Confidence_Intervals.pptConfidence_Intervals.ppt

Statistical inference: CLT,
confidence intervals, p-
values

Statistical Inference
The process of making
guesses about the truth
from a sample.
Sample
(observation)
Make guesses
about the whole
population
Truth (not
observable)
N
x
N
i
i
2
1
2
)
( 





N
x
N
i


 1

Population
parameters
1
)
(
ˆ
2
1
2
2






n
X
x
s
n
n
i
i

n
x
X
n
i
n



 1
̂
Sample statistics
*hat notation ^ is often used to indicate
“estitmate”

Statistics vs. Parameters
 Sample Statistic – any summary measure calculated from
data; e.g., could be a mean, a difference in means or
proportions, an odds ratio, or a correlation coefficient

E.g., the mean vitamin D level in a sample of 100 men is 63 nmol/L
 E.g., the correlation coefficient between vitamin D and cognitive
function in the sample of 100 men is 0.15
 Population parameter – the true value/true effect in the
entire population of interest
 E.g., the true mean vitamin D in all middle-aged and older
European men is 62 nmol/L

E.g., the true correlation between vitamin D and cognitive function
in all middle-aged and older European men is 0.15

Examples of Sample Statistics:
Single population mean
Single population proportion
Difference in means (ttest)
Difference in proportions (Z-test)
Odds ratio/risk ratio
Correlation coefficient
Regression coefficient
…

Example 1: cognitive
function and vitamin D
 Hypothetical data loosely based on [1]; cross-
sectional study of 100 middle-aged and older
European men.
 Estimation: What is the average serum vitamin D
in middle-aged and older European men?
 Sample statistic: mean vitamin D levels
 Hypothesis testing: Are vitamin D levels and
cognitive function correlated?
 Sample statistic: correlation coefficient between vitamin
D and cognitive function, measured by the Digit Symbol
Substitution Test (DSST).
1. Lee DM, Tajar A, Ulubaev A, et al. Association between 25-hydroxyvitamin D levels and cognitive performance in
middle-aged and older European men. J Neurol Neurosurg Psychiatry. 2009 Jul;80(7):722-9.

Distribution of a trait: vitamin
D
Right-skewed!
Mean= 63 nmol/L
Standard deviation = 33 nmol/L

Distribution of a trait: DSST
Normally distributed
Mean = 28 points
Standard deviation = 10 points

Distribution of a statistic…
 Statistics follow distributions too…
 But the distribution of a statistic is a theoretical
construct.
 Statisticians ask a thought experiment: how much
would the value of the statistic fluctuate if one
could repeat a particular study over and over
again with different samples of the same size?
 By answering this question, statisticians are able
to pinpoint exactly how much uncertainty is
associated with a given statistic.

Distribution of a statistic
 Two approaches to determine the
distribution of a statistic:
 1. Computer simulation

Repeat the experiment over and over again virtually!

More intuitive; can directly observe the behavior of
statistics.
 2. Mathematical theory

Proofs and formulas!

More practical; use formulas to solve problems.

Example of computer
simulation…
 How many heads come up in 100 coin
tosses?
 Flip coins virtually

Flip a coin 100 times; count the number
of heads.

Repeat this over and over again a large
number of times (we’ll try 30,000 repeats!)

Plot the 30,000 results.

Coin tosses…
Conclusions:
We usually get
between 40 and 60
heads when we flip a
coin 100 times.
It’s extremely unlikely
that we will get 30
heads or 70 heads
(didn’t happen in
30,000 experiments!).

Distribution of the sample mean,
computer simulation…
 1. Specify the underlying distribution of vitamin
D in all European men aged 40 to 79.

Right-skewed

Standard deviation = 33 nmol/L

True mean = 62 nmol/L (this is arbitrary; does not
affect the distribution)
 2. Select a random sample of 100 virtual men
from the population.
 3. Calculate the mean vitamin D for the sample.
 4. Repeat steps (2) and (3) a large number of
times (say 1000 times).
 5. Explore the distribution of the 1000 means.

Distribution of mean vitamin
D (a sample statistic)
Normally distributed! Surprise!
Mean= 62 nmol/L (the true
mean)
Standard deviation = 3.3 nmol/L

 Normally distributed (even though
the trait is right-skewed!)
 Mean = true mean
 Standard deviation = 3.3 nmol/L
 The standard deviation of a statistic is
called a standard error
 The standard error of a mean =
Distribution of mean vitamin
D (a sample statistic)
n
s

If I increase the sample size
to n=400…
Standard error = 1.7 nmol/L
7
.
1
400
33


n
s

If I increase the variability of
vitamin D (the trait) to SD=40…
Standard error = 4.0 nmol/L
0
.
4
100
40


n
s

Mathematical Theory…
The Central Limit Theorem!
If all possible random samples, each of size n, are
taken from any population with a mean  and a
standard deviation , the sampling distribution of
the sample means (averages) will:

 
x
1. have mean:
n
x

 
2. have standard deviation:
3. be approximately normally distributed regardless of the shape
of the parent population (normality improves with larger n). It all
comes back to Z!

Symbol Check
x
 The mean of the sample means.
x
 The standard deviation of the sample means. Also
called “the standard error of the mean.”

Mathematical Proof
(optional!)
If X is a random variable from any distribution with known
mean, E(x), and variance, Var(x), then the expected value
and variance of the average of n observations of X is:
)
(
)
(
)
(
)
(
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( 1
1
x
E
n
x
nE
n
x
E
n
x
E
X
E
n
i
n
i
i
n 




 

n
x
Var
n
x
nVar
n
x
Var
n
x
Var
X
Var
n
i
n
i
i
n
)
(
)
(
)
(
)
(
)
( 2
2
1
1





 


Computer simulation of the CLT
1. Pick any probability distribution and specify a mean and
standard deviation.
2. Tell the computer to randomly generate 1000 observations
from that probability distributions
E.g., the computer is more likely to spit out values with high
probabilities
3. Plot the “observed” values in a histogram.
4. Next, tell the computer to randomly generate 1000 averages-
of-2 (randomly pick 2 and take their average) from that
probability distribution. Plot “observed” averages in
histograms.
5. Repeat for averages-of-10, and averages-of-100.

Uniform on [0,1]: average of 1
(original distribution)

~Exp(1): average of 1

~Bin(40, .05): average of 1

~Bin(40, .05): 1000 averages of 2

~Bin(40, .05): 1000 averages of
5

~Bin(40, .05): 1000 averages of 100

The Central Limit Theorem:
If all possible random samples, each of size n, are
taken from any population with a mean  and a
standard deviation , the sampling distribution of
the sample means (averages) will:

 
x
1. have mean:
n
x

 
2. have standard deviation:
3. be approximately normally distributed regardless of the shape
of the parent population (normality improves with larger n)

Central Limit Theorem caveats
for small samples:
 For small samples:
 The sample standard deviation is an imprecise estimate
of the true standard deviation (σ); this imprecision
changes the distribution to a T-distribution.

A t-distribution approaches a normal distribution for large n
(100), but has fatter tails for small n (<100)
 If the underlying distribution is non-normal, the
distribution of the means may be non-normal.
More on T-distributions next week!!

Summary: Single population
mean (large n)
 Hypothesis test:
 Confidence Interval
n
s
Z
mean
null
mean
observed 

)
(
*
Z
mean
observed
interval
confidence /2
n
s




Single population mean
(small n, normally
distributed trait)
 Hypothesis test:
 Confidence Interval
n
s
Tn
mean
null
mean
observed
1



)
(
*
T
mean
observed
interval
confidence /2
,
1
n
s
n 




 1. Specify the true correlation coefficient
 Correlation coefficient = 0.15
 2. Select a random sample of 100 virtual
men from the population.
 3. Calculate the correlation coefficient for
the sample.
 4. Repeat steps (2) and (3) 15,000 times
 5. Explore the distribution of the 15,000
correlation coefficients.
Distribution of a correlation
coefficient?? Computer
simulation…

coefficient…
Normally distributed!
Mean = 0.15 (true correlation)
Standard error = 0.10

coefficient in general…
 1. Shape of the distribution

Normally distributed for large samples

T-distribution for small samples (n<100)
 2. Mean = true correlation
coefficient (r)
 3. Standard error 
n
r2
1

Many statistics follow
normal (or t-distributions)…
 Means/difference in means
 T-distribution for small samples
 Proportions/difference in proportions
 Regression coefficients
 T-distribution for small samples
 Natural log of the odds ratio

Estimation (confidence
intervals)…
 What is a good estimate for the true
mean vitamin D in the population
(the population parameter)?
 63 nmol/L +/- margin of error

95% confidence interval
 Goal: capture the true effect (e.g., the
true mean) most of the time.
 A 95% confidence interval should
include the true effect about 95% of
the time.
 A 99% confidence interval should
include the true effect about 99% of
the time.

Mean Mean + 2 Std error =68.6
Mean - 2 Std error=55.4
Recall: 68-95-99.7 rule for normal distributions! These is a 95%
chance that the sample mean will fall within two standard errors of
the true mean= 62 +/- 2*3.3 = 55.4 nmol/L to 68.6 nmol/L
To be precise, 95%
of observations fall
between Z=-1.96
and Z= +1.96 (so the
“2” is a rounded
number)…

 There is a 95% chance that the sample
mean is between 55.4 nmol/L and 68.6
nmol/L
 For every sample mean in this range,
sample mean +/- 2 standard errors will
include the true mean:

For example, if the sample mean is 68.6 nmol/L:

95% CI = 68.6 +/- 6.6 = 62.0 to 75.2

This interval just hits the true mean, 62.0.

 Thus, for normally distributed statistics,
the formula for the 95% confidence interval
is:
 sample statistic  2 x (standard error)
 Examples:
 95% CI for mean vitamin D:

63 nmol/L  2 x (3.3) = 56.4 – 69.6 nmol/L
 95% CI for the correlation coefficient:

0.15  2 x (0.1) = -.05 – .35

Simulation of 20 studies of
100 men…
95% confidence intervals
for the mean vitamin D
for each of the simulated
studies.
Only 1 confidence
interval missed the true
mean.
Vertical line indicates the true mean (62)

Confidence Intervals give:
*A plausible range of values for a
population parameter.
*The precision of an estimate.(When
sampling variability is high, the confidence
interval will be wide to reflect the
uncertainty of the observation.)
*Statistical significance (if the 95% CI does
not cross the null value, it is significant
at .05)

Confidence Intervals
point estimate  (measure of how confident we
want to be)  (standard error)
The value of the statistic in my sample
(eg., mean, odds ratio, etc.)
From a Z table or a T table, depending
on the sampling distribution of the
statistic.
Standard error of the statistic.

Common “Z” levels of confidenc
 Commonly used confidence levels are
90%, 95%, and 99%
Confidence
Level
Z value
1.28
1.645
1.96
2.33
2.58
3.08
3.27
80%
90%
95%
98%
99%
99.8%
99.9%

99% confidence intervals…
 99% CI for mean vitamin D:

63 nmol/L  2.6 x (3.3) = 54.4 – 71.6 nmol/L
 99% CI for the correlation coefficient:

0.15  2.6 x (0.1) = -.11 – .41

Testing Hypotheses
 1. Is the mean vitamin D in middle-
aged and older European men lower
than 100 nmol/L (the “desirable”
level)?
 2. Is cognitive function correlated
with vitamin D?

Is the mean vitamin D
different than 100?
 Start by assuming that the mean = 100
 This is the “null hypothesis”
 This is usually the “straw man” that we
want to shoot down
 Determine the distribution of statistics
assuming that the null is true…

Computer simulation
(10,000 repeats)…
This is called the null
distribution!
Std error = 3.3
Mean = 100

Compare the null
distribution to the observed
value…
What’s the
probability of
seeing a sample
mean of 63 nmol/L
if the true mean is
100 nmol/L?
It didn’t happen in
10,000 simulated
studies. So the
probability is less
than 1/10,000

Compare the null
distribution to the observed
value…
This is the p-value!
P-value < 1/10,000

Calculating the p-value with
a formula…
Because we know how normal curves work, we can exactly calculate the
probability of seeing an average of 63 nmol/L if the true average weight is 100
(i.e., if our null hypothesis is true):
2
.
11
3
.
3
100
63



Z
Z= 11.2, P-value << .0001

The P-value
P-value is the probability that we would have seen our
data (or something more unexpected) just by chance if
the null hypothesis (null value) is true.
Small p-values mean the null value is unlikely given
our data.
Our data are so unlikely given the null hypothesis
(<<1/10,000) that I’m going to reject the null
hypothesis! (Don’t want to reject our data!)

P-value<.0001 means:
The probability of seeing what you saw or something
more extreme if the null hypothesis is true (due to
chance)<.0001
P(empirical data/null hypothesis) <.0001

The P-value
 By convention, p-values of <.05 are often
accepted as “statistically significant” in the
medical literature; but this is an arbitrary
cut-off.
 A cut-off of p<.05 means that in about 5 of
100 experiments, a result would appear
significant just by chance (“Type I error”).

Summary: Hypothesis
Testing
The Steps:
1. Define your hypotheses (null, alternative)
2. Specify your null distribution
3. Do an experiment
4. Calculate the p-value of what you observed
5. Reject or fail to reject (~accept) the null
hypothesis

Hypothesis Testing
The Steps:
1. Define your hypotheses (null, alternative)
 The null hypothesis is the “straw man” that we are trying to shoot down.
 Null here: “mean vitamin D level = 100 nmol/L”
 Alternative here: “mean vit D < 100 nmol/L” (one-sided)
2. Specify your sampling distribution (under the null)
 If we repeated this experiment many, many times, the mean vitamin D
would be normally distributed around 100 nmol/L with a standard error
of 3.3 3
.
3
100
33 
3. Do a single experiment (observed sample mean = 63 nmol/L)
4. Calculate the p-value of what you observed (p<.0001)
5. Reject or fail to reject the null hypothesis (reject)

 Confidence intervals give the same
information (and more) than
hypothesis tests…

Duality with hypothesis
tests.
Null value
Null hypothesis: Average vitamin D is 100 nmol/L
Alternative hypothesis: Average vitamin D is not 100
nmol/L (two-sided)
P-value < .05
50 60 70 80 90 100

Duality with hypothesis
tests.
Null value
Null hypothesis: Average vitamin D is 100 nmol/L
Alternative hypothesis: Average vitamin D is not 100
nmol/L (two-sided)
P-value < .01
50 60 70 80 90 100

2. Is cognitive function
correlated with vitamin D?
 Null hypothesis: r = 0
 Alternative hypothesis: r  0
 Two-sided hypothesis
 Doesn’t assume that the correlation will
be positive or negative.

Computer simulation
(15,000 repeats)…
Null distribution:
Std error = 0.1
Mean = 0

What’s the probability of our
data?
Even when the true
correlation is 0, we get
correlations as big as 0.15
or bigger 7% of the time.

data?
This is a two-sided hypothesis
test, so “more extreme” includes
as big or bigger negative
correlations (<-0.15).
P-value = 7% + 7% = 14%

data?
Our results could have
happened purely due to a
fluke of chance!

Formal hypothesis test
 1. Null hypothesis: r=0
 Alternative: r  0 (two-sided)
 2. Determine the null distribution


Standard error = 0.1
 3. Collect Data, r=0.15
 4. Calculate the p-value for the data:
 Z =
 5. Reject or fail to reject the null (fail to reject)
5
.
1
1
.
0
15
.
0

 Z of 1.5 corresponds to a
two-sided p-value of 14%

Or use confidence interval to
gauge statistical
significance…
 95% CI = -0.05 to 0.35
 Thus, 0 (the null value) is a plausible
value!
 P>.05

Example 2: HIV vaccine trial
 Thai HIV vaccine trial (2009)
 8197 randomized to vaccine
 8198 randomized to placebo
 Generated a lot of public discussion about
p-values!

Source: BBC news, http://news.bbc.co.uk/go/pr/fr/-/2/hi/health/8272113.stm
51/8197 vs. 75/8198
=23 excess infections in the
placebo group.
=2.8 fewer infections per 1000
people vaccinated

Null hypothesis
 Null hypothesis: infection rate is the
same in the two groups
 Alternative hypothesis: infection rates
differ

Computer simulation
assuming the null (15,000
repeats)…
Normally distributed,
standard error = 11.1

Computer simulation
repeats)…
If the vaccine is
completely
ineffective, we
could still get 23
excess infections
just by chance.
Probability of 23
or more excess
infections = 0.04

How to interpret p=.04…
 P(data/null) = .04
 P(null/data) .04
 P(null/data)  22%
*estimated using Bayes’ Rule (and
prior data on the vaccine)
*Gilbert PB, Berger JO, Stablein D, Becker S, Essex M, Hammer SM, Kim JH, DeGruttola VG. Statistical
interpretation of the RV144 HIV vaccine efficacy trial in Thailand: a case study for statistical issues in
efficacy trials. J Infect Dis 2011; 203: 969-975.

Alternative analysis of the
data (“intention to treat”)…
 56/8202 (6.8 per 1000) infections in
the vaccine group versus 76/8200 (9.3
per 1000)

Computer simulation
repeats)…
Probability of 20
or more excess
infections = 0.08
P=.08 is only slightly
different than p=.04!

Confidence intervals…
 95% CI (analysis 1): .0014 to .0055
 95% CI (analysis 2): -.0003 to .0051
 The plausible ranges are nearly
identical!

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