Transcript The Aortic Stenosis Data: A Review

STAT 651
Lecture # 11
Copyright (c) Bani Mallick
1
Topics in Lecture #11

Comparing two population means

Review to date
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Book Sections Covered in Lecture #11

All previous chapters
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Lecture 10 Review: Robust
Inference via Rank Tests



Because sample means and sd’s are sensitive
to outliers, so too are comparisons of
populations based on them
Rank tests form a robust alternative, that can
be used to check the results of t-statistic
inferences
You are looking for major discrepancies, and
then trying to explain them
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Lecture 10 Review: Robust
Inference via Rank Tests


Typically called the Wilcoxon rank sum test
The algorithm is to assign ranks to each
observation in the pooled data set

Then apply a t-test to these ranks

Robust because ranks can never get wild
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Lecture 10 Review: Robust
Inference via Rank Tests



The rank tests give the same answer no
matter whether you take the raw data, their
logarithms or their square roots.
If you have data (raw or transformed) that
pass q-q plots tests, then Wilcoxon and t-test
should have much the same p-values
In this case, you can use the latter to get CI’s
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Lecture 10 Review: Inference for
Equality of Variances

SPSS uses what is called Levene’s test

From the SPSS Help file (slightly edited)

Levene Test

For each case, it computes the absolute
difference between the value of that case and
its cell mean and performs a t-test on those
absolute differences.
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Inference When the Variances Are
Not Equal


Generally, you should try to find a scale (log,
square root) for which the variances are
approximately equal.
If you cannot, then a small change is needed
in the confidence intervals and test statistics.
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
The Confidence Interval With
Unequal Variances
The CI for μ -μ
is given by
1
2
2
1
2
2
s s
X1 -X 2 ±t α/2 (n1 + n 2 -2)
+
n1 n 2
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
The t-test Statistic With Unequal
Variances
The t-statistic is defined by
t=
X1  X 2
2
1
2
2
s
s

n1 n 2
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The Aortic Stenosis Data: A Review

Let’s remember that the aortic stenosis data
is comparing healthy with sick kids

Aortic Valve Area (AVA) and Body Surface
Area (BSA) are noninvasive measures

The idea is to see if we can use these
measures to understand whether a kid is
stenotic, without having to do an invasive test

Or at least to be able to narrow down those
who need an invasive test
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The Aortic Stenosis Data: A Review

We will study the variable log(1 + AVA/BSA)

I’ll try to review much of what we have done
to this point by using these data

There was a kids with an enormous AVA: I’ll
delete this kid from all analyses

First comes the relative frequency histogram
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The Aortic Stenosis Data: A Review

First comes the relative frequency histogram

The height of each bar is the % of the
sample which lies in the interval for that bar
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The Aortic Stenosis Data: A Review:
what % have X > 1?
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The Aortic Stenosis Data: A Review

Next comes the boxplot to

Compare medians

Compare variability

Screen for massive outliers
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The Aortic Stenosis Data: A Review:
read off median, IQR, outliers
1.4
1.2
1.0
99
72
79
.8
.6
.4
.2
0.0
1
-.2
N=
70
55
Healthy
Stenoti
Health Status
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The Aortic Stenosis Data: A Review

Next comes the q-q plot to check for
approximate bell-shape

Why does it matter that the data are
approximately bell-shaped?

First: statistical inferences such as
confidence intervals are most accurate and
most powerful

Second: Probability calculations are most
accurate
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The Aortic Stenosis Data: A Review

For the healthy kids: what do you think?
Normal Q-Q Plot of Log(1 + (AVA to BSA Ratio))
1.4
1.2
Expected Normal Value
1.0
.8
.6
.4
.2
-.2
0.0
.2
.4
.6
.8
1.0
1.2
1.4
Observed Value
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The Aortic Stenosis Data: A Review

For the stenotic kids: what do you think?
Normal Q-Q Plot of Log(1 + (AVA to BSA Ratio))
1.0
.8
Expected Normal Value
.6
.4
.2
0.0
0.0
.2
.4
.6
.8
1.0
Observed Value
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The Aortic Stenosis Data: A Review

Next we do some simple summary statistics

Healthy: n = 70, mean = 0.84, median =
0.83, sd = 0.22, iqr = 0.31, se (of mean) =
0.026

Stenotic: n = 55, mean = 0.47, median =
0.46, sd = 0.17, iqr = 0.22, se (of mean) =
0.023

What does the IQR mean?
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The Aortic Stenosis Data: A Review

Healthy: n = 70, mean = 0.84, median =
0.83, sd = 0.22, iqr = 0.31, se (of mean) =
0.026

Degrees of freedom = 70 – 1 = 69

95% CI for population mean (from SPSS) =
0.79 to 0.89

Interpret this!
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The Aortic Stenosis Data: A Review

Healthy: n = 70, mean = 0.84, median =
0.83, sd = 0.22, iqr = 0.31, se (of mean) =
0.026

Estimate the probability that a healthy child
has a log(1+ASA/BSA) less that 0.46

Pr(X < 0.46): z-score = (0.46 – 0.84)/0.22
= -1.73

Table 1: Pr(X < 0.46) ~ 0.0418

Interpret this!
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The Aortic Stenosis Data: A Review

Note: for computing confidence intervals
for the population mean I use the
standard error

In computing probabilities about the
population I use the standard deviation

Which one is affected by the sample size?
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The Aortic Stenosis Data: A Review

Next I will compare the population means

We have concluded that the data are fairly
(although not exactly) normally distributed

We have also concluded that the variabilities
are not too awfully different, although the
stenotic kids appear to be less variable.

So, we will try out the t-test type inference
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The Aortic Stenosis Data: A Review

SPSS computes a confidence interval for the
difference in the population means
H0 :μ1 -μ 2 =0 vs Ha :μ1 -μ 2  0

We conclude the population means are
different if the confidence interval does not
include zero
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The Aortic Stenosis Data: A Review

SPSS
Group Statistics
Log(1 + (AVA
to BSA Ratio))
Health Status
Healthy
Stenotic
N
70
55
Mean
.8424
.4659
Std. Error
Std. Deviation
Mean
.2199 2.628E-02
.1734 2.338E-02
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The Aortic Stenosis Data: A Review

SPSS
Independent Samples Test
Levene's Tes t for
Equality of Variances
F
Log(1 + (AVA
to BSA Ratio))
Equal variances
ass umed
Equal variances
not as sumed
3.346
Sig.
.070
t-tes t for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
10.407
123
.000
.3765
3.618E-02
.3049
.4482
10.705
122.997
.000
.3765
3.518E-02
.3069
.4462
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The Aortic Stenosis Data: A Review

The 95% confidence interval is from 0.30 to
0.45

Interpret this!
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The Aortic Stenosis Data: A Review

The p-value is 0.000.

Interpret this! What would have happened if I
had done a 99% CI?
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The Aortic Stenosis Data: A Review

The p-value is 0.000.

Interpret this! What would have happened if I
had done a 99% CI?

It would not have covered zero, and I
would have rejected the null hypothesis
that the two population means are the
same.
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The Aortic Stenosis Data: A Review

Outliers can really mess up an analysis

We use the Wilcoxon rank sum test for this

There were no massive outliers, the data
were nearly normal, so we expect to get just
about the same p-value

It is 0.000, just as for the CI analysis
Test Statisticsa
Mann-Whitney U
Wilcoxon W
Z
Asymp. Sig. (2-tailed)
Log(1 + (AVA
to BSA Ratio))
305.500
1845.500
-8.055
.000
a. Grouping Variable: Health Status
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The Aortic Stenosis Data: A Review

Next we will use Levene’s test to compare
variability

SPSS has its version of Levene’s test, which is
a t-test on the absolute differences with the
sample mean: the p-value for this is 0.07

Weak evidence of differences in variability
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ANalysis Of VAriance




We now turn to making inferences when
there are 3 or more populations
This is classically called ANOVA
It is somewhat more formula dense than
what we have been used to.
Tests for normality are also somewhat more
complex
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33
ANOVA





The Analysis of Variance is often known as
ANOVA
We are going to consider its simplest form,
namely comparing 3 or more populations.
If there are two populations, we have
covered this in the first part of the course
Confidence intervals for the difference in two
population means
Wilcoxon rank sum test.
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34
ANOVA

Suppose we form three populations on the
basis of body mass index (BMI):

BMI < 22, 22 <= BMI < 28, BMI > 28

This forms 3 populations

We want to know whether the three
populations have the same mean caloric
intake, or if their food composition differs.
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ANOVA


Concho water snake data for males have four
separate subpopulations, depending on the
year class.
Are there differences in snout-to-vent lengths
among these populations?
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ANOVA

As in all analyses, we will combine graphical
analyses, summary statistics and formal
hypothesis tests and confidence intervals to
for a picture of what the data are telling us
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ANOVA



There is considerable controversy as to how
to compare 3 or more populations.
One possibility is to compare them 2 at a time
In the body mass example, there are 3 such
comparisons



Low BMI versus Middle BMI
Low BMI versis High BMI
Middle BMI versus High BMI
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ANOVA

In the Concho water snake example there are
6 such comparisons.

1 year olds versus 2 year olds

1 year olds versus 3 year olds
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1 year olds versus 4 year olds
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2 year olds versus 3 year olds
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2 year olds versus 4 year olds

3 year olds versus 4 year olds
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ANOVA


In general, if there are t populations, there
are t(t-1)/2 two-sample comparisons.
Special Note: t is a symbol used in the book
to denote the number of populations
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ANOVA




In general, if there are t populations, there
are t(t-1)/2 two-sample comparisons.
The controversy revolves around the concept
of Type I error
Specifically, if we do 6 different 95%
confidence intervals, what is the probability
that one or more of them do not include
the true mean?
Certainly, it is higher than 5%!
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41
ANOVA



If you do lots of 95% confidence intervals,
you’d expect by chance that about 5% will be
wrong
Thus, if you do 20 confidence intervals, you
expect 1 = 20 x 5% will not include the true
population parameter
This is a fact of life
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42
ANOVA



If you do lots of 95% confidence intervals,
you’d expect by chance that about will be
wrong
One school of thought is to stick to a few
major hypotheses (2-4 say), do 95% CI on
them, and label anything else as exploratory,
with possibly inflated Type I errors
I am of this school
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43
ANOVA



If you do lots of 95% confidence intervals,
you’d expect by chance that about 5% will be
wrong
Another school thinks that you should control
the chance of making any errors at all to be
5%
I am NOT of this school. Why not: am I just
crazy?
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44
ANOVA



Another school thinks that you should control
the chance of making any errors at all to be
5%
I am NOT of this school.
I worry about Type II error (power). To be
95% confident that every single one of my
conclusions is correct, I will not have much
power to detect meaningful changes or
differences.
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