Download sample mean - Columbus State University

Sampling Distributions
and the Central Limit
Theorem
Statistical Methods
The study of statistics has two major branches:
descriptive statistics and inferential statistics.
Statistics
Descriptive
Statistics
Inferential
Statistics
Inferential Statistics
1.
Involves:
 Estimation
 Hypothesis Testing
2.
Purpose
 Make Decisions
about Population
Characteristics
Population?
Inference Process
Estimates
& tests
Population
,  2, p
Sample
statistic
X , S X2 , pˆ
Sample to
estimate ,  2, p
Estimators
1. Certain random variables are used to estimate a
Population Parameter:
Estimate Population
Parameter...
with the Sample
Statistic
Mean

X
Proportion
p
p^
Variance
2
s2
Sampling Error
For instance, the sampling error of the mean is given by
X E
where E is the error in the estimate.
Recall that in real applications the value of the mean 
is not known. The purpose of this course is to get an
estimate of E in order to have an idea where  is.
The Problem is that
a sample statistic changes from sample to sample leading to
different estimates. For instance, X , the sample mean, has
different values for different samples of the same size, say, n
Sample 1
Sample 4
x4
x1
Sample 5
x5
Sample 3
x3
Sample k
Sample 2
xk
x2
The fact that the sample mean changes from sample to sample,
makes X a random variable.
The Random Variable X
Like any other random variable, the sample mean X , has its
own probability distribution. This distribution is called the
sampling distribution of the sample mean.
The sampling distribution of the sample mean consists of
the values of the sample means, x 1 , x 2 , x 3 , x 4 , x 5 ,
,xk ,
of all the samples of the same size, say, n.
Similar arguments apply to the distribution of values ( sampling
distribution) of the sample proportion, and the sample
variance.
Sampling Distribution
1. Is a theoretical probability distribution of a
random variable.
2. The random variable is a Sample Statistic
 Sample Mean, Sample Proportion, Sample Variance
3. Results from drawing all possible samples of a
Fixed Size
4. For example, the list of all possible [ X, P( X ) ]
pairs is the Sampling Distribution of the Sample
Mean
The Distributions We Need
 Sampling distribution of the sample mean
 Sampling distribution of the sample proportion
 The Central Limit Theorem (CLT) will help us
understand what these distributions look like.
 The sampling distribution of the sample variance
requires a different distribution and we will get to
it if time permits.
Sampling
Distribution of the
Sample Means
http://onlinestatbook.com/stat_sim/sampling_dist/index.html
Distribution of the Sample Mean
Step 1: Obtain a simple random sample of size n.
Step 2: Compute the sample mean.
Step 3: Assuming we are sampling from a finite
population, repeat Steps 1 and 2 until all simple
random samples of size n have been obtained.
Step 4: Construct a relative frequency distribution
of the sample means. This distribution represents
the sampling distribution of the sample mean.
Distribution of the Sample Mean
 The sampling distribution of the sample means
has its own mean and standard deviation. They
are denoted by
 and
μx  “Mean of the sample means”
 x  “Standard deviation of the
sample means”
 and what we know about them is the following:
Properties of the Distribution of X
1. The mean of the sample means,  , is equal to the population
X
mean. That is,
X  
2. The standard deviation of the sample means,  X , is equal to
the population standard deviation, , divided by the square
root of n.
X 

n
(if sampling with replacement)
 X is also called the standard error of the mean.
Properties of the Distribution of X
3. When sampling without replacement, the standard deviation of
the sample means,  X , is equal to the population standard
deviation,  , divided by the square root of n times a correction
factor that takes into account the population size. That is,
X 

n
where N is the population size.
N n
N 1
The Central Limit Theorem
If samples of size n  30 are taken from a population with
any type of distribution that has a mean  and standard
deviation ,

x

x
the sample means will have an approximately normal
distribution.
xx
x x
x x x
x x x x x

x
The Central Limit Theorem
If the population itself is normally distributed, with mean
 and standard deviation ,
x

the sample means will have a normal distribution for
any sample size n.
xx
x x
x x x
x x x x x

x
The Central Limit Theorem
In either case, the sampling distribution of sample mean has
a mean equal to the population mean.
μx  μ
Mean of the sample means
The sampling distribution of sample mean has a standard
deviation equal to the population standard deviation divided
by the square root of n.
σ
σx 
n
Standard deviation of the sample means or
Standard Error of the Mean
The Central Limit Theorem
Sampling Distribution of Sample Means
Example:
The population values {5, 10, 15, 20} are written on slips of
paper and put in a hat. Two slips are randomly selected, with
replacement.
a. Find the mean, standard deviation, and variance of the
population.
b. Graph the probability histogram for the population values.
c. List all the possible samples of size n = 2 and calculate the
mean of each.
d. Create the probability distribution of the sample means.
e. Graph the probability histogram for the sampling
distribution.
Sampling Distribution of Sample Means
Example:
The population values {5, 10, 15, 20} are written on slips of
paper and put in a hat. Two slips are randomly selected, with
replacement.
a. Find the mean, standard deviation, and variance of the
population.
Population
5
10
15
20
μ = 12.5
σ = 5.59
σ 2 = 31.25
Sampling Distribution of Sample Means
Example:
The population values {5, 10, 15, 20} are written on slips of
paper and put in a hat. Two slips are randomly selected, with
replacement.
b. Graph the probability histogram for the population values.
Probability Histogram
of Population of x
P(x)
Probability
0.25
x
5
10
15
Population values
20
This uniform distribution
shows that all values have
the same probability of
being selected.
Sampling Distribution of Sample Means
Example:
c. List all the possible samples of size n = 2 and calculate
the mean of each.
Sample
5, 5
5, 10
5, 15
5, 20
10, 5
10, 10
10, 15
10, 20
Sample mean,
5
7.5
10
12.5
7.5
10
12.5
15
x
Sample
15, 5
15, 10
15, 15
15, 20
20, 5
20, 10
20, 15
20, 20
Sample mean,
x
10
12.5
15
17.5
12.5
15
17.5
20
These means form the sampling distribution of the
sample means.
Sampling Distribution of Sample Means
Example:
d. Create the probability distribution of the sample means.
x
f Probability
5
1
0.0625
7.5
2
0.1250
10
3
0.1875
12.5 4
0.2500
15
3
0.1875
17.5 2
0.1250
20
1
0.0625
Probability
Distribution of
Sample Means
Sampling Distribution of Sample Means
Example:
e. Graph the probability histogram for the sampling
distribution.
Probability Histogram of Sampling Distribution
0.18
0.16
The shape of the graph is
symmetric and bell shaped.
0.14
Probability
0.12
0.1
0.08
0.06
0.04
0.02
0
5
7.5
10
12.5
Sample Means
15
17.5
20
Sampling Distribution of Sample Means
Example:
The population values {5, 10, 15, 20} are written on slips of
paper and put in a hat. Three slips are randomly selected,
with replacement.
a. List all the possible samples of size n = 3 and calculate the
mean of each.
b. Create the probability distribution of the sample means.
c. Graph the probability histogram for the sampling
distribution.
Notice: that 64 possible random samples (with replacement) of
size n = 3 can be chosen from the population {5, 10, 15, 20}.
Sample
5, 5, 5
5, 5, 10
5, 5, 15
5, 5, 20
5, 10, 5
5, 10, 10
5, 10, 15
5, 10, 20
5, 15, 5
5, 15, 10
5, 15, 15
5, 15, 20
5, 20, 5
5, 20, 10
5, 20, 15
5, 20, 20
Sample mean,
5
6.666
8.333
10
6.666
8.333
10
11.666
8.333
10
11.666
13.333
10
11.666
13.333
15
x
Sample
10, 5, 5
10, 5, 10
10, 5, 15
10, 5, 20
10, 10, 5
10, 10,10
10, 10,15
10, 10,20
10, 15, 5
10, 15,10
10, 15,15
10, 15,20
10, 20, 5
10, 20,10
10, 20,15
10, 20,20
Sample mean,
6.666
8.333
10
11.666
8.333
10
11.666
13.333
10
11.666
13.333
15
11.666
13.333
15
16.666
x
Sample
15, 5, 5
15, 5, 10
15, 5, 15
15, 5, 20
15, 10, 5
15, 10,10
15, 10,15
15, 10,20
15, 15, 5
15, 15,10
15, 15,15
15, 15,20
15, 20, 5
15, 20,10
15, 20,15
15, 20,20
Sample mean,
8.333
10
11.666
13.333
10
11.666
13.333
15
11.666
13.333
15
16.666
13.333
15
16.666
18.333
x
Sample
20, 5, 5
20, 5, 10
20, 5, 15
20, 5, 20
20, 10, 5
20, 10,10
20, 10,15
20, 10,20
20, 15, 5
20, 15,10
20, 15,15
20, 15,20
20, 20, 5
20, 20,10
20, 20,15
20, 20,20
Sample mean,
10
11.666
13.333
15
11.666
13.333
15
16.666
13.333
15
16.666
18.333
15
16.666
18.333
20
x
Sampling Distribution of Sample Means
Example:
b. Create the probability distribution of the sample means.
Probability Distribution of Sample Means
x
f
5
1
Probability
0.015625
6.666
3
0.046875
8.333
6
0.09375
10
10
0.15625
11.666
12
0.1875
13.333
12
0.1875
15
16.666
0.15625
0.09375
18.333
10
6
3
0.046875
20
1
0.015625
Sampling Distribution of Sample Means
Example:
c. Graph the probability histogram for the sampling
distribution.
Probability Histogram of Sampling Distribution
0.2
0.18
0.16
Probability
0.14
0.12
0.1
0.08
The shape of the graph is
symmetric and bell shaped.
0.06
0.04
0.02
0
5
6.666 8.333
10
11.67 13.33
Sample Means
15
16.67 18.33
20
Sampling Distribution of Sample Means
Comparison of the sampling distributions for sample sizes n = 2
and n = 3 with their normal curve.
Notice we have used the same scales on both graphs.
Sampling Distribution of Sample Means
Comparison of the two normal curves approximating the sampling
distributions for sample sizes n = 2 and n = 3
μx = μ = 12.5
σx =
σ
5.59

 3.95
n
2
σx =
σ
5.59

 3.22
n
3
The Mean and Standard Error
Example: The heights of fully grown magnolia bushes have a
mean height of 8 feet and a standard deviation of 0.7 feet.
Samples of 38 bushes are randomly selected from the population,
and the mean of each sample is determined. Find the mean and
standard error of the mean of the sampling distribution.
Mean
μx  μ
=8
Standard deviation or
Standard error
σ
σx 
n
=
0.7
= 0.11
38
Interpreting the CLT
Example continued: The mean of the sampling distribution is 8
feet ,and the standard error of the sampling distribution is 0.11
feet.
From the Central Limit Theorem,
because the sample size is greater than
30, the sampling distribution can be
approximated by the normal distribution.
x
7.6
μx = 8
8
8.4
σ x = 0.11
Finding Probabilities
Example continued: The mean of the sampling distribution is 8
feet ,and the standard error of the sampling distribution is 0.11
feet.
μx = 8
Find the probability that the mean
height of a randomly selected sample
of 38 bushes is less than 7.8 feet.
n = 38
σ x = 0.11
x
7.6
7.8
8
8.4
Finding Probabilities
μx = 8
n = 38
σ x = 0.11
z
P( X  7.8)
x
7.6
7.8
8
8.4
z
0
x - μx
σx
7.8 - 8
=
0.11
= -1.82
P( X  7.8)  P(Z  1.82)  0.0344
The probability that the mean height of the 38 bushes is less than
7.8 feet is 0.0344.
Another Example
The average on a statistics test was 78 with a standard deviation of
8. If the test scores are normally distributed, find the probability
that the mean score of 25 randomly selected students is between
75 and 79.
μx = 78
σx =
σ
8
=
= 1.6
n
25
P(75  X  79)
z1 =
x - μx 75 - 78
= -1.88
=
σx
1.6
z2 =
x - μ 79 - 78
=
σ
1.6
x
75
1.88
?
78 79
0 0.63
?
z
= 0.63
Example continued:
P(75  X  79)
75
1.88
?
78 79
x
z
0 0.63
?
P(75  X  79)  P(1.88  Z  0.63)  P(Z  0.63)  P(Z  1.88)
= 0.7357  0.0301 = 0.7056
Approximately 70.56% of all samples of 25 students will have
a mean score between 75 and 79.
Problem 9 in Section 6-5
Problem 9 in Section 6-5
Problem 9 in Section 6-5
Problem 10 in Section 6-5
Problem 10 in Section 6-5
Problem 19 in Section 6-5
Problem 19 in Section 6-5
Problem 19 in Section 6-5