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Unformatted text preview: 5.8 a. 5.14 For conﬁdence coefﬁcient .90 a : 10 and (1,2 __ 05 me Table W _
9 I F . I , A d
2.05 = 1.645. The conﬁdence interval is: PM I): B, — s
x j; zan— = 3.39 i moi = 3.39 i .0466 =9 (3.3434, 3.4366) 1/; W Since the sample size was so large, no assumptions are necessary We are 90% conﬁdent that the mean risk is between 3.3434 and 3.4366. Since: all values included in tile interval exceed 2.5. the rescarchers would conclude that students in these
grades exhibit an awareness of risk involved in bicycling. First. we must compute E and s. E=£x=3_0=5
n 6
Ex? _ Ex? 176 ._ (30}2 '
2 n 6 26_
S: = =.._—52
nel 6—1 5 .s = ‘63 = 2.2804 .90 = .10 and 0112 = .101’2 = .05. From Table VI, For conﬁdence coefﬁcient .90, o: 
1 = S. 135 = 2.015. The 90% conﬁdence interval is: Appendix B, with df = n e 1 = 6 i + 1051 = 5 i 2.015 12804 = 5 i 1.876 = (3.124, 6.376) — Jr? J3 For conﬁdence coefﬁcient .95. o: = 1 — .95 = .05 and 1212 = .05i2 = .025. From Table
VI, Appendix B, with df z n — 1 = 6 — l = 5, 11025 = 2.571. The 95% conﬁdence
interval is: E + (mi 3 5 4; 2.5712'2804 h 1/? J3 = 5 i 2.394 = (2.606, 7.394) For conﬁdence coefﬁcient .99, or = 1 — .99 = .01 and 010 = .011'2 = .005. From Table
V1, Appendix B. with df = n — 1 = 6  1 = 5,1035: 4.032. The 99%.confidence
interval is: E i 7.005.: = 5 i 4.032 2230“ = Jr? 1/3 a) For conﬁdence coefﬁcient .90, o: = 1 — .90 = .10 and 0112 = .1012 = .05. From
Table VI. Appendix B, with df = n — 1 = 25 — 1 = 24.1.05 =1.711. The 90%
conﬁdence interval is: 5 i 3.754 =9 (1.246, 8.754) } J; 105.2 = 5 1 1.711”304 = 5 i .780 = (4.220, 5.730)
,[n' ¢2s 5.22 b) For conﬁdence coefﬁcient .95, o: = 1 — .95 = .05 and (132 = .0572 = .025. From
Table VI, Appendix B, with (if = n — l = 25 ~ 1 = 24, {025 = 2.064. The 95%
conﬁdence interval is: E i 1.025% a 5 i 106422804 = 5 i .941 =~ (4.059, 5.941)
77 1/25 For conﬁdence coefﬁcient .99. o = 1 — .99 = .01 and of? = .011’2 = .005. From
Table VI, Appendix B, with df = n —— 1 = 25 — 1 = 24, {005 = 2.797. The 99%
conﬁdence interval is: 5 = 5 i 2.797 22804 = 5 i 1.276 = (3.724, 6.276) i+t ._
.oos‘l/F «.25— Increasing the sample size decreases the width of the conﬁdence interval. The population from which the sample was drawn is the Forbes 500 Biggest Private
Companies. Some preliminary calculations are: §= 2‘ =ﬂ=19427
n . 15
2x2  (2“? 1,430,756 — (2914):
52 a n _1” = 15 ‘1 15 = 61,761.633 s = 061361.638 = 248.5189 For conﬁdencc coefﬁcient .98, or = 1
Appendix B, with df = 77 — 1 = 1
is:  — .98 = .02 and 011’2 = .020 = .01. From Table VI
5 — 1 = 14,131 = 2.624. The 98% conﬁdence interva} 9 E 1 {mi a» 194.27 i 2.624% = =194.27 i 1/3 1/? 168.38 ==~ (25.89, 362.65) For conﬁdence coefﬁcient .98. or = l — .98 = .02 and (112 = .0212 = .01. From Table IV
Appendix B, em = 2.33. The conﬁdence interval is: p 4; mfg = .144 i 2.33'%@ = .144 i .023 = (.121, .167) We are 98% conﬁdent that the proportion of debit cardholders who have used their card in
making purchases over the Internet is between .121 and .167. 9 Since we would have less conﬁdence with a 90% conﬁdence interval than with a 98%
conﬁdence interval, the 90% interval would be narrower. 5.30 5.34 The point estimate ofp is f: = I101 = 35155 = .636. We must 'check to see if the sample size is sufﬁciently large: is: 36}, 4.5: 31:22 =.636i #113? =.636i .195=>(.441,.331) Since the interval is wholly contained in the interval (0, 1) We may assume that the normal
approximation is reasonable. For conﬁdence coefﬁcient, .99, or = .01 and on? = .0112 = .005. From Table IV, Appendix
B, zms = 2.575. The conﬁdence interval is: ,6 i 2305 E‘i = .636 i 2.575 ‘ ﬁg?! = .636 i .167 = (_469, .303)
H We are 99% conﬁdent that the true proportion of fatal accidents involving children is between
.469 and .803. The sample proportion of children killed by air bags who were not wearing seat belts or were
improperly restrained is 24135 = .686. This is rather large proportion. Whether a child is
kined by an airbag could be related to whether or not hefshe was properly restrained. Thus, the number of children killed by air bags could possibly be reduced if the child were properly
restrained. The population studied by Arthur Anderson is the set of all privately~held family ﬁrms with
revenues exceeding $1,000,000 per year. or the 3,900 observations, 1,911 had no Strategic business plan = f: = 191113.900 = .49. To see if the sample size is sufficiently large: faiaoﬁ=fsis ﬂ=§i3 3:.49i3 49(51) __..—.__ = .49 i .024
n n 3,900 m (.466, .5 14) Since the interval lies within the interval (0, 1), the normal approximation will be adequate. For conﬁdence coefﬁcient .90, o: = .10 and 01/2 = .1012 = .05. From Table IV, Appendix
B, 2:05 = 1.645. The conﬁdence interval is: *  p9 * 33:7 .49(.51)
+ __ w i 1.645 __ a .49 + 1.645 ____ .494; .013
p F 4'05 n p _ n _ 3,900 = = (.471, .503) We are 90% confident that the proportion of familyowned companies without strategic
business plans is between .477 and .503. The width of the interval construaed in part b is .503 — .477 = .026. An 80% conﬁdence
interval would be narrower because we have less conﬁdence. For conﬁdence coefﬁcient .80, or = .20 and on? = .2012 = .10. From Table IV, Appendix
B, 2.10 = 1.28. The conﬁdence interval is: ,5: “01% =4}? 4; 1.281%: = .49 451.2%:9333) =4 .49 i .010 =(.480, .500) The width of the interval constructed in part c is .500 w .480 = .020. This is, in fact,
narrower than the 90% conﬁdence interval. \ 5.38 a. To compute the needed sample size, use: (2:112qu
n = _.?._ where 22025 = 1.96 from Table IV, Appendix B. 2
Thus, n = (1'96) (2)93) = 96.04
.082 =97 You would need to take a sample of size 97. b. To compute the needed sample size, use: 2
n = M = .(1'96)2('5)('5) = 150 0625 = 151
32 .032 ' You would need to take a sample of size 151. 5.64 a. The point estimate for the fraction of the entire market who refuse to purchase bars is: 23
m , x
p:_:
n .094 b. To see if the sample size is sufﬁcient: 244 Since the interval above is contained in the interval (0, l), the sample size is sufficiently large. p i ﬂﬁ a .094 a 3r'0944'906) = .094 i .056 = (.038, .150) c, For conﬁdence coefﬁcient .95, or = 1 — .95 = .05 and 0:12 = .0542 = .025. From Table
IV, Appendix B, 2325 = 1.96. The conﬁdence interval is; p i 2025‘??? = .094 i 1.961% a .094 i .037 = (.057, .131) d. The best estimate of the true fraction of the entire market who refuse to purchase bars six months after the poisoning is .094. We are 95% conﬁdent the true fracrion of the entire
market who refuse to purchase bars six months after the poisoning is between .057 and . 1 3 1. e. For confidence coefﬁcient .95, or = .05 and 0:12 = .025. From Table IV, Appendix B,
2325 = 1.96. From part a, a good approximation for p is .094. Also, B = .02. 2
(7.qu = (l.96)2(.094)(.906) = any a 818 The sample size is n =
a2 .022 You would need to take a = 818 samples. 1. . c. 1.96 d. 1.28 5.3 a. 28 :i: .784 b. 102 i .65 c. 15 :1: .0588 d. 4.05 I“ .163 e. no 3”.9 i .65 h. 33.9 i .32 c. width is halved 5.7 (1. claim probablyr not true 5.9 (3526,4921) 5.11 a. 66.83 i 4.69
c. (41.009.49.602) 5.13 a. 2.228 b. 2.228 c. —1.812 d. 1.225 e. 4.032 5.15 a. 97.94 41 4.24 b. 97.94 i 6.24
5.17 a. 2.886 :t 4.034 b. .408 I .256 5.19 a. 49.3 I 8.6 h. 99% confident that the mean amonnt removed from all soil specimens using the poison is between 40.70% and 57.90%. 5.21 184.99 :1: 133.93 5.23 a. 22.46 :1: 11.18 :1. validity
is suspect 5.25 a. yes b. no c. yes d. no 5.27 a. yes b. .46 i .065 5.29 b. .29 :1: .028 5.31 a. .24 b. .24 :1: .181
5.33 .85 :‘t .002 5.35 308 5.37 a. 68 h. 31 5.39 34 5.41 a. .226 i .00? h. .014 c. 1.680 5.43 1,692 5.45 43; 171; 385 5.47 no 5.49 a. —1.225 b. 3.250 c. 1.860 d. 2.898 5.51 a. 32.5 :t 5.16 13. 23,964 5.53 a. (298.6, 582.3}
5.55 a. .876 .4: .003 5.57 a. men:2.4 i .979; women: 4.5 '1 .755 b. men: 9.3 :1: 1.185; women: 6.6 _ 1.138
5.59 a. 12.2 :I: 1.645 b. 167 5.61 a. 191 5.63 b. 3.256 3: .348 5.65 154 5.1 a.
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This note was uploaded on 09/11/2008 for the course ECON 317 taught by Professor Safarzadeh during the Spring '07 term at USC.
 Spring '07
 Safarzadeh

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