Download Distribution of the Sample Mean - MATH 130, Elements of Statistics I

Distribution of the Sample Mean
MATH 130, Elements of Statistics I
J. Robert Buchanan
Department of Mathematics
Fall 2015
Experiment (1 of 3)
Suppose we have the following “population”:
4
3
1
4
8
5
0
7
1
6
3
5
2
8
7
7
3
9
9
7
4
3
3
6
9
0
2
4
1
0
5
1
0
7
7
3
4
2
2
0
If we take random samples of size 10, we might obtain:
S1 = {2, 8, 3, 1, 5, 7, 4, 1, 3, 9}
S2 = {8, 7, 6, 7, 4, 4, 0, 2, 2, 0}
S3 = {6, 0, 0, 7, 4, 0, 7, 0, 6, 1}
Experiment (2 of 3)
The samples were:
S1 = {2, 8, 3, 1, 5, 7, 4, 1, 3, 9}
S2 = {8, 7, 6, 7, 4, 4, 0, 2, 2, 0}
S3 = {6, 0, 0, 7, 4, 0, 7, 0, 6, 1}
and there corresponding means were:
x 1 = 4.3
x 2 = 4.0
x 3 = 3.1
Experiment (2 of 3)
The samples were:
S1 = {2, 8, 3, 1, 5, 7, 4, 1, 3, 9}
S2 = {8, 7, 6, 7, 4, 4, 0, 2, 2, 0}
S3 = {6, 0, 0, 7, 4, 0, 7, 0, 6, 1}
and there corresponding means were:
x 1 = 4.3
x 2 = 4.0
x 3 = 3.1
Observation: since the samples are chosen randomly the
mean calculated from the sample is a random variable.
What is the distribution of this random variable?
Experiment (3 of 3)
One way to determine the distribution of the sample mean for
samples of size 10 from this population of size 40, would be to
list all the possible samples; however, since
40 C10
= 847, 660, 520
this is impractical.
Perhaps we can get an idea from a smaller population.
Example (1 of 3)
Suppose we have a population {1, 2, 3, 4, 5, 6}, we can list all
the samples of size n = 2 with their corresponding sample
means.
Sample
{1, 2}
{1, 5}
{2, 4}
{3, 4}
{4, 5}
x
1.5
3.0
3.0
3.5
4.5
Sample
{1, 3}
{1, 6}
{2, 5}
{3, 5}
{4, 6}
x
2.0
3.5
3.5
4.0
5.0
Sample
{1, 4}
{2, 3}
{2, 6}
{3, 6}
{5, 6}
x
2.5
2.5
4.0
4.5
5.5
Example (2 of 3)
The frequency distribution of the sample means is below.
x
1.5
3.0
4.5
Freq.
1
2
2
x
2.0
3.5
5.0
Freq.
1
3
1
x
2.5
4.0
5.5
Freq.
2
2
1
Question: what is the probability of a random sample of size
n = 2 having a sample mean 2.5 ≤ x ≤ 4.0?
Example (3 of 3)
Rel. Freq.
x
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
Relative Frequency
0.0667
0.0667
0.1333
0.1333
0.2000
0.1333
0.1333
0.0667
0.0667
0.20
0.15
0.10
0.05
1
2
3
4
5
6
x
Example
Suppose a population of 100 people yields the following ages:
14
6
59
64
39
12
8
34
27
4
16
18
17
33
56
60
65
73
53
43
26
42
60
87
58
42
82
21
35
64
58
53
36
66
63
66
39
62
58
49
31
27
39
35
12
28
28
20
3
54
41
41
63
39
37
23
79
43
28
17
12
45
52
10
11
32
32
23
86
61
50
27
19
15
3
51
5
36
83
39
35
44
59
30
31
69
40
16
40
66
15
55
32
4
43
41
23
46
61
30
Question: what are the population mean and standard
deviation?
Population Age Distribution
Age distribution for the population:
20
15
10
5
20
40
µ = 39.3
60
σ = 20.9
80
100 Samples of size 30
If we create 100 random samples of size n = 30 and calculate
the sample means we get:
36.4
35.5
36.8
36.7
36.1
36.9
37.4
36.3
34.6
35.9
37.5
36.6
35.7
35.6
36.5
36.2
35.5
36.6
35.7
34.8
36.4
37.2
37.3
37.5
37.8
37.1
36.8
35.5
36.3
36.4
38.2
37.3
36.7
38.1
36.0
36.5
37.5
37.3
36.4
35.8
37.5
37.4
37.5
36.6
37.4
37.3
37.0
36.6
36.5
36.6
36.5
35.7
36.3
36.3
36.
35.4
37.1
35.6
34.8
35.4
36.6
36.6
36.1
38.0
36.8
36.3
35.9
35.7
35.7
35.2
35.7
35.7
36.4
36.4
36.1
38.2
37.4
34.8
34.8
35.5
35.8
35.4
37.
35.5
36.5
38.1
36.6
34.9
34.5
36.1
36.3
36.4
37.6
36.7
35.9
36.7
37.5
35.4
35.5
36.7
Distribution of the Sample Means
Distribution of x:
15
10
5
35
36
µ = 36.4
37
σ = 0.9
38
Increasing the Sample Size n
As the sample size increases the mean of the sample means
approaches the population mean.
x
38
36
34
32
30
n
20
40
60
80
Law of Large Numbers
Theorem (Law of Large Numbers)
As additional observations are added to a sample, the sample
mean x approaches the population mean µ.
What Happens to the Standard Deviation?
As the sample size increases the standard deviation of the
sample means decreases, but does not disappear.
ӏx
2.5
2.0
1.5
1.0
0.5
n
20
40
60
80
Result
Theorem
Suppose that a simple random sample of size n is drawn from a
large population with a mean µ and a standard deviation σ. The
sampling distribution of x will have mean µx and standard
deviation
σ
σx = √ .
n
The standard deviation of the sampling distribution of x is
called the standard error of the mean and is denoted σx .
Example
For the population of 100 ages µ = 39.3 and σ = 20.9.
1. What is the standard error of the mean for samples of size
n = 20?
2. What is the standard error of the mean for samples of size
n = 25?
3. What is the standard error of the mean for samples of size
n = 30?
Example
For the population of 100 ages µ = 39.3 and σ = 20.9.
1. What is the standard error of the mean for samples of size
n = 20?
σ
20.9
σx = √ = √ = 4.7
n
20
2. What is the standard error of the mean for samples of size
n = 25?
3. What is the standard error of the mean for samples of size
n = 30?
Example
For the population of 100 ages µ = 39.3 and σ = 20.9.
1. What is the standard error of the mean for samples of size
n = 20?
σ
20.9
σx = √ = √ = 4.7
n
20
2. What is the standard error of the mean for samples of size
n = 25?
20.9
σx = √ = 4.2
25
3. What is the standard error of the mean for samples of size
n = 30?
Example
For the population of 100 ages µ = 39.3 and σ = 20.9.
1. What is the standard error of the mean for samples of size
n = 20?
σ
20.9
σx = √ = √ = 4.7
n
20
2. What is the standard error of the mean for samples of size
n = 25?
20.9
σx = √ = 4.2
25
3. What is the standard error of the mean for samples of size
n = 30?
20.9
σx = √ = 3.8
30
Normal Distributions
Theorem
If a random variable X is normally distributed, the distribution of
the sample mean x is normally distributed.
PDF
0.06
Sample
0.05
0.04
0.03
0.02
Population
0.01
20
40
60
80
X
Central Limit Theorem
Theorem (Central Limit Theorem)
Regardless of the shape of the population distribution, the
sample distribution of x becomes approximately normal as the
sample size increases.
Central Limit Theorem
Theorem (Central Limit Theorem)
Regardless of the shape of the population distribution, the
sample distribution of x becomes approximately normal as the
sample size increases.
Remark: if the sample size is greater than 30, generally the
distribution of x can be treated as normal.
Illustration
The random variable X is not normally distributed, but the
means of samples of size 30 randomly sampled from this
distribution are nearly normally distributed.
60
1.0
50
0.8
40
0.6
30
0.4
20
0.2
10
1
2
3
4
5
0.6
0.8
1
1.2
1.4
1.6
Application
The average speed of winds in Honolulu, HI is µ = 11.3 mph.
The standard deviation of the wind speeds is σ = 3.5 mph.
Assuming that the wind speeds are normally distributed,
1. find the probability that a single wind speed reading will
exceed 13.9 mph,
2. describe the sampling distribution of the means for
samples of size n = 9,
3. find the probability that the mean of 9 wind speed readings
will exceed 13.9 mph.
Solution
Probability that a single wind speed reading will exceed 13.9
mph,
P(X > 13.9) = P(Z > 0.74) = 1 − 0.7704 = 0.2296.
Solution
Probability that a single wind speed reading will exceed 13.9
mph,
P(X > 13.9) = P(Z > 0.74) = 1 − 0.7704 = 0.2296.
The sampling distribution of means for samples of size 9 is
normally distributed with a mean of µx = 11.3 mph and the
standard deviation of
3.5
σx = √ = 1.2 mph.
9
Solution
Probability that a single wind speed reading will exceed 13.9
mph,
P(X > 13.9) = P(Z > 0.74) = 1 − 0.7704 = 0.2296.
The sampling distribution of means for samples of size 9 is
normally distributed with a mean of µx = 11.3 mph and the
standard deviation of
3.5
σx = √ = 1.2 mph.
9
Probability that the mean of 9 wind speed readings will exceed
13.9 mph,
P(X > 13.9) = P(Z > 2.23) = 1 − 0.9871 = 0.0129.
Application
The average salary for a registered nurse is $45,900 and the
standard deviation in salaries is $7790. Suppose a sample of
salaries of 50 registered nurses is collected. What is the
probability that the sample mean is between $43,000 and
$47,000?
Solution
The mean of samples of size 50 is µx = 45, 900 and the
standard error of the mean of samples of size 50 is
7790
= 1102.
σx = √
50
P(43, 000 < X < 47, 000) = P(−2.63 < Z < 1.00)
= 0.8413 − 0.0043
= 0.8370