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answer key of chapter 9 - 9.14 11 Some preliminary...

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Unformatted text preview: 9.14 11. Some preliminary calculations are: 23: = 6167 2}: = 135.8 11: 24 2.11 =1.641.115 Exy = 34.765 x 1" SSH. = 240’ _ L: 398') = 34,765 — _.m_(6167)(135'8) = —129.94167 24 '2 *1 I SS);r 2 Ex“ — (£21) (6167)2 = 1.641.115 - 24 = 56.451958 __ —129.94167 8er 56452958 3"” H 5 II = —.002301769 z —.0023 135-_8 — (—.002301769)[_6::7] = 6249792065 4 6.25 to c: [I \f-ll | U» NI || 24 The least squares line is)“: = 6.25 — 00231 b. 30 = 6.25. Since x = 0 is not in the observed range. 30 has no interpretation other than being the y-intercept. .173I z —.0023. For each additional increase of 1 part per million of pectin, the mean sweetness index is estimated to decrease by .0023. c. 51 = 6.25 — 0023(300) = 5.56 9.16 a. A straight line model relating an NFL team’s current value to its operating income is: y=130+81x+6 15. Ex 2 590.4 2.1- = 11,560 2.0 = 251.8294 222 = 16,782.94 E = E3 = % = 19.68 §= 2?. = “1560 = 385.3333 11 71 30 55“ = XI." _ 032(2)") = 251,829.. _ 590.423.5610) 251.8294 — 227.5008 = 24,328.15 H 2 2 H Ex?” — (Ex) = 16,782.94 — 593%“ H 16,782.94 w 11.619.072 = 5.163.838 U} (I) II 1 ss . 61 = _£-‘_ = ﬂ = 4711340673 = 4.711 ssxx 5.163.838 A 130 f — [31} = 385.3333 M 4.711314067309655) = 385.3333 — 92.?1918444 = 292.6141156 =: 292.614 The fined regression line is 33 = 292.614 + 4.711x c. 31 = 4.711. When Operating income increases by l millon dollars, the mean current value 1. estimated to increase by 4.711 million dollars. This is meaningful for values of operating income between 3.4 and 56.1 million dollars. 5’0 = 292.614. This has no meaning since x = 0 is not in the observed range. 9.18 a. From the printout, the least squares line is: 3:: 301348 + 301815;; b. & c. Using MINITAB, the seattergram with the ﬁtted regression line is: Regression Plot v 2 some . aemsx 3-5:; was as It appears that the least squares line could be an appropriate characterization of the relationship between y and x over the 22-year period. However, one might argue that a curvilinear line might be more appropriate. d. For .r = \$15, the predicted value ofy would be: 3? = 30.1348 + 3.0181505) = 35.40? cents per ganen. e. r ”5‘4 -.-'1 -_e LETS-PE“: 5-3 To would have {he smallest 5‘ because the width of the data noires i” are smaiiest. 9.26 8. Using MINITAB, the scattergram of the data is: 15000 E1 10090 E U 5900 D 3.0 3.5 4.0 Slay b. Ex = 44.71 2y 2 131,670 Ex); = 493.1177 21:2 = 167.4615 23:2 = 1,514,402,100 E = Ex = ﬂ = 3.7258333 y = £2 = ”“670 = 10,9725 :1 12 n 12 8813. = Ex); — w = 493.1177 — w 12 4493,11111 — 490.580.4715 = 2.537.225 2 2 ss = 2x2 4 (2:) = 167.4615 - 441'? 2 167.4615 — 1665820083 = .8794917 1 ss 61 u _*1 = E3335. = 2884876571 = 2884.877 SSH .8794917 12:: c: 11 i — 31} = 10,9725 — 2884.876571(3.7258333) = 10.9725 - 10,748.56929 = 23393071 = 233.931 The ﬁtted regression line is; = 233.931 + 2884.871: _ 2 1'12 _ 131,6702 e. 88” — 2y — n _ 1,514,402,100 — T = 1,514,402,100- 1,444,749,075 = 69,653.025 SSE SS” - 3188.0, = 69,653.025 - 2,884.8765?1(2,537.225) = 69,653.025 -- 7,319,580.958 = 62,333,444.04 52 = SSE = 62,333,444.04 n w 2 12 — 2 s = 1,152 = 1/6,233,344.404 = 24966667 We would expect to see most of the hospital charges to fall within 25 or 2(\$2,496.666?) = \$4,993.3333 of the least squares line. = 6,233,344.404 9.38 a. 9.42 a. Forx = 4, = 223.931 + 2,884.87?(4) = 11,763.439 Md)- 5? j; 25 = 11,763.439 i 4,993.3333 = (6.770.106, 16,756.772) Only one state (California) had an average hospital charge more than 2 standard errors from the least squares line. Thus, 11 out of 12 or 111’ 12 or .917 of the states had average hospital charges within 2 standard errors of the least squares line. There appears to be a somewhat positive linear relationship. If there was very little snowfall in an area, then the erosion will not be typical. Thus, it seems reasonable to remove these data points. For conﬁdence level .90, a = .10 and 1511'2 = .1032 = .05. From Table VI, Appendix B, with df = n — 2 = 47 — 2 = 45, {05 = 1.684. The conﬁdence interval is: [31 i (05531 = 1.39 i 1.684(.06)= 1.39 :l: .101 =(1.289, 1.491) We are 90% conﬁdent that the change in the mean McCool winter-adjusted rainfall erosivity index for each one unit change in the once-in-S-year snowrnelt runoff amount is between 1.289 and 1.491. r 1| too i . 1: ao—j 0' .' . ' " - 23m: 9042 r/IJ-jﬂl'l— 4 . 0 ° C . 4o- . ' _ j . tommmsosomac From the printout, the least squares line is )7 = 44.13 + 23662:. From the printout, s = 19.40 The standard deviation 5 represents the spread of the manager success index about the least squares line. Approximately 95% of the manager success indexes should lie within 25 = 2(19.40) = 38.8 of the least squares line. Refer to the scattergram in part a. The number of interactions with outsiders might contribute some information in the prediction of managerial success, but it does not look like a very strong relationship. To determine if the number of interactions contributes information for the prediction of managerial success, we test: H0161=0 Ha: 61¢ 0 9.46 n - 2 = 19 - 2 = 17. From Table VI, A pendtx B, {025 = 2.110. The rejection region is r > 2.1100” < -2.110. Since the observed value of the test statistic does not fall in the rejection region (r = 1.27 9 2.110), H0 is not rejected. There is insufﬁcient evidence to indicate the number of interactions contributes information for the prediction of managerial success at or = .05. For conﬁdence coefﬁcient .95, or = l - .95 = .05 and ai2 = .05/2 = .025. From Table VI, Appendix B, with df = 1?, {025 = 2.110. The 95% conﬁdence interval is: = .2366 i 2.110(.1865)= .2366 :t .3935 = (—.1569, .6301) We are 95% conﬁdent the change in the mean manager success index for each additional interaction with outsiders is between —.1569 and .6301. Some preliminary calculations are: 2x=0 Ex2=10 2.9120 12 Eyz = 70 M H ssh. = xy — XXX” 20 ~ L12) — 20 - n 5 2 2 0 ss— 2-():’A‘-)=10—c_=10 X). Ex n 5 1 .e 2 ssW — yz — (23“) = 70 — __12 = 412 -- n 5 55 r = 9 = _20__ = 9853 Jssnssyy «10(412) :2 = .98532 = .9709 Since r = .9853, there is a very strong positive linear relationship between I and 3;. Since 3'2 = .9709, 97.09% of the total sample variability around i is explained by the linear relationship between x and y. 6« o 5—; I i ‘— a- 2-. I 'l — —-—|—-—v——-—g——-px 2 l . ‘1 2 3 .1- b. Some preliminary calculations are: Ex=0 £x2=10 Exy=-15 2y = 16 2y? = 74 _ DD a _ 0116) _ _ SSxy—Exy— n #15—_5_— IS 2 02 ss— 2—(2")=10-_=10 xx Ex n 5 2 2 _ 2 y) _ 15 _ SSyy—Ey—(En —74—_5._—22.8 SS _ r= ___xy = _.._15 = —.9934 ,fssxxss” 1/10(22.8) r2 = (+9934)2 = .9868 J' a G a 5 4 3 2 I 1 2 1 1 2 3 I Since 1" = —.9934, there is a very strong negative linear relationship between it and y. Since r2 = .9868, 98.68% of the total sample variability around y is explained by the linear relationship between x and y. (3. Some preliminary calculations are: Ex=18 2x2=52 2xy=36 £y=14 Ey2=32 _ x y_ 1814 _ SSH—Exy—EHE —36— (7-)— 2 58” = 2x2 — (2:? e 52 — g e 5.71428571 2 2 SSW-Eyz— (E?) =32-%=4 s r: _S xy = ._._.—._.O = 0 1!. 1(33 H35” 5 71428571{4) 72=Oz=0 Since r = 0, this implies that x and y are not related. Since r2 = 0, 0% of the total sample variability around i is explained by the linear relationship between I and y. 1’ 4 3 o a 2i 1-: Ex By M M HM II M Q ll {3 II 4; M “6 N H _ 2x2)? = _ 15(4) = 3513, — Ely - —;1-— 12 T 2 2 2 x} _ -15 = 5%» Jr. T 26 2 2 _ 2 - (E?) _ _ 4 = ssyy _ 2)) n _ 6 T 28 r = “-SSI—xy = —-—----—O =0 [SSHSSyy d26(2.8) R=y=o Since r = 0, this implies that x and y are not related. Since r2 = 0. 0% of the total sample variability around Y is explained by the linear relationship between x and y. 9.52 a. No. From the printout, r2 = R-square = .7483. 74.83% of the total sample variability around the sample mean fertility rate is explained by the linear relationship between the fertility rate and the contraceptive prevalence. b. Yes. If the contraceptive use increases by 15%, the mean fertili ty rate decreases by an estimated 0546105) = .819 which is approximately 1. 9.54 a. r = .679. Because this value is near .5, there is a moderate positive linear relationship between the fraction of documents retrieved using Mediine and the number of terms in the search query. b. r2 = .6692 = .461. 46.1% of the total sample variability around the sample mean fraction of documents retrieved using Medline is explained by the linear relationship between the fraction of documents retrieved using Medline and the number of terms in the search query. 9.58 a, b. The scattergram is: c. 8813 = ssyy — 813s”, = 33.6 — 34318766618) = 594344473 :2 = ”55“; = 29133— “? = 342930591 5 = {342930591 = .8619 _ 31 I = __ = 3 l 10 ' - ‘ 1 (x1: ' if The form of the conﬁdence interval 15 y ;|; {6:25 .. + T n H For ID = 6.; = —.414 + 843(6) = 4.644 For conﬁdence coefﬁcient .95, a = .05 and on? = .025. From Table VI, Appendix B. with df = n — 2 = 10 — 2 = 8, {025 = 2.306. The conﬁdence interval is: - 2 4.644 4; 2.306(.8619) 1—10 + (3.3% =4.644 i 1.118 = (3.526. 5.762) = —.414 + .843(3.2) = 2.234 E» d. For xp = 3.2, The conﬁdence interval is: _ 2 2.284 4; 2.306(.8619)L_10 + Eggs—'1) = 2.284 i .629 = (1.655, 2.913) For IF = 0,3» = —.414 + 843(0) = —.414 The conﬁdence interval is: ﬂ 1 (0 , 3.1)2 .414 :r 2.306(.8619) E + “T?— = —.414 i 1.717=(—1.585, .757) 9.60 The width of the conﬁdence interval for the mean value of y depends on the distance xp is from E. The width of the interval for xp = 3.2 is the smallest because 3.2 is the closest to 3E = 3.1. The width of the interval for xp = 0 is the widest because 0 is the farthest from E = 3.1. . ss 61 _13’ = E = .875 ssxx 32 56 = .7" -1§1f = 4 — 875(3) = 1.375 The least squares line i332 = 1.375 + .875x. The least squares line is: \- / y-1.375+.8?5 SSE = 35),). — 31881), 2 26 — 875(28) = 1.5 52: SSE = 1- =.1875 8—2 10—2 5 = 1.7.1875 = .4330 1 (x - if The form of the conﬁdence interval is § i 36725 J; + p n SS H For xp = 2.5,} = 1.375 + .875(2.5) = 3.5625 For conﬁdence coefﬁcient .95, a = .05 and tit/2 = .025. From Table VI, Appendix B, with df = n — 2 = 10 — 2 = 8, {025 = 2.306. The conﬁdence interval is: 1 + (2.5 - 3)2 _ 3.5625 .3279= 3.2346, 3. 10 32 =’ i ( 8904} 3.5625 : 2.306(.4330}[ For l:p = 4,9 = 1.375 + 875(4) : 4.875 For conﬁdence coefﬁcient .95, o: = .05 and all = .025. From Table VI, Appendix B, with df = n — 2 = 10 — 2 = 8. {025 = 2.306. The prediction interval is: 1 2 3.875 i 2.306(.4330}J>1 + 1_*0 + 5% = 3.875 4; 1.062 = (2.813. 4.937) ...
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