Stat 571
Final Exam Solution
December 20, 2009
1.(a) true.
(b) false: independence is important no matter how much data there is. The t-test is robust to a lack of
normality, but not to a lack of independence.
(c) true.
(d) false. The CI width is a descri
Second Midterm Exam
Stat 571
1.(a) (8 points) power = P (X 18 or 22| = 23)
1823
22
= P (Z
or 23 ) which is P (Z
64/16
November 24, 2009
assumption is that the number of acorns is normally distributed (which we know is not true) or
that the sample size i
First Midterm Exam
Stat 571
October 13, 2009
approximated by a normal N (38.4, 5.112 ) distribution (np = 38.4 and nq = 81.6 are both > 5).
We observe z = (58 38.4)/5.11 = 3.835 (or
X 2 = 14.71 with a chi-square test) so that the
p-value is p = 2 0.00006
p(1 p)
which equals 0 because 1
n
p = 0. Actually, there are errors bars at this
point, they extend the horizontal line a bit to the
left.
be found to make the samples look normally distributed. If the transformed data have very different variances, use
Stat 571
Second Midterm Exam Solutions
December 2, 2008
1(a) (1) (4 points) Each rat must be independent of every
other rat - this cant really be veried from the given
information, but theres no particular reason to think
it is invalid. (2) Expected count
Stat 571
First Midterm Exam
October 7, 2008
5.(a) mean= 8 (center of symmetry). We know there
is 95% of the area whithin 1.96sd of the mean,
2.(a) false. Counter-example: A = cfw_0, 1, 2 (2 or fewer
with 0.025 on each side. This is the area between
red-ey
Stat 571
Second Midterm Exam
1(a) df= 10+(13 1) 2 = 20 since one zero is dropped in
the second sample. With a 2-sided test .02 < p < .05
and we conclude that the two groups have dierent
variances (reject 1 = 2 ).
November 20, 2007
binomial with a normal d
Stat 571
Final Exam - Solution
1.(a) 37 24/66 = 13.45
(b) X 2 = 1.5365 + 0.055 + + 2.943 = 8.8705. Using df=2
we get a p-value slightly above 0.01. We reject the null
hypothesis: there is evidence that mole rats from dierent
colonies choose their tubers d
Second Midterm Exam
(c) Mann-Whitney test. Ranks sum to 52 for variety A
and 53 for variety B. Variety B is the sample with
fewer data points, so T = 53, and T = 6 (6 +
8 + 1) 53 = 37 so T = 37. From the table, the
critical value of T is 29 at level = .05
Stat 571
First Midterm Exam
October 10, 2006
ii. IPcfw_I |B = 0.40 is not the same as IPcfw_I =
0.66, so B and I are not independent.
1. We rst nd z such that IPcfw_Z < z = 0.98, i.e
IPcfw_Z > z = 0.02. We get z = 2.05. It gives
x 25
= 2.05 i.e. x = 2
Stat 571
Final Exam - Solution
December 16, 2006
(b) s2 = 1.326 associated with df=24. For LSD we get dL =
e
2.797 1.326 2/7 = 1.721. dQ = 4.91 1.326 1/7 =
2.317 for Tukey. For Bonferroni, we need to know t.005/3,24 =
t.0016,24 but the value = .0016 is no