−nx= 0.
Since xis constant, Pn
i=1 x(xi−x) = xPn
i=1(xi−x) = 0 as well.
21. From the answer to Exercise 20, we know that Pn
i=1(xi−x) = 0, Pn
i=1 x(xi−x) = 0, and
Pn
Page 391
22. σ2
b
β1
=
n
X
Pn
σ2
i=1(xi−x)2
23.
σ2
b
β0
=
n
X
n−x(xi−x)
Pn
σ2
n
X
(xi−x)
i=1(xi−x)2
Page 392
SECTION 8.1 393
Chapter 8
Section 8.1
1. (a) The predicted strength is 26.641 + 3.3201(8.2) −0.4249(10) = 49.62 kg/mm2.
2. (a) b
β1= 3.3201, sb
β1= 0.33198, n= 20.
There are 20 −3 = 17 degrees of freedom. t17,.025 = 2.110. A 95% confidence interval is 3.3201 ±
2.110(0.33198), or (2.620,4.021).
(d) b
β2=−0.4249, sb
β2= 0.12606, n= 20.
Page 393
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