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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 exclusiveterritory 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 logodds 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 onethird 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. osoooaaciumeia
MumehOﬂPNDOD 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 pvalue 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, pvalue = 0.02900
alternative hypothesis: true location shift is greater than 0 Warning message:
In wilcox.test.defau1t(speciesl, speciesE, alternative = "greater")
cannot compute exact pvalue with ties EX 15.29 (a) The stemandleaf 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: 13.17.2 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(radon105)
Wilcoxon signed rank test data: radon  105
V = 31, pvalue = 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—andleaf 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)
> stemChaitifactory)
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, pvalue = 9 28e06
alternative hypothesis: true location shift is less than 0 Warning message:
In wilcox.test.default(haiti, factory, alternative = "less") :
cannot compute exact pvalue 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 chisquare 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)) KruskalWallis rank sum test data: list(beetle31, beetlesZ, beetles3, beetles4)
KruskalWallis chi—squared = 16.9755, df = 3, pvalue = 0.000715 ...
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