Stat/For/Hort 571
Midterm II, Fall 2004
Brief Solutions
1. (a) TRUE — From the tables, P(T≤
1.533) = 1 −P(T≥1.533) = 0.90 (Check
the directionality!!) Similarly, P(Z≤
1.533) = 1 −P(Z≥1.533) ≈0.938. Alter-
natively, since the tdistribution is known
to have fatter tails than the Z, the prob-
ability that tis above 1.533 (in the right
hand tail) is smaller than the correspond-
ing probability for Z. Considering the di-
rectionality, the conclusion follows. (A pic-
ture might be helpful!!)
(b) FALSE — The given ’reasoning’ essen-
tially is a ’dressed-up’ version of the argu-
ment that scientific significance can out-
weigh statistical significance. In practice,
one needs both. A claim about what might
have occurred means practically nothing.
Inference rests on the data actually ob-
tained. (Issues regarding df are not rel-
evant here.) The most one can say for X
is that if the potential scientific importance
is great enough, perhaps a new experiment
with larger nmight be conducted.
2. This is a two-independent sample situation.
(a) Let YAand YBbe the concentrations on
A and B respectively and let µAand µB
be the respective population means. The
needed summary statistics are ¯yA= 6.1,
s2
A= 0.8733, ¯yB= 4.9, and s2
B= 0.74.
Then, due to the balanced data, s2
p= (s2
A+
s2
B)/2 = .8067. The CI for µA−µBis
¯yA−¯yB±t∗qs2
p(1/nA+ 1/nB). Since s2
p
has 6 df, the appropriate t-value is 1.943.
Thus, (−0.03 < µA−µB<2.43).
(b) The null hypothesis is written: H0:µA−
µB= 0.3. We notice that the value of 0.3
is contained within the CI above. Thus,
we can conclude that the p−value > 0.10
for the given test.
3. The appropriate formula for the pooled variance
is: s2
p= (Pk
i=1(ni−1)s2
i)/(N−k) where kis the
number of treatments and Nis the total number
of observations. You are given the standard de-
viations so you must square them. s2
p= 157.02.
There are N−k= 46 degrees of freedom asso-
ciated with this variance. This can be thought
of as the addition of the number of df for each
variety (9 + 9 + 6 + 5 + 8 + 9).
4. (a) The random variable, Yis the number of
tigers with the bacteria present. A rea-
sonable model (given the available infor-
mation) is Y∼B(15, p). The hypothe-
ses can be written: H0:p= 0.05 vs
HA:p > 0.05. Evidence against the null
is obtained for ’large’ observed values of
Y. Using the basic definition of p-values,
p−val =P(Y≥2). This can be written
as 1 −(P(Y= 0) + P(Y= 1)). Using the
binomial formula results in p−val = 0.171.
This means that there is no meaningful ev-
idence against the claim that the rate of
occurrence of this bacteria is 0.05 or less.
(b) The random variable Yis the number of
cats with bacteria present. A reasonable
model is Y∼B(60, p). H0:p= 0.15 vs
HA:p≥0.15. Since np(= 9) and n(1 −
p)(= 51) are both larger than 5, the nor-
mal approx may be used. Let ˆp=Y/60.
Then, under H0, ˆpNA ∼N(.15,(.0461)2).
α=P(ˆp≥0.25) = P(Z≥2.17) = .015.
5. (a) The general form for a CI for the difference
between two proportions is: ( ˆpA−ˆpB)±
Zα/2∗qpA∗(1−pA)
nA+pB∗(1−pB)
nB. Because
pA= 0.4 and pB= 0.6, both expressions
of the form p(1 −p) are the same. Since
Zα/2affects all CIs equally (given the same
1−α), we need to minimize q.24
nA+.24
nB.
This is 0.163 for choice (1) and 0.107 for
choice (2). Choice (2) has smaller width.
(b) By looking at the expression above, the
CI width is minimized when 1
nA+1
nBis
as small as possible with nA+nB= 100.
Based on part(a), the width is smaller for
the values of nAand nBthat are closer
together. Thus the ’natural’ conjecture
is that the CI width can be minimized if
nA=nB= 50. This is the correct answer.
It is primarily for this reason that most
scientific studies comparing 2 (or more)
groups have equal sample sizes.
Grade Distribution
90-99:22
80-89:41
70-79:26 median = 79
60-69:17
50-59:13
below: 8