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
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
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
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
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
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
95% condence interval for Ye is (3.144, 3.508).
Thus, at 95 % condence, a plausible range for
the mean rate of oxygen consumption is from
3.144 to 3.508 ml/g/hr. Since the value of 3
does not fall within this range, it is not a plausible value with 95% c
Stat/For/Hort 571
Midterm II, Fall 2004
Brief Solutions
1.
of as the addition of the number of df for each
variety (9 + 9 + 6 + 5 + 8 + 9).
4.
(a) The random variable, Y is the number of
tigers with the bacteria present. A reasonable model (given the avai
Stat./For./Hort. 571
Final Exam, Fall 2005
Brief Solutions
(b) At this level of signicant, there is no evidence that sulde concentration is higher at wells C, D, and E, but
there is evidence at well F, all compared to the mean
sulde concentration of wells
Stat./For./Hort. 571
Midterm I, Fall 2005
Brief Solutions
1. This is a paired two-sample inference problem. Let Y1 =
aphid number at the bottom and Y2 = aphid number at the
top of a plant. Let D = Y2 Y1 denote the dierence of
aphid numbers with D = E (D).
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
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/For/Hort 571
Clayton and Zhu
October 16, 2012
Midterm I
Name:
For the section that you attend please indicate:
Instructor:(circle one) Clayton
Zhu
TA: (circle one) Cook
Jiang
Zhang
Instructions:
1. This exam is open book. You may use textbooks, noteb
(b) Since X N (67, 900/16) = N (67, 7.52 ),
Stat/For/Hort 571
Midterm I, Fall 2012
Brief Solutions
=
1.
(a) Here, n = 14 and for the median we need p to
be 0.5; for the IQR we need to set p equal to
0.25 and 0.75. Thus, for the median n p = 7
and the med
Stat/For/Hort 571
Midterm II, Fall 2012
Brief Solutions
1.
(a) For testing H0 : 1 = 2 vs. HA : 1 = 2 ,
compute the pooled variance estimate
s2 = (9 252 + 9 122 )/18 = 384.5.
p
The observed test statistic is
58 39
t=
= 2.17
384.5 2/10
on 18 df. The p-value
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
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
Stat./For./Hort. 571
Midterm I, Fall 2005
Brief Solutions
1. This is a paired two-sample inference problem. Let Y1 =
aphid number at the bottom and Y2 = aphid number at the
top of a plant. Let D = Y2 Y1 denote the dierence of
aphid numbers with D = E (D).
Stat./For./Hort. 571
Final Exam, Fall 2005
Brief Solutions
(b) At this level of signicant, there is no evidence that sulde concentration is higher at wells C, D, and E, but
there is evidence at well F, all compared to the mean
sulde concentration of wells