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832 Chapter 12
81. a, .686, no i Sues ee
Ze Si DF ss MS F
b. We find f = 28.6 > 2.62 = Foo1,16.196, 80 there is a Sonuce PE) SS SCE
significant relationship at the .001 level. Regression 2 5 2.5 0.625
¢. With all other predictors held constant, the estimated Error 1 4 4.0
difference in y between class A and not is 364. In Total 3.009
terms of $/ft?, the effect is multiplicative. Class A tt
buildings are estimated to be worth 44% more With f =.625 < 1995 = Fos91, there is no
dollars per square foot, with all other predictors held significant relationship at the .05 level.
constant. . ;
d. The difference in (c) is highly significant because the 93. By = J, s= VO (y SY /(n— 1),
two-tailed P-value is .00000013 coo = 1m, FE torsn1s/Vn
83. a. 48.31, 3.69 95. a. ee yy =F,
b. No, because the interaction term will change. aa pie ee
c. Yes, f = 18.92, P-value < .0001. =o Ha
d. Yes, 1 = 3.496, P-value = .003 < .01 we nae ae
b. §, =I f= les =Jai = m+ lyemtn
©. (21.6, 41.6) aE AC Ca le a al
f. There appear to be no problems with normality or SLOW) + Un Oi- hy =
curvature, but the variance may depend on x, VSSE/(m+n—2) er, = 4/(m +n)
5) wane d. By = 128.17, B, = 14.33 f= 121, 1=1,...,3;
b. With = 5.03 > 3.69 = Foss, there is a significant J = 135.33, i=4,...,6 5
relationship at the .05 level SSE OG kms: Bn 28
c. Yes, the individual hypotheses deal with the issue of 95% CI for B; (2.09, 26.58)
whether an individual predictor can be deleted, not the 97, Residual = Dep Var ~ Predicted Value
effectiveness of the whole model. Std Error Residual = [MSE — (Std Error Predict)*]*
de 6.25330 CG aL) Student Residual = Residual/Std Error Residual
e. With f = 3.44 < 4.07 = Fos... there is no reason to
reject the null hypothesis, so the quadratic terms can 101. a. Hy = 1/n+ (x) — X)(x) —¥)/E(ae — 8)?
be deleted. VF.) = 97[L/n + (i — 3) /BQ% — 8°] 5
87. a. The quadratic terms are important in providing a good Bev ¥) = oh — Wn (i) (eG 3))|
Ele heddie ¢. The variance of a predicted value is greater for an x
a) that is farther from x
Bi DROS ELS (S90, 77), d. The variance of a residual is lower for an x that is
89. a. rey = 843 (.000), rea = 621 (.001), ma = 843 farther from ©
(.000) Here the P-values are given in parentheses to €. Itis intuitive that the variance of prediction should be
three decimals: higher with increasing distance. However, points that
b. Rating = 2.24 + 0.0419 IBU — 0.166 ABV. Because are farther away tend to draw the line toward them,
the two predictors are highly correlated, one is so the residual naturally has lower variance.
fedundant, 103. a. With f= 12.04 >9.55=Foi27, there is a
Stearn este: significant relationship at the .O1 level
e. The regression is quite effective, with R? = .872. The ae anensine ee
To test Ho: fi=0 vs. Hy fi #0, =
ABV coefficient is not significant, so ABV is not
g u 2.96 > to25.7 = 2.36, So reject Hy at the .05 level.
needed. The highly significant positive coefficient &
: The foot term is needed.
for IBU and negative coefficient for its square show : .
To test Ho: Po=0 vs. Hy fo #0, I=
that Rating increases with IBU, but the rate of increase
howertenithes tac: 0.02 < f25.7 = 2.36, so do not reject Hy at the .0S
TEER SMES abet level. The height term is not needed.
i A et i b. The highest leverage is .88 for the fifth point. The
a height for this student is given as 54 inches, too low
stiaxwe|i — 1] y= 141, to be correct for this group of students. Also this
Port 0 value differs by 8” from the wingspan, an extreme
root 4 difference.
400 6 15 ¢. Point 1 has leverage .55, and this student has height
0 4 olp=|2 bp=| 5 75, foot length 13, both quite high.
bd 4 Fi Point 2 has leverage .31, and this student has height
66 and foot length 8.5, at the low end.
0 1 Point 7 has leverage .31 and this student has both
a_|| 2 ~_|-1 height and foot length at the high end.
Gm ~¥i SSE = 4, MSE = 4 fe
ced te d. Point 2 has the most extreme residual. This student
3 1 has a height of 66” and a wingspan of 56" differing
a. (122, 13.2) by 10”, so the extremely low wingspan is probably
r sf . wrong.
e Fo ste set of Hit Bis 9 a He he, x“ e. For this data set it would make sense to eliminate
= 5 < tors = 12.7, s 0 Doe a y sect ’
the .05 level. The x, term does not play a significant pointsc2eand:2/hecauserthey seemstobexwrong:
aie However, outliers are not always mistakes and one
needs to be careful about eliminating them,
--- Trang 846 ---
Chapter 13 833
105. a. 507% —_b..7122 3. Do not reject Hy because 7? = 1.57 < 7.815 = 7455
¢. To test Ho: fy = Ovs. Hy: B; #0, wehavet = 3.93, | A
with P-value .0013. At the .01 level conclude that 5+ Because 7” = 6.61 with P-value .68, do not reject Ho.
there 15,4 useful linear relationship 7. Because 7° = 4.03 with P-value > .10, do not reject Ho.
d. (1.056, 1.275)
ej=10l4 y-y=-214 9. a. [0, .223), [.223, .510), [510, .916), [.916, 1.609).
[1.609, 00)
107. —36.18, (—64.43, -7.94) b. Because 7° = 1.25 with P-value > .10, do not reject
109. No, if the relationship of y to x is linear, then the Ho.
relationship of y* to x is quadratic. IL. a. (—00, ~.967), [-.967, —.431), [-.431, 0), [0, 431),
IIL. a. Yes 431, .967), [.967, 00)
b. §=98.293 y-j= a7 b. (—2c,.49806), [.49806, .49914), [.49914, .50), [.50,
esa 155 50086), [-50086, .50194), [.50194, 00)
a. 794 ¢. Because 7° = 5.53 with P-value > .10, do not reject
e. 95% CI for B1: (.0613, .0901) Ho.
f. The new observation is an outlier, and has a major 43, Using j —.0843, 2 — 280.3 with P-value < 001, so
impact: reject the independence model.
The equation of the line changes from i
y = 97.50 + 0757 x to y = 97.28 + .1603 x 15. The likelihood is proportional to 07°5(1 — 0)°°7 from
changes from .155 to .291 which 0 = 3883. This gives estimated probabilities