Stat 401 A – HW 12 answers
1) Bat echolocation. (1 pt each part)
a) F = 0.4276, p = 0.66
Note: This is the test of Type in a model with Type as a factor/class variable and log mass. Type has 2 df
because there are three groups.
b) (Intercept)
-0.2024476
ebats
-1.2680677
birds
-1.3783902
logmass
0.5897821
massebat
0.2148750
massbird
0.2455883
Notes: massebat is the product of logmass and the ebat indicator, massbirds is the produce of logmass
and the bird indicator.
c) Intercept: -0.202, slope: 0.590 (or 0.5898)
Note: non-echolocating bats is the reference group (all indicators = 0, so that group is described by the
Intercept and overall slope terms. These are the same coefficients you get when subsetting the data
only non-echolocating bats and fitting that line. The standard errors are not the same (you should be
able to figure out and explain why).
d) Intercept: -1.470, slope: 0.805.
Note: Computed as Intercept = -0.2024476 + -1.2680677 = -1.470515 and
slope = 0.5897821 + 0.2148750 = 0.8046
e) t = 0.961, p = 0.35.
Note: The difference in slopes between e-bats and n-bats is given by the massebat term. The test of
massebat = 0 is a test whether the e-bat line and the n-bat line have the same slope.
f) F = 0.6718 p = 0.53.
Note: The far-from-significant test for unequal slopes supports dropping the interaction from the model,
assuming that all three groups have the same slope and analyzing the data using last week’s model.
Notes: For pt. a, e, & f, many students got that wrong for looking at the wrong test. I think most ran the
right code to get the correct output but looked at/reported the wrong test.
2) Palm Beach votes
a) 2 pt. Bush2000 Nader2000 Browne2000 Perot96 Buchanan96p TotalReg
b) 2 pt. Bush2000 Nader2000 Browne2000 Clinton96 Perot96 Buchanan96p TotalReg
c) 1 pt. Bush2000
d) 2 pts. Yes, the model with Bush2000 Perot96 Buchanan96p is a reasonable alternative to the model
with Bush2000. The BIC for the suggested model is 0.467 units less than the best model. Models with
BIC (or AIC) within 2 units of the best are reasonable alternatives.
e) 2 pts. 592 votes.
Note: computed by fitting the model with one predictor variable: Bush2000, predicting log votes in Palm
Beach County (prediction = 6.38), then exponentiating to get a predicted number of votes.
f) 1 pt. No, you can not just compare the observed value to the prediction. The predicted value has
uncertainty that can not be ignored. If that uncertainty is large, quite a large deviation may be quite
ordinary.
Note: I gave ½ credit for saying “no” and ½ credit for the correct resoning
g) 2 pts. You could compute a prediction interval for Buchanan in Palm Beach Co and see whether the
recorded number is within that prediction interval.
Note: The prediction interval is the appropriate one here because you are predicting a single value (one
county, one election). I did not ask you to compute the interval. If you did, you should get something
close to (251, 1399).
h) 2 pts. You could compute intervals for all reasonable models and see whether there is a consensus.
Note: Making conclusions when there are multiple reasonable models is an active area of statistical
research.
Note: For g & h, you needed to use a prediction interval. Most lost points because they reasoned using a
confidence interval.