Download Lecture 29 - Sampling Distribution Mean

Sampling Distribution
of a Sample Mean
Lecture 28
Section 8.4
Fri, Oct 28, 2005
Sampling Distribution of the
Sample Mean

Sampling Distribution of the Sample Mean–
The distribution of sample means over all
possible samples of the size n from that
population.
With or Without Replacement?



If the sample size is small in relation to the
population size (< 5%), then it does not matter
whether we sample with or without replacement.
The calculations are simpler if we sample with
replacement.
In any case, we are not going to worry about it.
Example



Suppose a population consists of the numbers
{1, 2, 3}.
If we select samples of size n = 3, find the
sampling distribution ofx.
Draw a tree diagram showing all possibilities.
The Tree Diagram
1
1
2
3
1
2
2
3
1
3
2
3
1
2
3
1
4/3
5/3
1
2
3
1
2
3
1
2
3
1
2
3
4/3
5/3
2
5/3
2
7/3
4/3
5/3
2
1
2
3
1
2
3
1
2
3
2
7/3
8/3
5/3
2
7/3
2
7/3
8/3
1
2
3
7/3
8/3
3
5/3
2
7/3
The Sampling Distribution

Of the 27 possible samples (equally likely), we
find that
1 sample has an average of 1.
 3 samples have an average of 4/3 = 1.333.
 6 samples have an average of 5/3 = 1.667.
 7 samples have an average of 2.
 6 samples have an average of 7/3 = 2.333.
 3 samples have an average of 8/3 = 2.667.
 1 sample has an average of 3.

The Sampling Distribution

The sampling distribution ofx is
x
1
1.333
P(X = x)
1/27
3/27
1.667
2
2.333
6/27
7/27
6/27
2.667
3
3/27
1/27
Probability Histogram
7/27
6/27
5/27
4/27
3/27
2/27
1/27
1
2
3
Probability Histogram
7/27
6/27
5/27
4/27
3/27
2/27
1/27
1
2
3
Experiment


We may simulate this on the TI-83 using the
function randInt.
The expression
randInt(1, 3, 3)
will select 3 random numbers from the set {1, 2,
3}, with replacement, and put them in a list.
Experiment


For example, randInt(1, 3, 3) gives
{3, 3, 2}
which has a sample mean of 2.667.
Generate a list of 100 such sample means, each
from a sample of size 3.
Experiment
50
40
30
20
10
1
2
3
Fit a Normal Curve
50
40
30
20
10
1
2
3
Observations


The distribution appears to be centered at .
The distribution appears to be approximately
normal (maybe).
Experiment


Now generate a list of 100 sample means, each
from a sample of size 30.
Draw the distribution.
Experiment
50
40
30
20
10
1
2
3
Fit a Normal Curve
50
40
30
20
10
1
2
3
The Central Limit Theorem


Begin with a population that has mean  and
standard deviation .
For sample size n, the sampling distribution of
the sample mean is approximately normal with
Mean  
Variance 
2
n
Standard deviation 

n
The Central Limit Theorem



The approximation gets better and better as the
sample size gets larger and larger.
For many populations, the distribution is almost
exactly normal when n  10.
For almost all populations, if n  30, then the
distribution is almost exactly normal.
The Central Limit Theorem


If the original population is exactly normal, then
the sampling distribution of the sample mean is
exactly normal for any sample size.
This is all summarized on pages 536 – 537.
Example

Suppose the heights of all adult males has
Mean 68 inches.
 Standard deviation 3.6 inches.

Example

The set of sample means from all possible
samples of size n = 4 will have
Mean 68 inches.
 Standard deviation 3.6/4 = 0.18 inches.


But the distribution will (probably) not be
normal.
Example

The set of sample means from all possible
samples of size n = 100 will have
Mean 68 inches.
 Standard deviation 3.6/100 = 0. 36 inches.


And the distribution will be normal.
Example

If we collect a sample of 100 adult males from
this population, what is the probability that their
average height will be at least 67 inches?
Example

The sampling distribution ofx
Is normal (because n 30),
 Has mean 68,
 Has standard deviation 0. 36.


Therefore, P(x > 67) is given by
normalcdf(67, E99, 68, 0.36),
which is 0.9973.
Let’s Do It!


Let’s Do It! 8.9, p. 541 – Probability of
Accepting the Shipment.
Let’s Do It! 8.10, p. 543 – Mean Grocery
Expenditure.
Estimating the Population Mean


Example 8.12, p. 543 – Estimating the
Population Mean Grocery Expenditure.
The sampling distribution ofx is approximately
normal with
x = $60.
 x = $35/100 = $3.50.


Based on the Empirical Rule, 95% of all samples
have a mean within $7.00 of $60, that is,
between $53 and $67.
Estimating the Population Mean




Suppose that we do not know the true
population mean for grocery expenditures, but
our sample mean is $58.
What is our best estimate of the true mean?
Point estimate: $58.
Interval estimate: Allow 2 standard deviations
either way: $58  $7.