Stat/For/Hort 572 — Midterm I, Spring 2006 — Solutions
1. (a) Consider the four model assumptions: correct model, independence, equal variance, and normal distri-
bution. The linear line relationship app ears inadequate and the equal variance assumption may not be
satisfied.
(b) H0: no LOF versus HA: LOF of the SLR model. From the R output, the observed f= 9.0348. Compare
to Fdistribution with df = (12, 14), the p-value is less than 0.001. Thus reject H0at 5% level and there
is very strong evidence of a lack of fit of the SLR model.
(c) From the R output, for x∗= 18, ˆyest = 6.76, s.e.(ˆyest) = 0.7084. Since t0.025,26 = 2.056, the 95%
confidence interval is ˆyest ±t0.025,26 ×s.e.(ˆyest), which is 6.76 ±1.46 or [5.30,8.22].
(d) H0: no LOF versus HA: LOF of the quadratic regression model. Since SS Pure Error is 18.601 on 14
df and SS Error is 126.64 on 25 df, SS LOF is 126.64 −18.601 = 108.039 on 25 −14 = 11 df. By the
additional sum of squares principle, the observed f=108.039/11
18.60/14 = 7.39. Compare to Fdistribution with
df = (11, 14), the p-value is less than 0.001. Thus reject H0at 5% level and there is very strong evidence
of a lack of fit of the quadratic regression model.
2. (a) The parameter b3is the slope difference between the two regression lines corresponding to farms A and
B respectively. From the R output, use either the observed t=−0.608 on 6 df or f= 0.3694 on (1, 6) df.
The p-value is 0.5656. Do not reject H0at 5% level and there is no evidence of a nonzero b3.
(b) The parameter b0is the intercept of the regression line for farm A, which represents the expected weight
gain for a zero level of diet supplement. Since ˆ
b0= 1.888, s.e.(ˆ
b0) = 0.3299, t0.05,6= 1.943, the 90%
confidence interval is ˆ
b0±t0.05,6×s.e.(ˆ
b0), which is 1.89 ±0.64 or [1.25,2.53].
(c) The test of interest is H0: [b2=b3= 0|b0, b1] versus HA: not H0. The additional sum of squares is
10.04 + 0.0106 = 10.0506 on 2 df and the SSE of the full model is 0.1718 on 6 df. By the additional sum
of squares principle, the observed f=10.0506/2
0.1718/6= 175.51. Compare to Fdistribution with df = (2, 6),
the p-value is less than 0.001. Thus reject H0at 5% level and there is very strong evidence that the two
regression lines are not equal.
(d) The full model is y=b0+b1w1+b2w2+ewhich has SSE = 0.0106 + 0.1718 = 0.1824 on 7 df. The
reduced model is y=b0+b1w1+ewith additional sum of squares 10.04 on 1 df. By the additional sum
of squares principle, the observed f=10.04/1
0.1824/7= 385.31. Compare to Fdistribution with df = (1, 7), the
p-value is less than 0.001. Thus reject H0at 5% level and there is very strong evidence of a nonzero b2.
That is, although there is no evidence of slope difference, there is strong evidence of intercept difference
between the two regression lines for farms A and B.
3. (a) From the R output, the correlation between yand each individual xis the highest for x3. This implies
that the third model y=b0+b1x3+ehas the largest R2and thus the smallest SSE. The df of SSE is
the same for all three models. Thus the third model has the smallest MSE.
(b) H0: the third observation is not an outlier versus HA: not H0. Because of the way x4is coded, use the
observed t-value 2.983 on 7 df. The p-value is 0.02043 for one comparison and thus the exp eriment-wise
p-value is 12 ×0.02043 = 0.2452. Do not reject H0at 5% level and there is no evidence that the third
observation is an outlier.
(c) According to the full model fit, the t-value for b1(1.885) is the smallest and is less than 2. Thus eliminate
x1is the first step. Now fit the model with x2and x3. Since the smallest t-value is 16.905 and is more
than 2, stop. The model selected by backward elimination is
y=b0+b2x2+b3x3+e
Grade Distribution
100:2
90-99:15
80-89:18
70-79:17 mean = 78, median = 80
60-69:3
<60:10
1