hw9 - STAT 3022 TA: Jinghan Meng Homework 9 Solutions...

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Unformatted text preview: STAT 3022 TA: Jinghan Meng Homework 9 Solutions 14.10. (a) £11 2 i—gg é 0.7606 for exclusive-territory firms. (1)) f7; = g 2 0.5357 for other firms. (c)odd51 = 2 3.1765 and odds;; = TEE]: i 1.1538. (d) log(oddsl) : 1.1558 and 171 A log(oddsz) : 0.1431. (Be sure to use the natural logarithm for this computanon.) 14.11. (a) n... = $6 = 0.2123 for women and ,0,“ = g 4 0.1076 for men. (b) odds... = 7% i 0.2704 and oddslrn = Tihfi; é 0.1205. (c) log(oddsw) 211.3079 and log(oddsm) a —2.1158. 14.12. (a) 170 = log(oddsz) : 0.1431 and b] = log(oddsl) — log(odd52) é 1.0127. (b) The fitted model is log(odds) = 0.1431 + 1.0127x. (c) The odds ratio is odds. /oddsz = eh 5 2.7529. 14.13, (a) b0 2 log(oddsm) & —2.1158 and b; = l.og(oddsw) — log(oddsm) : 0.8079. (b) The fitted model is log(odds) = —2.1158 + 0.807%. (c) The odds ratio is oddsw/oddsm = ab: 2 2.2432. 14.35. For women, the preponioni odds, and log-odds for testing'positive are ,6. = % 2 0.1414, oddsr = A1“; e 0.1646, and log(odd5f) 2 —1.8040. For men, these numbers are fim = % é 0.3388, oddsl.l1 = TE: é 0.5124. and log(oddsm) : —0.6686. The fitted logistic regression model depends on whether we use x = 1 to indicate male or female (the exercise does not specify which to use). If x = 1 for males, we have: log(odds) = —1.8040 + 1.13551: and odds ratio 3.1126 If x = 1 for females, we have: log(odds) = —0.6686 — 1.1355: and odds ratio 0.3213 The odds of teSILing positive for alcohol are about 3 times higher for men than for women. (0r, those odds are about one-third as high for women as for men.) 15.9. (a) See table. (1)) For Story 2, W = 1—St0ry 1 -r Story 2 3 + 9 + 4 + 7 + 10 = 38- Under H03 Child Progress LScore Rank Licorc Rank 2 (5)111) = 27 5 high 055 4.5 0.80 W 2 ' high 0.57 6 0.82 p. h' h 0.72 3.5 0.54 O‘w = ./——u(5)(f;(l]) : 4.7371 ‘3 high 0.70 7 0.79 (c) z = a 2.19; with the continuity high 0.84 10 0.89 low 0.40 3 0.77 low 0.72 8.5 0.49 low 0.00 1 0.66 low 0.36 2 0.28 low J_ 0.55 4.5 ]_0.38 correction, we compute W i 2.09, which gives P = P(Z > 2.09) = 0.0183. (d) See the table. osoooaaciu-meia MumehOfl-PNDOD H STAT 3022 TA: Jinghan Meng EX 15.13(b) Hg: There is no difference in the number of species on logged and unlogged plots vs. Ha: Unlogged plots have a greater variety of species. First, I conduct the test by hand. Unlogged 22 22 15 13 19 19 18 20 21 13 13 15 Rank 20.5 20.5 9.5 5 10.5 16.5 14 18 19 5 5 9.5 Logged 17 18 18 15 12 4 14 15 10 Rank 12 14 14 9.5 3 1 7 9.5 2 Sum of ranks for unlogged group is W : 159. m = 12, 112 = 9, N = 21, so W has the mean my = nfiN + 1)/2 = 132, and standard deviation 0w = ,xn1n2(N+ 1)/12 = 1407. So the p-value is 159 — 132 14.07 The W'ilcoxon test using R has P = 0.029. We conclude that the observed dif— ference is significant, unlogged plots really do have a greater number of species. P(W > 159) = P(.~. > )= 0.027 9x15.13 <— read.table("D:/ch15/ex15_013.txt", header=T) attach(ex15.13) speciesl <— species[group==1] species2 <— speciesEgroup==2J wilcox.test(speciesl,species2,alternative="greater") VVVVV Wilcoxon rank sum test with continuity correction data: speciesl and species2 W = 81, p-value = 0.02900 alternative hypothesis: true location shift is greater than 0 Warning message: In wilcox.test.defau1t(speciesl, speciesE, alternative = "greater") cannot compute exact p-value with ties EX 15.29 (a) The stem-and-leaf plot is given below by R. (b) First, I conduct the Wileoxon signed rank test to the differences between observations and 105 by hand. The. differences between observations and 105 are: -13.1-7.2 6.4 17.3 0.4 -10.0 -1.2 -5.4 -8.4 14.3 -0.2 -3.3. Then the Wilcoxon signed rank statistic is the sum of the ranks of the positive differences ‘er+ = 31. n. = 12, and W has the mean pw+ = n(n + 1}/4 = 39, and standard deviation aw+ = ,/n(n +1)(2n + 1)/24 = 12.75. So the p—value is 31 — 39 + w = 2P(W < 31) — 2P(z < 12.75 ) 0.53 There is no evidence that the median is other than 105. Below are R codes to do this test. ST¥¥T 3022 TVX:Jinghan hdeng > ex15.29 <— read.table("D:/ch15/ex15_029.txt", header=T) > attach(ex15 29) > stem(radon,scale=2) The decimal point is 1 digit(s) to the right of the | 9 l 2 9 l 578 10 | 024 10 l 55 11 l 1 11 | 9 12 | 2 > wilcox.test(radon-105) Wilcoxon signed rank test data: radon - 105 V = 31, p-value = 0.5693 alternative hypothesis: true location is not equal to 0 EX 15.30 If we compute Haiti content minus factory content (so that a negative difference means that the amount of vitamin C decreased), we find that the mean change is -5.33 and the median is -6. The stem—and-leaf plot is right— skewed. The Wilcoxon test given by R has P = 9.28 x 10—06. The differences are systematically negative, so vitamin C content is lower in Haiti. > ex15 30 <- read.table{"D:/ch15/ex15_030.txt", header=T) > attachCex15.30) > stemChaiti-factory) The decimal point is at the I -14 | 0 —12 I 0000 —10 | -8 I 0000 -6 I 0000000 -4 | 0000 -2 | 0 -0 I 0 0 | 0 2 | 00 4 | 0 6 | 8 I 0 STAT 3022 TA: Jinghan Meng > vilcox.test(haiti,factory,alternative="less") Wilcoxon rank sum test with continuity correction data: haiti and factory w = 117.5, p-value = 9 28e-06 alternative hypothesis: true location shift is less than 0 Warning message: In wilcox.test.default(haiti, factory, alternative = "less") : cannot compute exact p-value with ties EX 15.34 (a) We test Ho : a1 = a; = #3 = #4 vs. Ha: Not all n,- are equal. For Kruskal—Wallis, He says that the distribution of the trapped insect count is the same for all board colors; the alternative is that the count is systematically higher for some colors. (b) In the order given, the median are 46.4, 15.5, 34.5 and 15 insects; it appears that yellow is most effective, green is in the middle, and white and blue are least effective. First, I conduct the Kruskal—Wallis test by hand. Lemon Yellow 45 59 48 46 38 47 20 24 23 21 17 22 White 21 12 14 17 13 17 12.5 3 5.5 9.5 4 9.5 21 14 7 2 11 12.5 5.5 1 Green 37 32 15 25 39 41 16 15 7 14 18 19 16 11 20 8 Blue Rank Then the sum of ranks for each group: Lemon Yellow, R] = 127; White, R2 = 44; Green, R3 = 89; Blue, R4 = 40. 111 = n; = r13 = T14 = 6 and N = 24. So the Kruskal—Wallis statistic is 12 1272 «.142 892 402 ff== ——————— ———— -- +--- -w~ —— 3 25 ==1 .95 (24)(25)( 6 + 6 5 + a) “(l 6 H has chi-square distribution with 3 degrees of freedom. So the p—value is PW2 > 16.95) = 0000?. So we have strong evidence that color affects the insect count [that is, the difference we observed is statistically significant). > ex15.34 <— read.table("D:/ch15/ex15_034.txt", header=T) > attach(ex15.34) > beetlesl <— beetles[color=="Lemon"] > beetles2 <- beetles[color=="White"] STYyT 3022 TUK:Jinghan hieng > beetleSB <— beetlesEcolor=="Green"] > beetles4 <— beetles[color=="Blue"] > kruskal.test(list(beetlesl,beetles2,beetle53,beetles4)) Kruskal-Wallis rank sum test data: list(beetle31, beetlesZ, beetles3, beetles4) Kruskal-Wallis chi—squared = 16.9755, df = 3, p-value = 0.000715 ...
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hw9 - STAT 3022 TA: Jinghan Meng Homework 9 Solutions...

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