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Kreyszig Advanced Engineering Mathematics 9 solution manual.pdf


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im25.qxd

9/21/05

2:06 PM

Page 405

Instructor’s Manual

405

From this and (7) we get
354 100
k1 4.31829.
82 000
Also q0 993 and K 0.2. The answer is
CONF0.95{4.1
1
4.5}.
14. Multiplying out the square, we get three terms, hence three sums,

(x

j

x )2


x

x





2x
2
xj2
n
2
j

j

nx 2

xi

1
xj n(
n



2

xj)

and the last of these three terms cancels half of the second term, giving the result.

SOLUTIONS TO CHAP. 25 REVIEW QUESTIONS AND PROBLEMS,
page 1092
0.392 and
22. From Table 25.1 in Sec. 25.3 we obtain k 1.96 • 4/ 400
CONF0.95{52.6

53.4}.
24. x 26.4, k 2.576 • 2.2/ 5 2.534 by Table 25.1, Sec. 25.3. This gives the answer
CONF0.99{23.86

28.94}.
26. n 1 3 degrees of freedom, F(c1) 0.025, c1 0.22, F(c2) 0.975, c2 9.35
from Table A10 in App. 5; hence k1 0.7/0.22 3.182, k2 0.7/9.35 0.075 by
Table 25.3 in Sec. 25.3. The answer is
CONF0.95{0.075
2
3.182}.
28. k 2.26 • 0.157/ 10
0.112; CONF0.95{4.25

4.49}
30. n 41 by trial and error, because F(c) 0.975 gives c 2.02 (Table A9 with 40
d.f.), so that L 2k 2sc/ n
0.099 by Table 25.2.
32. From Table A10 in App. 5 and Table 25.3 in Sec. 25.3 we obtain
c _1 ( 253
2.58)2 88.8,
1

2

c2 _12( 253
2.58)2 170.9,
so that k1 244/c1 2.76 and k2 244/c2 1.43. The answer is
CONF0.99{1.42
2
2.75}.
34. The test is two-sided. We have 2/n 0.025, as before. Table A8 gives
c 15.0
P(X
c)15.0 ( ) 0.975,
0.025


c 15.31

and 15.0 0.31 14.69 as the left endpoint of the acceptance region. Now
x 14.5 14.70, and we reject the hypothesis.
36. The test is right-sided. From Table A9 with n 1 14 degrees of freedom and
P(T c) 0.01,
0

thus

P(T
c) 0.99
0