Download Lecture 28 - Sampling Distribution Mean 2

Sampling
Distribution of a
Sample Mean
Lecture 28
Section 8.4
Wed, Mar 5, 2008
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 if n  30, with

Mean of x   x  
Variance of x   x 
2
2
n
Standard deviation of x   x 

n
The Central Limit Theorem
The approximation gets better and better
as the sample size gets larger and larger.
 That is, the sampling distribution “morphs”
from the original distribution to the normal
distribution.

The Central Limit Theorem
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
Also, 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 a population consists of the
numbers {6, 12, 18}.
 Using samples of size n = 1, 2, 3, 4, and 5,
find the sampling distribution ofx.
 Draw a tree diagram showing all
possibilities.

The Tree Diagram (n = 1)

n=1
6
mean = 6
12
mean = 12
18
mean = 18
The Sampling Distribution (n = 1)


The sampling distribution ofx is
The parameters are

= 12
 2 = 24
x
P(x)
6
1/3
12
1/3
18
1/3
The Sampling Distribution (n = 1)

The shape of the distribution:
density
1/3
6
8
10
12
14
16
18
mean
The Sampling Distribution (n = 1)

The shape of the distribution:
density
1/3
6
8
10
12
14
16
18
mean
The Tree Diagram (n =mean2)
6
12
18
6
6
12
9
18
12
6
9
12
12
18
15
6
12
12
15
8
18
The Sampling Distribution (n = 2)


The sampling distribution ofx is
The parameters are

= 12
 2 = 12
x
P( x)
6
1/9
9
2/9
12
3/9
15
2/9
18
1/9
The Sampling Distribution (n = 2)

The shape of the distribution:
density
3/9
2/9
1/9
6
8
10
12
14
16
18
mean
The Sampling Distribution (n = 2)

The shape of the distribution:
density
3/9
2/9
1/9
6
8
10
12
14
16
18
mean
The Tree Diagram (n = 3)
6
6
12
18
6
12
12
18
6
18
12
18
mean
6
12
18
6
12
18
6
12
18
6
12
18
6
12
18
8
10
12
10
12
14
8
10
12
6
12
18
6
12
18
6
12
18
12
14
16
10
12
14
12
14
16
6
12
18
14
16
18
6
8
10
10
12
14
The Sampling Distribution (n = 3)


The sampling distribution ofx is
The parameters are

= 12
 2 = 8
x
P(x)
6
1/27
8
3/27
10
6/27
12
7/27
14
6/27
16
3/27
18
1/27
The Sampling Distribution (n = 3)

The shape of the distribution:
density
9/27
6/27
3/27
6
8
10
12
14
16
18
mean
The Sampling Distribution (n = 3)

The shape of the distribution:
density
9/27
6/27
3/27
6
8
10
12
14
16
18
mean
The Sampling Distribution (n = 4)


The sampling distribution ofx is
The parameters are

= 12
 2 = 6
x
P(x)
6
1/81
7.5
4/18
9
10/81
10.5
16/81
12
19/81
13.5
16/81
15
10/81
16.5
4/81
18
1/81
The Sampling Distribution (n = 4)

The shape of the distribution:
density
20/81
16/81
12/81
8/81
4/18
6
8
10
12
14
16
18
mean
The Sampling Distribution (n = 4)

The shape of the distribution:
density
Normal curve
20/81
16/81
12/81
8/81
4/18
6
8
10
12
14
16
18
mean
The Sampling Distribution (n = 5)


The sampling distribution ofx is
The parameters are

= 12
 2 = 4.8
x
P(x)
P(x)
6
1/243
0.004
7.2
5/243
0.021
8.4
15/243
0.062
9.6
30/243
0.123
10.8
45/243
0.185
12
51/243
0.210
13.2
45/243
0.185
14.4
30/243
0.123
15.6
15/243
0.062
16.8
5/243
0.021
18
1/243
0.004
The Sampling Distribution (n = 5)

The shape of the distribution:
density
50/243
40/243
30/243
20/243
10/243
6
8
10
12
14
16
18
mean
The Sampling Distribution (n = 5)

The shape of the distribution:
density
Normal curve
50/243
40/243
30/243
20/243
10/243
6
8
10
12
14
16
18
mean
Bag A vs. Bag B
There are two bags, Bag A and Bag B.
 Each bag contains 20,000 vouchers with
values from $10 to $60.
 Their distributions are shown on the
following slide.

Bag A vs. Bag B
= 1000 vouchers
Bag A
10 20 30 40 50 60
= 1000 vouchers
Bag B
10 20 30 40 50 60
Bag A vs. Bag B

Use the TI-83 to compute the mean and
standard deviation of each population
(Bag A and Bag B).
Bag A vs. Bag B

The Bag A population:

= 23.5
  = 14.24

The Bag B population:

= 46.5
  = 14.24
Bag A vs. Bag B

The hypotheses:
 H0:
The bag is Bag A.
 H1: The bag is Bag B.
Suppose that we sample 100 vouchers
(with replacement).
 Decision rule: Reject H0 if the average of
the 100 vouchers is more than 35.

Bag A vs. Bag B
Find the sampling distribution ofx if H0 is
true.
 Find the sampling distribution ofx if H1 is
true.

The Two Sampling Distributions
H0
15
20
25
30
35
40
45
50
55
15
20
25
30
35
40
45
50
55
H1
The Two Sampling Distributions
N(23.5, 1.424)
H0
15
20
25
30
35
40
45
50
55
45
50
55
N(46.5, 1.424)
H1
15
20
25
30
35
40
The Two Sampling Distributions
H0
15
20
25
30
35
40
45
50
55
15
20
25
30
35
40
45
50
55
H1
Bag A vs. Bag B
What is ?
 What is ?
 How reliable is this test?

Example
Suppose a brand of light bulb has a mean
life of 750 hours with a standard deviation
of 120 hours.
 What is the probability that 36 of these
light bulbs would last a total of at least
26000 hours?
