Download Lecture 9

Today’s lesson (Chapter 12)
• Paired experimental designs
• Paired t-test
• Confidence interval for E(W-Y)
Paired Design
• Find two experimental units that are more
like each other than randomly selected
units.
– Two animals from the same litter
– Before and after measurements on the same
subject.
• Apply the A treatment to one unit randomly
selected and the B treatment to the other.
Analysis of Paired Design
• There are n pairs of experimental units.
• For unit i, Wi represents the A treatment
observation from the i-th pair.
• For unit i, Yi represents the B treatment
observation from the i-th pair.
• Calculate the difference Di=Wi-Yi.
• Do a one-sample t-test on the differences.
Analysis of Paired Design
• That is, null hypothesis is E(W-Y)=0.
• Alternative hypothesis may be left, right, or
two-sided.
• Test statistic is the mean of the differences.
• Estimated standard error of the mean
difference is the standard deviation of the D
values divided by the square root of n.
• Standardize the test statistic as usual.
Example Problem 1
• A research team evaluated a medicine to
determine whether it lowered a blood
component. They measured B, the amount
of the component in six patients before a
protocol using the medicine was followed
and then measured A, the amount of the
component after the administration of the
medicine.
Example Problem 1
• They wished to test the null hypothesis that
E(A)=E(B) against the alternative that
E(A)<E(B). Their experimental results are
given in the following table. Which of the
following is a correct decision?
• Usual options: reject at 0.01 level, accept at
0.01 and reject at 0.05, accept at 0.05 and
reject at 0.10, accept at 0.10.
Data for Problem 1
Patient
1
2
3
4
5
6
Before
treatment
300
340
200
300
320
290
After
treatment
280
310
160
270
310
240
Solution of Problem 1
• Recognize that this problem requires a
paired t-test (before and after comparison).
• Compute the six differences A-B:
– -20, -30, -40, -30, -10, -50.
• Compute mean difference
– sum of differences is -180
– mean difference is -30
Solution of Problem 1
• Compute standard deviation of the six
differences
– six deviations from mean are
• -20-(-30)=10, 0, -10, 0, 20, -20
– check that they sum to zero
– find the squared deviations from the mean
• 100, 0, 100, 0, 400, 400
Solution of Problem 1
• Compute standard deviation of the six
differences (continued)
– sum the six squared deviations
• 1000
– find the degrees of freedom (pairs-1=5)
– find the variance (sum of squared deviations
per degree of freedom) = 1000/5 =200
– take the square root of the variance = 14.1
Solution of Problem 1
• Compute the estimated standard error of the
mean difference; that is, the standard
deviation over the square root of the number
of pairs
– 14.1/60.5=5.77
• Compute the t-statistic (standard score value
of the test statistic).
– T=(-30-0)/5.77=-5.20
Solution of Problem 1
• Decide on the side of the test: left sided!
• Determine the degrees of freedom: # pairs1=5.
• Stretch the critical values:
– -2.326 to -3.365, -1.645 to -2.015, and -1.282 to
-1.476.
• Make your decision.
– -5.20 is to the left of -3.365; reject at 0.01 level.
Most Fundamental Design
Advice
• Pair what you can, randomize what you
cannot.
• That is, always used a paired design when
possible.
• There is a generalization of a pair. It is
called a block.
• ADVICE: Block what you can, randomize
what you cannot.
Why this advice?
• Var(W-Y) equals var(W)+var(Y)2cov(W,Y).
• ASS-U-ME var(W)=var(Y)=σ2
• Then, cov(W,Y)=ρσ2.
• Var(W-Y)=σ2+σ2-2ρσ2=2σ2(1-ρ).
• When there is a large positive correlation
within the units, the variance of the
difference is small.
Extension to Hedging Strategies
• It is true that Var(W+Y) equals
var(W)+var(Y) +2cov(W,Y).
• Same assumptions.
• Then Var(W+Y)=σ2+σ2+2ρσ2=2σ2(1+ρ).
• When W and Y are negatively correlated,
variance of W+Y is reduced. That is, risk is
lessened.
Problem 2
• Based on the data given in problem 1, what
is a 99 percent confidence interval for the
difference in expected values E(A-B)?
Solution to Problem 2
• Find the degrees of freedom for your
estimate of the standard deviation: #pairs-1,
here 5.
• Stretch the normal factor for this level of
confidence (2.576) for the correct df: 4.032.
• Use the estimated standard error for the
unknown standard deviation of the mean.
– Left endpoint is -30-4.032(5.77)=-53.3
– Right endpoint is -30+4.032(5.77)=-6.7
Using SPSS to get paired t-test
• Statistics, compare means, paired t-test.
Example Computer Problem
• Two statistical procedures exist to
determine the estimated location of a gene.
• Which procedure comes closer to the
correct location of the gene?
• Use the results of a simulation study to
answer the question.
Data of Study
• Genetic model specified (recessive, all
families affected by the same genetic
pattern--no heterogeneity).
• Use simulation to generate the results of
100 independent studies.
• Apply two statistics (maximum hlod and
Kong and Cox correct to the Nonparametric
Linkage (NPL) statistic of Genehunter) to
each study.
Design of the Comparison
• For each of the 100 replicates, calculate the
difference D of the distance from the
maximum hlod analysis and the distance
from the maximum NPL statistic.
• Use paired t-test to test the null hypothesis
that the expected distance using the
maximum hlod is the same as the expected
distance using the maximum NPL.
• Closer is better.
Results
• Maximum hlod is significantly closer than
the maximum NPL for this genetic model.
Summary
• Paired t-test design
• Discussion of why paired t-test likely to be
better.
• Block what you can, randomize what you
cannot.
• Illustrated computations of the paired t-test
on both test and confidence interval.