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7° changes from .794 to 616 1400, .3555, .3385, .1433, 0227 and expected counts
21.00, 53.32, 50.78, 21.49, 3.41. Because 3.41 <5,
113. a. The paired 1 procedure gives ¢ = 3.54 with a two- combine the last two categories, giving 7 = 1.62 with
tailed P-value of .002, so at the .01 level we reject the P-value > .10, Do not reject the binomial rddel,
hypothesis of equal means. .
b. The regression line is y = 4.79 + .743x, and the test 17. 4 = 3.167 which gives 7° = 103.9 with P-value < .001,
of Ho: Bi = 0 vs. Hs: Bi 4 0, gives t= 7.41 with a so reject the assumption of a Poisson model.
P-value of <.000001, so there is a significant Z % .
relationship. However, prediction is not perfect, 19+ 1 = 4275. 02 = .2750 which gives 7° = 29.3 with P-
with 1 = .753, so one variable accounts for only value: <,-001, so-reject the:model,
75% of the variability in the other. 21. Yes, the test gives no reason to reject the null hypothesis
117. a. linear of a normal distribution,
b. After fitting a line to the data, the residuals show a 93. The p-values are both 243.
lot of curvature.
¢. Yes. The residuals from the logged model show some 25. Let pj: = the probability that a fruit given treatment i
departure from linearity, but the fit is good in terms matures and pj =the probability that a fruit given
of R® = .988. We find & = 411.98, f = —.03333. treatment j aborts, so Hg: pia = Pia for i = 1, 2, 3,4, 5.
d. (58.15, 104.18) We find 7° = 24.82 with P-value < .001, so reject the
null hypothesis and conclude that maturation is affected
119. a. The plot suggests a quadratic model. ipfledt sano.
b. With f = 25.08 and a P-value of < .0001, there is a
significant relationship at the 0001 level 21. If p;; denotes the probability of a type j response when
¢. Ch: (3282.3, 3581.3), PI: (2966.6, 3897.0). Of course, treatment i is applied, then Hy: p1j = p>; = Pay = Pay for
the PI is wider, as in simple linear regression, j=1, 2, 3, 4. With 7? = 27.66 > 23.587 = Zoso,
because it needs to include the variability of a new reject Ho at the .005 level. The treatment does affect the
observation in addition to the variability of the mean. response.
d. Cl: (3257.6, 3565.6), PI: (2945.0, 3878.2). These are > >
slightly wider than the intervals in (c),which is 29+ With 7° = 64.65 > 13.277 = 7, 4, reject Ho at the .001
appropriate, given that 25 is slightly closer to the level. Political views are related to marijuana usage. In
Haan gndthe vertex. particular, liberals are more likely to be users.
e. With 1=~6.73 and a two-tailed P-value of 34, Compute the expected -~—=counts-—oby
<.0001, the quadratic term is significant at the en = Bin = nb. Phan 22", — For the
.0001 level, so this term is definitely needed. aH a - lial
y 7 statistic df = 20.
121. a. With f=2.4 <586=Fosis4, there is no 2 ,
detec relationship at the 05 level 33. a. With 7? = 681 < 4.605 = Zio2, do not reject
b. No, especially when & is large compared to n pce readies te LO level,
ake ene b. With 7 = 6.81 > 4.605 = 795, reject independence
¢. 9565
at the .10 level.
. 677
Chapter 13 35. a. With 7 = 6.45 and P-value .040, reject independence
at the .05 level.
1. a. reject Hy bs do not reject Hye. donot reject. © b. With 2=~2.29 and P-value .022, reject
Hy d.donot reject Hy independence at the .05 level.
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834 Chapter 14
¢. Because the logistic regression takes into account the 5. We form the difference and perform a two-tailed test of
order in the professorial ranks, it should be more Hs: 1 = 0 at level .05. This gives s, = 72 and because
sensitive, so it should give a lower P-value. it does not satisfy 14 <5, < 64, we reject Ho at the
d. There are few female professors but many assistant .05 level.
professors, and the assistant professors will be the .
professors of the frtures 7. Because s, = 162.5 with P-value .044, reject Ho: 1 = 75
in favor of H,: 1 > 75 at the .05 level.
37. With 7 = 13.005 > 9.210= 73,5, reject the null
hypothesis of no effect at the .01 level. Oil does make a9 With w = 38, reject Ho at the .05 level because the
difference (more parasites). rejection region is {w > 36).
39. a. Ho: The population proportion of Late Game H- Test Ho: a — #2 =1 vs. Hat fy — fa > 1. After
Leader Wins is the same for all four sports; H,: The subtracting 1 from the original process measurements,
proportion of Late Game Leader Wins is not the same we get w = 65. Do not reject Ho because w < 84.
for all four sports. With 7? = 10.518 > 7.815 = oss 13, b, Test Hot str — yl2 = 0 vs. Hat sh — #2 <0. With a
reject the null hypothesis at level .0S. Sports differ in Povaue Of 002 We rejéct Hp at the .01 level.
terms of coming from behind late in the game.
b. Yes (baseball) 15, With w = 135, z = 2.223, and the approximate P-value
5s is .026, so we would not reject the null hypothesis at the
41. With 2 = 197.6 > 16.812 = 74,6, reject the null ‘0! level.
hypothesis at the .01 level. The aged are more likely to
die in a chronic-care facility. 17. (11.15, 23.80)
43. With 77 = 763 < 7.779 = 794, do not reject the 19. (—.585, .025)
hypothesis of independence at the .10 level. There is no
evidence that age influences the need for item pricing. 24+ (16, 87)
45. a. No, #2 =9.02 > 7815 = Bes. 29. a. (4736, .6669)
b. With 7? = .157 < 6.251 = 7g 5, there is no reason to b. (4736, 6669)
say the model does not fit. 33. Fora two-tailed test at level .05, we find that s, = 24 and
GF aviienp=y =..=9~10 va cHe anlleavons because 4 < s, < 32, we do not reject the hypothesis of
p: # 10, with df = 9. equal means.
b. Ho: py = 01 for i and j = 0,1,2,...,9 vs. Ha: at least 35, a. y — 0207; Bin(20, .5)
one pi; # .01, with df = 99. b. ¢ = 14; because y = 12, do not reject Ho,
¢. No, there must be more observations than cells to do a
valid chi-square test. 37. With K = 20.12 > 13.277 = fy 4, reject the null hypo-
. The results give no reason to reject randomness. thesis of equal means at the 1% level. Axial strength does
seem to (as an increasing function) depend on plate
length.
Chapter 14 39. Because f. = 6.45 < 7.815 = 7s 3, do not reject the null
hypothesis of equal emotion means at the 5% level.
1. For a two-tailed test of Ho: 1 = 100 at level .05, we
find that s, = 27 and because 14 < s, < 64, we do not 41. Because w’ = 26 < 27, do not reject the null hypothesis
reject Ho. at the 5% level.
3. Fora two-tailed test of Hp: 1 = 7.39 at level .05, we find
that s,=18 and because s, does not satisfy
21 < sy < 84, we reject Ho.