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FEM3004 Chapter 3 Sampling Distribution

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Chapter 3

Sampling Distributions
Introduction
• Parameters are numerical descriptive measures for populations.
• For the normal distribution, the location and
shape are described by and 
• For a binomial population, the important
parameter is the binomial proportion p.
• Often the values of parameters that specify the exact form of a
distribution are unknown.
• You must rely on the sample to learn about these parameters.
Sampling Distributions
•Numerical descriptive measures calculated
from a sample are called statistics.
statistics

•Statistics vary from sample to sample and


hence are random variables.

•The probability distributions for statistics are


called sampling distributions.
distributions
Example
Population: x
Population:3,3,5,5,2,2,11 Possible
Possiblesamples
samples
3,3,5,5,22 10 / 3  3.33
Draw
Drawsamples
samplesof
ofsize
sizenn==33 3,3,5,5,11 9/3  3
without
withoutreplacement
replacement 3,3,2,2,11 6/3  2
5,5,2,2,11 8 / 3  2.67
p(x)
1/4
x Prob.
x
2 3 3.33 1/4
Mean
Mean of of xx 3 1/4
11 11 11 11
EE((xx))33.33
.334   34   24   2.674 22.75
   3   2   2 . 67
 4  4  4  4
.75 2 1/4
Population
Population MeanMean 2.67 1/4
3  5  2 1
 3  5  2  122.75
.75
44
Example
Toss a fair die n = 1 time. What is the distribution
of the number x?
x p(x)
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
Flat
Uniform 6 1/6
Example
Toss a fair die n = 2 times. Sampling distribution of x
average on the two numbers?
1 2 3 4 5 6
1 1 1.5 2 2.5 3 3.5
2 1.5 2 2.5 3 3.5 4
3 2 2.5 3 3.5 4 4.5
4 2.5 3 3.5 4 4.5 5
5 3 3.5 4 4.5 5 5.5
Mound
shaped 6 3.5 4 4.5 5 5.5 6
Example
Toss a fair coin n = 3 times. Sampling distribution of x
the average of the three numbers?

Approximately
Normal
Example
Sampling Distributions
Central
CentralLimit
LimitTheorem:
Theorem:IfIfrandom
randomsamples
samplesof ofnn
observations
observationsare
aredrawn
drawnfrom
fromaanon-normal
non-normalpopulation
populationwithwith
meanand
mean andstandard deviation,,then,
standarddeviation then,when
whennnisislarge,
large,the
the
sampling
samplingdistribution
distributionof
ofthe
thesample mean x isisapproximately
samplemean approximately
normally
normallydistributed,
distributed,with meanand
withmean andstandard
standarddeviation
deviation
/ n
..The
Theapproximation
approximationbecomes
becomesmore
moreaccurate
accurateas
asnn
becomes
becomeslarge.
large.
How Large is Large?
 If the sampled population is normal,
normal then the
sampling distribution of x will also be normal, no
matter what the sample size.
 If the sampled population is symmetric,
symmetric the
distribution becomes approximately normal for
relatively small values of n.
 If the sampled population is skewed,
skewed the sample
size must be at least 30 before the sampling
distribution of x becomes approximately normal.
Sampling Distribution of Sample Mean

A random sample of size n is selected from a


population with mean  and standard deviation 
he sampling distribution of the sample mean x will
have mean and standard deviation  / n .
If the original population is normal, the sampling
distribution will be normal for any sample size.
If the original population is non-normal, the sampling
distribution will be normal when n is large (CLT).

The standard deviation of x-bar is sometimes called the


STANDARD ERROR (SE).
Finding Probabilities for
Sample Mean

IfIfthe
the sampling
sampling distribution of x isisnormal
distribution of normal or
or
approximately
approximately normal
normal standardize
standardize or
or rescale
rescale the
the
interval
interval of
of interest
interest in
in terms
terms of
of
x
z
/ n

Find
Find the
the appropriate
appropriate area
area using
using Table
Table 3.
3.
Example: A random
12
12 10
10)
sample of size n = 16 PP((xx 12 )  P ( z
12)  P ( z   )
from a normal 88// 16
16
distribution with  = 10  PP((zz 11)) 11..8413
and  = 8. 8413..1587
1587
Example
A soda filling machine is supposed to fill cans of
soda with 12 fluid ounces. Suppose that the fills are
actually normally distributed with a mean of 12.1 oz and
a standard deviation of 0.2 oz.
What is the probability that the average fill for a 6-pack
of soda is less than 12 oz?

P(x  12) 
x   12  12.1
P(  )
/ n .2 / 6
P( z  1.22)  .1112
Sampling Distribution of Sample
Proportion
The Central Limit Theorem can be used to
conclude that the binomial random variable x is
approximately normal when n is large, with mean np
and variance npq.
x
The sample proportion, pˆ  n is simply a rescaling
of the binomial random variable x, dividing it by n.
From the Central Limit Theorem, the sampling
distribution of p̂ will also be approximately
normal, with a rescaled mean and standard deviation.
The Sampling Distribution of the
Sample Proportion
A random sample of size n is selected from a
binomial population with parameter p
he sampling distribution of the sample proportion,
x
pˆ 
n
pq
n
will have mean p and standard deviation
If n is large, and p is notp̂too close to zero or one, the
sampling distribution of will be approximately
normal. (np>5, nq
The standard >5) of p-hat is sometimes called
deviation
the STANDARD ERROR (SE) of p-hat.
Finding Probabilities for
the Sample Proportion

IfIfthe
the sampling
sampling distribution of p̂ isisnormal
distribution of normal or
or
approximately
approximately normal
normal standardize
standardize or
or rescale
rescale the
the
interval
interval of
of interest
interest in
in terms of z  pˆ  p
terms of
pq
n

Find
Find the
the appropriate
appropriate area
area using
using Table
Table 3.
3.
Example: A random ..55..44
sample of size n = 100 PP((pˆpˆ ..55))  PP((zz  ))
..44(.(.66))
from a binomial
population with p = .4. 100
100
 PP((zz  22..0404)) 11..9793
9793..0207 0207
Example
The soda bottler in the previous example claims
that only 5% of the soda cans are underfilled.
A quality control technician randomly samples 200
cans of soda. What is the probability that more than
10% of the cans are underfilled?
nn==200
200 P ( pˆ  .10)
S:
S:underfilled
underfilledcan
can .10  .05
 P( z  )  P ( z  3.24)
pp==P(S)
P(S)==.05
.05 .05(.95)
qq==.95
.95 200
np
 1  .9994  .0006
np==10
10 nq
nq==190
190
This would be very unusual,
OK to use the normal
if indeed p = .05!
approximation
Key Concepts
I. Statistics and Sampling Distributions

1. Sampling distributions describe the possible values of a


statistic and how often they occur in repeated sampling.

2. Sampling distributions can be derived mathematically,


approximated empirically, or found using statistical
theorems.

3. The Central Limit Theorem states that sums and


averages of measurements from a non-normal population
with finite mean  and standard deviation  have
approximately normal distributions for large samples of
size n.
Key Concepts
II. Sampling Distribution of the Sample Mean
1. When samples of size n are drawn from a normal population
with mean  and variance 2, the sample meanx has a
normal distribution with mean  and variance 2n.
2. When samples of size n are drawn from a non-normal
population with mean  and variance 2, the Central Limit
Theorem ensures that the sample mean x will have an
approximately normal distribution with mean  and variance
2n when n is large (n  30).
3. Probabilities involving the sample mean can be calculated
by standardizing the value ofx using x
z  x
z
// nn
Key Concepts
III. Sampling Distribution of the Sample Proportion
1. When samples of size n are drawn from a binomial
population with parameter p, the sample proportion
‘p hat’ will have an approximately normal
distribution with mean p and variance pq n if np 
5 and nq  5.
2. Probabilities involving the sample proportion can
be calculated by standardizing the value p̂using

pˆpˆpp
zz
pq
pq
nn

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