HOMEWORK #5 due 12-6
Table 1 - Times and Weights for Motorized Carts
A 42 104 A 38 79 A 47 75 A 44 95 A 51 102
A 44 107 A 54 110 A 39 98 A 44 106 A 56 101
A 56 120 A 43 88 A 50 99 A 59 122 A 52 99
B 56 107 B 42 85 B 49 98 B 54 106 B 44 88
B 48 110 B 40 93 B 46 104 B 45 87 B 44 86
B 44 101 B 46 86 B 46 87 B 62 121 B 55 80
C 51 87 C 47 92 C 62 97 C 57 117 C 43 85
C 66 120 C 59 101 C 52 115 C 57 107 C 46 99
C 53 109 C 54 99 C 46 91 C 41 72 C 55 105
Each triple of values in Table 1 denotes the cart type (one of three
values A,B,C), the delivery time for that
load (Y), and the weight of the
load (X).
Carttype for
cart brand and Weight for the weight.) What is the P-value
for the ANOVA? What is the model R2?
proc
corr with by Carttype.) Do the within-cart-type
correlation coefficients vary? Also, construct a scatterplot of time
versus weight with cart type as the plotting symbol. Do these
conclusions affect your answer to part (ii)?
2. Variables AA, BB, and CC were measured for 32 subjects. Of these subjects, 12 later developed Condition X while the remaining 20 did not develop Condition X. The data are listed in Table 2.
Table 2 - Covariates AA BB CC for 32 subjects
that later either developed or did not develop Condition X
Developed Condition X Did NOT develop Condition X
Subj AA BB CC Subj AA BB CC
1 69 83 51 13 36 55 39
2 51 74 32 14 50 69 44
3 27 68 33 15 36 59 28
4 55 85 46 16 31 26 44
5 27 99 34 17 31 49 47
6 44 68 38 18 32 45 50
7 49 88 57 19 40 59 33
8 28 64 66 20 49 51 42
9 32 58 46 21 38 70 47
10 47 81 39 22 46 63 26
11 35 77 31 23 46 64 47
12 30 69 62 24 67 94 43
25 47 60 56
26 56 62 45
27 39 64 27
28 52 71 24
29 33 62 52
30 57 63 48
31 39 78 23
32 48 70 55
with the property that L(data)>0 predicts Condition X. Assume
SAS's default assumptions for proc discrim that the
variables AA BB CC in each group have joint normal distributions with the
same covariance matrix in each group and begin with a prior belief
of 0.50 that a randomly chosen subject has Condition X.
(Hints: (a) See plogistic.sas on the Math475 Web
site. (b) Look for Linear Discriminant Function in the
SAS output for proc discrim followed by Coefficient
Vector = COV(-1)Xbar_j or something similar. The coefficients in
the linear discriminant function are the differences between the
covariate coefficients for the two groups. The cutoff value is given by
the difference between the Constant values.)
simple tells SAS
to do this, as in (for example) proc discrim data=xnotx simple
....;. Alternatively, you can use proc means.) Which
covariates seem to be the most divergent between the two groups as judged
by the within-group means and standard deviations?
Resubstitution analysis.) If you enter
crossvalidate on the proc discrim command line,
then SAS will also do a crossvalidation procedure in which each
subject is classified on the basis of the discrimination rule defined by
the other subjects, NOT INCLUDING that subject itself. (The
resubstitution compares each subject with the rule for all subjects,
including the subject itself, which influences the rule about what group
that subject should belong to.) How does the number of misclassified
subjects change under this crossvalidation?
3. For the data in Table 2,
plogistic.sas on the Math475 Web
site.)