answer key of chapter 8

# answer key of chapter 8 - 8.12 a Let p = proportion of all...

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Unformatted text preview: 8.12 a. Let p = proportion of all children who recognize Joe Camel. x 15+46 ‘52—: __ ‘=1—‘=1~.735=.265 n 28+55 q p = .735 To see if the sample is sufﬁciently large: ﬁt 3o§=~13i 3J3€ =5: 3Iﬂ =>.':354_r =>.735i.145 n n 83 = (.590. .880} Since the interval lies within the interval (0, 1), the normal approximation will be adequate. For conﬁdence coefﬁcient .95, o: = .05 and 0:!2 = .OSQ = .025. From Table IV, Appendix B, 3025 = 1.96. The 95% conﬁdence interval is: .a. '” 1 p i :025‘ {E = .735 i 1.96 = .735 i .095 = (.640, .830) - n We are 95% conﬁdent that the proportion of all children who recognize Joe Camel is between .640 and .830. Let p1 = proportion of children under the age of 6 who recognize Joe Camel and p2 = proportion of children age 6 and over who recognize Joe Camel. i .rl 15 .n. A =r=m=.536 =1— =1—.536=.464 A x A A p2=£=‘f_6=.836 q2=1—p3=l—.836=.164 n2 :5 To see if the samples are sufﬁciently large: F‘—— T A A p q i — p1i3dﬁlmp1 3JL=~p1 3 ﬂasssisjm n1 :21 28 => .536 i .283 => (.253, .819) H- H- ‘ A , ,. P29 A ‘ ' —-—- Pg 1 3% =p2 3‘ n 2 3102 . 3JP242 = 336 i 3J.336(.164) 2 J"12 55 => .836 -_|- .150 => (.686, .986) H H Since both intervals lie within the interval (0, I), the normal approximation will be adequate To determine if the recognition of Joe Camel increases with age, we test: (I31 - r32) - 0 = (.536 - .836) — 0 .. 1 g 1 J.735(.265)[L -— i] qun—l ' 3;} 28 55 The test statistic is z = -2.93 The rejection region requires a = .05 in the lower taii of the z-distribution. From Table IV". Appendix B, z_05 = 1.645. The rejection region is z < —l.645. Since the observed value of the te5t statistic falls in the rejection region (2 = —2.93 < —1.645), H0 is rejected. There is sufﬁcient evidence to indicate that the recognition of Joe Camel increases with age at O! = .05. 8.14 a. Let p1 = proportion of females who have a food craving and p2 = prOportion of males who' have a food craving. Some preliminary eajculations are: 15 = x1 + I: : 600(.97) + 400(.67) = .35 “1 ‘1‘ n2 + To determine if the proportion of women who have food cravings exceeds the proportion of males who have food cravings, we test: Ha: p1 - p2 > 0 . _ _ 0 _ v The test statistic is 3 = = = 13.02 ,— 1 1 it. .1] + a} "1 Ir12 The rejecrion region requires or = .01 in the upper tail of the z-disttibution. From Table IV, Appendix B, em = 2.33. The rejection region is z > 2.33. "0'" Since the observed value of the test statistic falls in the rejection region (2 = 13.02 > 2.33), H0 is rejected. There is sufﬁcient evidence to indicate the preportion of females who have food cravings exceeds the proportion of males who have food cravings at or = .01. b. This study involved 1,000 McMaster University students. It is very dangerous to generalize the results of this Study to the general adult population of North America. The sample of students used may not be representative of the population of interest. 8.20 For probability .95, at = 1 —~ .95 = .05 and 0:12 = .0512 = .025. From Table IV, Appendix B. cogs = 1.96. Since we have no prior information about the proportions, we use p1 = p2 = .5 I“ get a conservative estimate. (ample + p242) = (1.96)2(.5(1 — .5) + .50 - 5)) 1.9203 n = = = = 1 “2 a2 .022 -0004 8.22 For conﬁdence coefficient .90, n: = .10 and on? = .1012 = .05. From Table IV, Appendix B, a”, = 1.645. Since we would expect the percentages in 2000 and 2001 to be fairly similar to the percentage in 1999, we will use 191999 = .92 to estimate both pzom and p200]. n] = “2 = (zaizfovm + 9292) = 1.6452(.92(.08) + .92(.08)) = 32 .03.2 442.6 = 443 We would need to sample 443 adult Americans in each year. 8.26 The hypotheses of interest are: H0: p1 = .25,p2 = .25, p3 = .50 H3: At least one of the probabilities differs from the hypothesized value E021) = npw = 320(.25) = 80 13022) = npgao = 320(.25) = 80 15013) = np3’g = 320(.50) = 160 _ . 2 2 2 . . . 2 g [at Evil)? = (73 — 80) + (60 — 80) + (182 — 160) The test statistic ts X E __.._._..._80 —80 .—....160 = 8.075 8.30 a. If Bon Appetit readers do not have a preference for their least favorite vegetable, then the vajues of pi, p2, p3, and p4 should all be the same. Since there are four categories, then p1 = P2 2P3 3 P4 = ~25- b. To determine if the Ben Appett'r readers have a preference for at least one of the vegetables as “least favorite”, we teat: H0? P1=P2=P3=P4 = 25 Ha At least one p!- ;é .25 c. Some preliminary calculations: rt=£nf=46+76+44+34=200 Emf) = rtpi‘o = 200{.25) = 50, t' = 1, 2, 3, or 4 n. - Em)? The test statistic is x2 = [I—' E Em) (46 - 50)2 + (76 — 50)2 + (44 — 50}2 + (34 — 50)2 — = 19.68 50 50 50 The rejection region requires or = .05 in the upper tail of the X2 distribution with (if = k ~ 1 = 4 - l = 3. From Table VII, Appendix B, X235 = 13.81473. The rejection region is x2 > 7.81473. Since the observed value of the test statistic falls in the rejection region (x2 2 19.68 > 7.81473), H0 is rejected. There is sufﬁcient evidence to indicate the Bon Appetit readers have a preference for at least one of the vegetables as “least favorite” at O: = .05. d. We must assume that: Sample is random Sample size is sufﬁciently large (every cell has an expected value of at least 5). 8.34 8.40 a. C. Some preliminary calculations are: 13019 = npm = 19,115(.2010) = 3,842.115 501;) = npz‘o : 19,115(.1010) = 1,930.615 5013) = np3‘0 = 19,115(.0542) = 1,036.033 £014) 2 71,043 : 19,115(.1601)= 3,060.3115 6015) = up“) = 19,115(.2865) = 5.4764475 Etna) = 11196.0 = 19,115(.1233) 2 2,366.43? 5647) = npm = 19,115(.0732) = 1,399.218 To determine if there is evidence to conclude that the purchases of the household panel is representative of the population of households. we test: H0: p1 = .201,p2 = .101,p3 = .0542,p4 = .1601,p5 : 2865,1326 = .1238, . p7 = .0732 Ha: At least one p,- does not equal its hypothesized value It a E(n-)]2 The test statistic is x2 = _[_‘._.__'_ E E(n,-) = (3,165 — 3,842.115)2 + (1,892 - 1330.615)2 + (726 — 1.036.033)2 3,842.115 1,930.615 1,036.033 + (4,079 — 3.0603115)2 + (6,206 — 15,476.447’5)2 + (1,622 — 2,366.43?)2 3,060.3115 5,476.4475 2,366.43? 2 + (1,420 — 1,399,218) = 880521 1,399.218 The rejection region requires or 2 .05 in the upper tail of the x2 distribution with df = k — 1 = 7 — 1 = 6. From Table V11, Appendix B, x335 = 12.5916. The rejection region is x2 > 12.5916. Since the observed value of the test statistic falls in the rejection region (x2 = 880.521 > 12.5916), H0 is rejected. There is sufﬁcient evidence to indicate that the purchases of the household panel is not representative of the population of households at o: = .05. We must assume that: l. A multinotnial experiment was conducted. This is generally satisﬁed by taking a random sample from the population of interest. 2. The sample size n will be large enough so that, for ever}r cell, the expected cell count, £02,), will be equal to 5 or more. From Table VII, Appendix B, with df = 6, the p-value = P(')(2 > 880.521) < .005. Some preliminary calculations are: 6011,) = :1; = lit—(31:1) = 47.007r 50131): \$ = 56.774 60:12) = w = 57.180 3012) = M = 69.062 601,3) : Egg? = 49.313 320123) = w : 60.164 60131) = 99g?) = 30.219 60233) = 99332) = 32.023 501,2) = 99%?” = 36.759 ' 439 To determine if the row and column classiﬁcations are dependent, we test: H0: The row and column classiﬁcations are independent H3: The row and column classiﬁcations are dependent r n .12 inn — E on The test statistic is x2 = L U A ( U)! Hat-j) _ (40 — 47.007): + (72 * 57.180)2 + (42 — 49.813)2 + (63 - 567' : 47.007 57.180 49.813 56.77! + (53 - 69.062)2 + (70 - 60.164)2 + (31 — 30.219)2 69.062 60.164 30.219 _ 2 A 2 + (38 36.759) + (30 32.023) = 12.36 36.759 32.023 The rejection region requires 0: = .05 in the upper tail of the 1:2 distribution with df = (r — 1)(c w 1) = (3 ~ 1)(3 - 1) = 4. From Table VII, Appendix B. x1205 = 9.48773. The rejection region is x2 > 9.48773. Since the observed value of the test statistic falls in the rejection region 08 = 12.36 > 9.487731. H0 is rejected. There is sufﬁcient evidence to indicate the row and column classiﬁcation are dependent at}; = .05. 8.44 21. Some preliminary calculations are: ~ tic-1 57(52) - J“1‘32 51(34) E =__F=_=34.465 E =_e= _._=22.535 (all) n 36 (“12) n 86 a T C A r C £0121): ii = 2952) = 17.535 50122) = .33. = 29(34) 2 11.465 rt 86 it 86 To determine if manufacturing ﬁrms were more likely to be involved with TQM than service ﬁrms, we test: H0: Type of ﬁrm and TQM are independent Ha: Type of ﬁrm and TQM are dependent .. _ E“ .. The test statistic is X2 = E E M 12%.) Z (34 - 34.465)2 + (23 - 22.535)2 + (18 — 17.535}2 + {11 - 11.465)2 = 047 34 465 22 535 ' 17 535 11 465 ' The rejection region requires a = .05 in the upper tail of the x2 distribution with df = (r — 1)(c — 1) 2 (2 # 1)(2 — 1) = 1. From Tabie VII, Appendix B, 3805 = 3.84146. The rejection region is x2 > 3.84146. Since the observed value of the test statistic does not fall in the rejection region ()3 = .047 :~ 3.84146), H0 is not rejected. There is insufﬁcient evidence to indicate that the type of ﬁrm and TQM are dependent at o: = .05. There is no evidence to indicate that manufacturing firms are more liker to be involved with TQM than service ﬁrms. b. The p-value is POE > .047). From Table VI]. Appendix B, with df = 1, .10 < 191')? > .047) < .90. We must assume: 1. The n observed counts are a random sample from the population of interest. We may then consider this to be a multinomial experiment with r X c = 2 X 2 = 4 possible outcomes 2. The sample size, n, will be large enough so that, for every cell, the expected cell count. any), will be equal to 5 or more. ...
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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.

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answer key of chapter 8 - 8.12 a Let p = proportion of all...

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