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
Assignment 13
Exercise 8.4 For the hypotheses considered in Examples 8.12 and 8.13, the sign test
d
is based on the statistic N+ = #cfw_i : Zi > 0. Since 2 n(N+ /n 1 ) N (0, 1)
2
under the null hypothesis, the sign test (with continuity correction) reject
Assignment 12
Exercise 7.12 (a) Derive the Jereys prior on 2 for a random sample from N (0, 2 ).
Is this prior proper or improper?
Sketch of solution:
Letting = 2 , we dierentiate the logdensity twice to nd that I () = 1/(22 ). Therefore, the Jereys prior
Assignment 11
Exercise 7.1 In this problem, we explore an example in which the MLE is not consistent. Suppose that for (0, 1), X is a continuous random variable with
density
f (x) = g (x) +
1
h
()
x
()
,
(7.4)
where () > 0 for all , g (x) = I cfw_1 < x
Assignment 10
Exercise 4.8 Prove that (4.13) implies both (4.12) and (4.14) (the forward half of
the Lindeberg-Feller Theorem). Use the following steps:
(a) Prove that for any complex numbers a1 , . . . , an and b1 , . . . , bn with |ai | 1
and |bi | 1,
n
Assignment 9
Exercise 5.5 Let Xn binomial(n, p), where p (0, 1) is unknown. Obtain condence intervals for p in two dierent ways:
d
(a) Since n(Xn /n p) N [0, p(1 p)], the variance of the limiting distribution
P
depends only on p. Use the fact that Xn /n p
Assignment 8
Exercise 4.7 Use the Cramr-Wold Theorem along with the univariate Central Limit
e
Theorem (from Example 2.12) to prove Theorem 4.9.
Sketch of solution: This proof is actually given in the course notes, just
before Exercise 4.8: Let X Nk (0, )
Assignment 7
Exercise 4.3 Use the Continuity Theorem to prove the Cramr-Wold Theorem, Thee
orem 4.12.
d
Hint: a Xn a X implies that a
Xn (1)
a
X (1).
Sketch of solution: As we pointed out in class, the only tricky part of the
d
d
Cramr-Wold Theorem is sh
Assignment 6
Exercise 1.41 Kolmogorovs inequality is a strengthening of Chebyshevs inequality
for a sum of independent random variables: If X1 , . . . , Xn are independent random
variables, dene
k
(Xi E Xi )
Sk =
i=1
to be the centered k th partial sum fo
Assignment 5
a.s.
Exercise 3.2 The diagram at the end of this section suggests that neither Xn X
qm
nor Xn X implies the other. Construct two counterexamples, one to show that
qm
qm
a.s.
Xn X does not imply Xn X and the other to show that Xn X does not
a.
Assignment 4
Exercise 2.10 The goal of this Exercise is to construct an example of an independent
P
sequence X1 , X2 , . . . with E Xi = such that X n but Var X n does not
converge to 0. There are numerous ways we could proceed, but let us suppose
that fo
Assignment 3
Exercise 1.43 The complex plane C consists of all points x + iy , where x and y are
real numbers and i = 1. The elegant result known as Eulers formula relates
the points on the unit circle to the complex exponential function:
expcfw_it = cos
Assignment 2
Exercise 1.20 According to the result of Exercise 1.16, the limit (1.21) implies that
the relative dierence between n=1 (1/i) and log n goes to zero. But this does
i
not imply that the dierence itself goes to zero (in general, the dierence ma
Statistics 410, Fall 2011
Solutions to homework 9, due 11/23/11
12.22 Let be the mean thickness of the plastic sheets being manufactured.
(a) H0 : = 1.2mm.
(b) Ha : = 1.22mm.
(c) The signicance level is = .05.
(d) The test statistic is 1.2275, and we expe
Statistics 410, Fall 2011
Solutions to homework 8, due 11/16/11
11.11 (a) This corresponds to being 1 standard deviation above the mean.
So .15866.
(b) For the sample total, the standard deviation is n = 2 5 = 10.
This is then two standard deviations over
Statistics 410, Fall 2011
Solutions to homework 5, due 10/26/11
6.1 (a) = .5. = 0.
(b) = .0625, = 0.
(c) = .525 , = 0.
(d) = .5100 , = 0.
6.2 (a) is the probability of no aces in 24 rolls if the die is fair. This
5 24
is 6 = .0126.
(b) = 0. That is, if th
Statistics 410, Fall 2011
Solutions to homework 4, due 10/19/11
4.16 U is the dosage level (1 through 7). Each level has probability 1/7,
so the histogram is like Figure 4.7 except height 1/7 and with 7 bars
instead of 6.
b.
1
1
1
1
= E (X ) = 1 + 2 + +