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```• Chapter 7, Sample Distribution
–A sampling distribution is a distribution of all of the possible values
of a statistic (say sample mean) for a given size sample selected from
a population.
• Sample Distribution of the Mean is an
Unbiased Estimate of the Population Mean
–If all possible samples of a certain size, n, are selected from a
population, the mean of these sample means (the grand mean) would
be equal to the population mean.
μX  μ
• Assume there is a population of size N=4
• Random variable, X, is age of individuals
• Values of X: 18, 20, 22, 24 (years)
• The population mean and standard deviation are:
μ

σ
X
i
N
18  20  22  24
 21
4
 (X  μ)
i
N
P(x)
2
 2.236
18
A
20
B
22
C
24
D
Uniform Distribution
Now consider all possible samples of size n=2 drawn from this
population
1st
Obs
2nd Observation
18
20
22
24
1st 2nd Observation
Obs 18 20 22 24
18 18,18 18,20 18,22 18,24
18 18 19 20 21
20 20,18 20,20 20,22 20,24
20 19 20 21 22
22 22,18 22,20 22,22 22,24
22 20 21 22 23
24 24,18 24,20 24,22 24,24
24 21 22 23 24
16 possible samples
(sampling with
replacement)
16 Sample Means
• Sampling Distribution of All Sample Means
P(X)
Notice:
18
19 20 21
22 23 24
(no longer uniform)
μX
X


i
N
σX 
( X
18  19  21    24
 21
16

i
 μ X )2
N
(18 - 21) 2  (19 - 21) 2    (24 - 21) 2

 1.58
16
Population
N = 4, μ  21
Sample Means
Distribution
n = 2, μ X  21
σ  2.236
σ X  1.58
• Why std. Dev. Of the means distribution is smaller than
that of the population?
Reasons:
• Different samples of the same size from the same
population will yield different sample means
• A measure of the variability in the mean from sample to
sample is given by the Standard Error of the Mean:
σ
σX 
n
•Note that the standard error of the mean decreases and the
distribution becomes less dispersed as the sample size
increases (see page 235)
• If a population is normally distributed with mean μ and
standard deviation σ, the sampling distribution of X is also
normally distributed with
μX  μ
σX 
σ
n
• Z-value for the sampling distribution of
Z
where:
( X  μX )
σX
( X  μ)

σ
n
X
= sample mean
σ
= population mean
μ
n
X is calculated:
= population standard deviation
= sample size
• If population is not normally distributed, we can apply
the Central Limit Theorem which proves that:
– …sample means from the population will be
approximately normal as long as the sample size is
large enough.
• What is large enough?
• For most distributions, n > 30 will give a sampling
distribution that is nearly normal
• For fairly symmetric distributions, n > 15
• For normal population distributions, the sampling
distribution of the mean is always normally distributed
• (See page 238 for distributions of the population and
samples)
Application:
A brand name breakfast cereal company produces 5000 boxes
of serial per day. Each box is suppose to have 368 grams
of cereal with an average dispersion of 15 grams.
•
Set up the information in terms of population
distribution.
•
Questions:
1. What percent of individual boxes will have less than 365
grams?
2. If a sample of 25 boxes are selected what
is the probability that the sample mean is
less than 365?
3.If all possible samples of size 25 are taken,
what interval around the population mean
will contain 95% of all sample means?
4.What is the probability that a sample mean
will be within the above estimated interval?
• All material in this chapter assumed
sampling with replacement. Apply the
Finite Population Correction (fpc) if:
– the sample is large relative to the
population (n is greater than 5% of N)
and…
– Sampling is without replacement
N n
• The fpc factor is
N 1
```
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