Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Stat 571
Final Exam Solution
December 20, 2009
1.(a) true.
(b) false: independence is important no matter how much data there is. The ttest is robust to a lack of
normality, but not to a lack of independence.
(c) true.
(d) false. The CI width is a descri
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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 pvalue slightly above 0.01. We reject the null
hypothesis: there is evidence that mole rats from dierent
colonies choose their tubers d
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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. Counterexample: A = cfw_0, 1, 2 (2 or fewer
with 0.025 on each side. This is the area between
redey
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Second Midterm Exam
(c) MannWhitney 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
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Stat 571
Second Midterm Exam
1(a) df= 10+(13 1) 2 = 20 since one zero is dropped in
the second sample. With a 2sided 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
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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 chisquare test) so that the
pvalue is p = 2 0.00006
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Outline
1
Signicance Tests
with known variance
with unknown variance
2
Signicance tests with binomial data
Using a ztest
Using a chisquare test
Using the binomial distribution
3
Tests on one sample: Recap
1
Signicance Tests
Now we view sample data as ou
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Outline
1
OneWay Anova
Twosample case reconsidered
General case of multiple independent samples
Assumptions and Model
Levenes test: are variances equal?
2
Comparisons among Means
3
Inference with Multiple Comparisons
Comparisonwise and Experimentwise
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Outline
1
Analysis of categorical data
Goodness of t: t of data to a claim
Test of independence
Assumptions
2
Simple Linear Regression
Estimating the slope and intercept
Testing the slope: Ftest
Correlation
Back to the Ftest
Testing the slope and interc
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Outline
1
The binomial distribution
Denition
Formula for probability calculations
Expectation and variance
Proportions
2
The normal distribution
Denition
The standard normal distribution
The general normal distribution
3
Calculations with R (binomial and
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Statistics 571
Statistical Methods for Bioscience I
Lecture 1: Cecile Ane
Lecture 2: Nicholas Keuler
Department of Statistics
University of WisconsinMadison
Fall 2009
Outline
1
Course Information
2
Introduction to Statistics
3
Descriptive Statistics
4
Sha
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Outline
1
Probability Model
Experiments and events
Rules for probability calculations
Independence
2
Random Variables
Denition and examples
Probability distribution of a RV
Expectation
Variance and Standard deviation
Independence
Notion of chance
What is
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
Outline
1
Condence intervals
Normal distribution, known variance
Normal distribution, unknown variance
Unspecied distribution
Binomial distribution
2
SD versus SE
3
On assumptions
4
Power analysis to determine sample size
Rejection region
Power
Sample siz
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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
Statistical Methods for the Bioscience I  Fall 2009
STATISTICS 371

Fall 2011
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