Day 6 – Distribution of the Sample Mean
Population
Sample
μ = Mean
X = sample mean
 = SD
s = sample SD
o In many investigations the data of interest can take
on many possible values. Examples we will
consider are attachment loss and DMFS.
o With this type of data it is often of interest to
estimate the population mean, μ.
o A common estimator for μ is the sample mean, X .
o In this lecture we will focus on the sampling
distribution of X .
Distribution of the Sample Mean
 X is a random variable. Its value is
determined partly by which people are
randomly chosen to be in the sample.
Many possible samples, many possible X ’s
mean = 1.78
0
2
4
6
8
10
mean = 1.55
0
mean = 1.6
0
2
4
6
8
2
4
6
8
10
0
2
4
6
8
6
8
10
0
10
2
4
6
8
10
0
2
4
6
8
10
0
0
2
4
6
8
10
4
6
8
10
0
2
4
6
8
10
0
2
4
6
8
10
0
2
4
6
8
10
4
6
8
10
0
2
4
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10
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2
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0
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4
6
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10
mean = 1.45
0
mean = 1.59
0
2
mean = 1.7
mean = 1.38
mean = 1.61
0
2
mean = 1.73
mean = 1.44
mean = 1.66
mean = 1.64
10
2
mean = 1.6
mean = 1.67
mean = 1.62
mean = 1.7
0
4
mean = 1.56
mean = 1.53
0
2
mean = 1.45
2
4
6
8
10
mean = 1.72
0
2
4
6
8
We only see one!
 We will have a better idea of how good our
one estimate is if we have good knowledge of
how X behaves; that is, if we know the
probability distribution of X .
10
Central Limit Theorem
 An important result in probability theory tells
us that averages (like X ) all have a certain
type of probability distribution, the Normal
distribution.
Probability distribution of
no matter what!*
X
the Normal distribution
,
μ
 The Central Limit Theorem says that the
distribution of X will be Normal, no matter
what distribution the original data have.*
 The size of the sample needs to be reasonably
large for the Central Limit Theorem to hold.
* well, almost all the time anyways
Example: Mean attachment levels
 By computer simulation we randomly chose
samples of 100 patients from the mean attachmentlevel data.
 For each sample of 100, we computed the sample
mean.
 The histogram below presents the distribution of
1000 of such sample means.
means from samples of 100 patients
1.2
1.4
1.6
1.8
2.0
average mean attachment level (mm)
 Note the shape is similar to Normal distribution
More about distribution of
X
 The average value of X will be μ.

 The standard deviation of X is n .
o σ is the standard deviation in the
population.
o n is the number of people in the sample
o The standard deviation of X is called the
standard error of the mean, or SE (X ) .
o The standard error ↓ as n ↑
o The standard error ↑ as σ ↑
Central Limit Theorem Example
mean =15.28
sd =5.27
60
mean =15.2
sd =11.78
0
0
200
20
40
600
Frequency
1000
80
Distribution of 4893 current smokers
0
20
40
60
5
10
20
25
30
35
means from samples of size 5
80
mean =15.2
sd =1.68
0
0
20
20
40
40
60
60
mean =15.02
sd =2.59
100
80
Cigarettes/day
15
10
15
20
means from samples of size 20
25
10
12
14
16
18
20
means from samples of size 50
As size of samples increases,
 variability of means decreases
 distributions of means look more like Normal distributions
Properties of the Sample Mean (Summary)
1. Unbiased:
E (X )  
Expected value of the sample mean is the true
population mean
2. Standard Error:
SE ( X ) 

n
SE (X ) is the standard deviation of the sampling
distribution of X .
3. Approximate Normality:
By the Central Limit Theorem, for large n, the
distribution of X should be approximately
Normal.
In summary, 1- 3 say

2
For large n, X  N  , n

Example: Rosner Birthweight data
Rosner Bi rthwei ght Data (T abl e 6.2)
μ = 112 oz
 = 20.6 oz
mean = 112
200
190
180
170
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
std. dev = 20.6
birthweight (oz)
The histogram above shows the distribution of birthweights at a
Boston hospital. Estimate the probability that the mean
birthweight of the next 20 babies born will be greater than 120 oz.


 X  112 120  112 
P( X  120)  P

  PZ  1.74  0.0409
20
.
6
20
.
6


20
20 

Now, assuming that the birthweights follow a normal distribution
(which they don’t exactly but it does look close), calculate the
percentage of births that exceed 120 oz.
 X  112 120  112 
P( X  120)  P

  PZ  0.39   0.3483
20
.
6
20
.
6


Note that we could not do the second calculation the same way
with the smoking data as the original distribution is not Normal.
But we could do the first!
Law of Large Numbers
Remember that for large n, X  N
 , n.
2
 For any n, the distribution of X is centered around μ.
 The variation of X decreases as n increases.
Thus, as the sample size gets large, the sample mean, X ,
is forced to be closer and closer to the population mean μ.
 Larger sample sizes give better estimates of μ.
The same is true for the sample standard deviation, s.
 As the sample size increases, s should get closer to the
population standard deviation σ.
Standard Error versus Standard Deviation
 Standard Deviation describes the variability of
a population or a sample.
 Standard Error describes the variability of an
estimator that is usually a function of the whole
sample.
These terms are sometimes used interchangeably,
which is incorrect.
Confidence Intervals for the Mean
 If n is large enough, we can use the result that
X 
 n
≈ N(0,1) to a construct confidence interval
for μ.
 However, this would result in a formula that
requires one to use σ, which, in usual practice, will
not be known.
 We will estimate the population standard deviation,
σ, with s, the sample standard deviation.
 Substituting the random variable s for σ will alter
the distribution of the Z score slightly.
 The distribution of the statistic
T
X 
s n
is called a “t” distribution with n -1 degrees of
freedom and will be denoted by tn-1.
t Distribution
 The t distribution is similar in shape to the standard
Normal distribution, but more spread out. Putting s
into the statistic adds more variability.
 The percentiles of t distributions are greater than
the respective N(0,1) percentiles.
 t distributions with lower degrees of freedom
(which correspond to smaller sample sizes) are
more spread out, and thus have higher percentiles.
 t distributions with higher degrees of freedom are
more similar to the Normal distribution.
N(0,1) pdf
t(4) pdf
97.5 %ile of N(0,1)
97.5 %ile of t(4)
0.0
1.96
2.78
Confidence Interval for the Mean
If X is Normal or n is large, then
T
X 
s
n
has a t distribution with n-1 degrees of freedom.


X 

P  tn1, .975 
 tn1, .975   0.95
s n


where tn-1, .975 denotes the 97.5th percentile of a tn-1
distribution.
from this, some algebraic maneuvering gives

P X  tn1, .975  s
n    X  tn1, .975  s

n  0.95 .
Which says that we are “95% confident” that μ lies
in the interval
X  t
n1, .975
s
n , X  tn1, .975  s

n .
This 95% confidence interval is a random quantity
that will “cover” the true population mean 95% of the
time.
Example: Chewing gum study
Group A was comprised of n=25 children. The
sample mean of the change in DMFS was – 0.72.
The sample standard deviation, s, is 5.37.
The 97.5th percentile of the t24 distribution is t24,.975 =
2.06 (can look up in Table 4 in coursepack or use
Excel), thus the 95% confidence interval is
 0.72  2.06  5.37
25 ,
which gives the interval (-2.92, 1.48).
“The probability is 95% that true mean change in
DMFS for this treatment is contained in the interval
(-2.92, 1.48)”
Note: the true mean, , is not a random quantity. It is
the confidence interval that may vary.
See coursepack Figure 8.1 for a pictorial description
of this idea.
General formula: 100(1-α)% confidence interval for μ,
X  tn1,1 / 2  s
n to X  tn1,1 / 2  s
n,
where t n 1,1 / 2 is the 1-α/2th percentile of a tn-1 distribution.
Example:
Suppose n = 30, for a 95% confidence interval, α =
0.05. We use tn 1,1 / 2  t 29,0.975  2.05 , the 97.5th
percentile, in the formula for a 95% confidence
interval.
α is meant to indicate the “error” we are willing to live
with. That is, when estimating the mean with a 95%
confidence interval, we are allowing an α = 5% chance of
missing the true mean.
It is standard to use the 1-α/2th percentile because we want
to split the error evenly to either side of the interval.
Example: Find the 99% confidence interval for  for
the change in DMFS in gum group A.
Acceptable error in this case would be α=1%, so we
use the 100(1-.01/2)% = 99.5th percentile.
From table 4, t24,.995 = 2.80, thus the 99% CI is
 0.72  2.80  5.37
25 ,
which gives the interval (-3.73, 2.29).
Note: the 99% confidence interval is larger than the
95% confidence interval. It needs to be wider to have
a better chance of covering .