ECON1310
Introductory Statistics for Social Sciences
LECTURE 6
Sampling Distribution of the Sample Mean
& Sampling Distribution of the Sample Proportion
1 hour of online YouTube videos on Blackboard
to be viewed to complete Lecture 6
Feedback (from Lecture 5)
All the fish at a fish farm are two years old and
their lengths are known to be normally distributed.
The mean length of all the fish is 16cm and
standard deviation of 2cm. Find the corresponding
Z-value for a fish of length 15cm?
a) 0.5
b) 0
c) -1
d) -0.5
Feedback (from Lecture 5)
From the previous question, what is the probability
a seagull can dive into a fish farm pond and catch
a single fish with a length less than 15cm long?
a) -0.1915
b) 0.3085
c) -0.3085
d) 0.1915
2Lecture 6 ECON1310
4
Today’s Topics:
1. Sampling Distribution of the Sample Mean
2. Sampling Distribution of the Sample Proportion
Note: See Blackboard for the videos to complete
the second half of this lecture relating to point 2
above.
Lecture 6 ECON1310 5
A Sampling Distribution
= the distribution of all possible values
of a statistic, using the same sample
size, selected from a population.
Sampling Distribution of the
Sample Mean
Example: Collect 20 fish in a bucket, measure each fish length,
then calculate the average length of the fish in the bucket. Repeat
until all possible samples of 20 fish are collected from the pond.
This is a large pond!!
. . .
20n
Bucket 1
1X 2X
Bucket 2 Bucket 3 Bucket n
nX3X
20n20n 20n
3(average length of fish in a bucket)
Sampling Distribution of the Sample Mean
Historical data available (σ is known)
X

Frequency
X ~ N (µ,
n

)
1.0 1.5 2.0 2.5 3.0 3.5 4.0
If the length of fish in the pond
are normally distributed, the
average lengths of fish in the
buckets is exactly normally
distributed.
That is, if f(x) is normally
distributed, is exactly
normally distributed.
)X(f
Here, we assume
that σ is known
from historical data
X
Lecture 6 ECON1310 8
Standard Error of the Mean
 Different samples of the same size from the same
population will yield different sample means
 A measure of the variability of the mean from sample to
sample is given by what is called the Standard Error of
the Mean,
 Note that the standard error of the mean decreases as the
sample size increases.
n
σ
σ
X

X
σ
Lecture 6 ECON1310 9
A Key Point:
If a population variable is normally distributed
with mean μ and standard deviation σ, the
sampling distribution of is also exactly
normally distributed with:
and
μμ
X

n
σ
σ
X

X
4Lecture 6 ECON1310 10
Z-value for Sampling Distribution of the
Sample Mean
where: = sample mean
= population mean
= population standard deviation
n = sample size
X
μ
σ
)
n
σ
(
μ)X(
σ
)μX(
Z
X
X 


Sampling Distribution Properties
Population variable is
normally distributed.
Sampling distribution of
the sample mean is
exactly normally
distributed AND
has the same mean as
the population variable.
x
x
μ
xμ
μμ
X

12
Sampling Distribution Properties
As n increases,
decreases. xσ
Large sample size.
eg: n=50
Small sample
size. eg: n=3
x
μ  xμ
5Sampling Distribution of the Sample Mean
Historical data available (σ is known)
(average length of a
bucket of fish)
f(z)
Z
µz = 0
σz = 1
X

n
x 
X
)X(f










n
Z



 XX
X
X
Z1
Z2
area =
x
1X 2X
14
Example 1:
Sampling Distribution
of the Sample Mean
A two year old fish species has been farmed for many
years under the same conditions. The fish lengths have
been found to be normally distributed with an average
length of 10cm and a standard deviation of 0.5cm (same as
in Example 1, Lecture 5).
Q1. What is the probability a pelican can catch 4 fish in its
bill from the fish pond so the average length of the fish it
catches (sample average) is between 10cm and 11cm?
15
P( 10 < < 11 )
= P( 0 < Z < 4 ) = 50%
Shaded areas are
equal
4
25.0
1
)
4
0.5
(
10)(11
)
n
σ
(
μ)X(
Z 




x
Z
0 4
Area = 0.499997
= 50% (Table A.5)
Area = 0
10 11 x
Solution Example 1
Sampling Distribution of the Sample Mean
6Summary: Transformation Z formulae
General formula (on exam formulae sheet)
X
Z




errorstandard
parameter
population
statistic
sample
Z


)
n
σ
(
μ)X(
Z


Lecture 6 ECON1310 17
Example 2.
From previous data, calculations have found a customer
waiting for a telephone banking enquiry to be answered
will have a variance of 32.6 minutes2. It is also known
only 8% of all customers placed on hold wait longer than
12.5 minutes.
a. Find the mean waiting time for a customer placed on
hold (assuming waiting time is approximately normally
distributed).
b. If a random sample of 20 customer calls is selected,
what is the probability(%) the sample average waiting time
is less than 5 minutes. Use the answer from part a.
Solution Example 2a.
a. Find µ given σ2 =32.6 min2
Z
1.41
0.42 = value IN Table A.5
0
0.08 (NOT in Table A.5)
= 0.5 – 0.42
X
12.5µ = ?
0.08
X
Z




 
12.5 1.41*5.70964
4.4494
4.45 .
X
Z
X Z
twodec pl


 


 
 


719
P( < 5.0 )
= P( Z < 0.43 )
= 0.1664 + 0.50
= 0.6664
= 66.64%
)(43.0
430793.0
)
20
5.70964
(
4.45)(5
)
n
σ
(
μ)X(
Z
tablezforpldectwo





x
Z
0 0.43
Area = 50%
Area = 0.1664 (Table A.5)
4.45 5.0 x
Solution Example 2b.
b. Find P( < 5.0)x
Lecture 6 ECON1310 20
The problem:
The population variable of interest may NOT be
normally distributed (eg: skewed), or the shape of
the population variable distribution is unknown.
The question:
Under what conditions will the sampling distribution
of the sample mean be normally distributed?
Central Limit Theorem (CLT)
 a key theoretical concept for inferential statistics
Population variable distribution is
skewed and NOT normal.
Sampling distribution of the sample mean
becomes closer to normal as n increases.
Central Tendency
Variation
x
x
Larger
sample
size
Smaller
sample size
If the Population Variable is not Normally
Distributed.
Sampling distribution
properties of the
sample mean.
μμx 
n
σ
σx 
xμ
μ
yet
822
As n increases
Central Limit Theorem (CLT) states:
As the sample size gets large enough…….
the sampling distribution of
becomes approximately normally
distributed regardless of the shape
of population variable distribution.
x
x
23
How large is large enough?
 For most population distributions, the CLT
states if n ≥ 30, the sampling distribution of
sample means will be approximately
normally distributed.
 For a population variable that is normally
distributed, the sampling distribution of
sample means is always exactly normally
distributed, for any size n.
Lecture 6 ECON1310 24
Example 3.
Suppose a population variable has mean weight
of μ = 8kg and standard deviation σ = 3kg.
Suppose a random sample of size n = 36 is
selected and it is NOT known if the population
variable is normally distributed.
What is the probability that the sample mean is
between 7.8kg and 8.2kg?
925
Example 3:
Solution
Even if the population variable is not normally
distributed, the Central Limit Theorem can be
applied since n ≥ 30.
… so the sampling distribution of is
approximately normally distributed
… with mean = 8 kg = µ
…and standard deviation
x
xμ
kg0.5
36
3
n
σ
σx 
Solution Example 3 (continued)
Z
Standardised
Normal Distribution
0μz 
Population
Distribution
? ?
?
?
?
?
??
?
?
?
?
8μ 
x
Sample
8μ
X

7.8 8.2
Sampling distribution of the
sample mean is approximately
normally distributed.
Xn ≥ 30
27
P( 7.8 < < 8.2 )
= P( -0.4 < Z < 0.4 )
= 2 * 0.1554
= 0.3108
4.0
5.0
0.2
)
36
3
(
8)(8.2
)
n
σ
(
μ)X(
Z 




x
Z
0 0.4
Area = 0.1554 (Table A.5)
8 8.2 x
Solution Example 3.
Sampling Distribution of the Sample Mean
7.8
-0.4
Area =
0.1554
10
Sampling Distribution of the
Sample Proportion
Now watch all the videos to complete the
second half of the lecture. You can also print
copies of the video slides.
You NEED to watch all the lecture videos for the
second half of the lecture so as to understand the
tutorial questions, the related CML quiz questions,
as well as understand the next lecture.
Lecture 6 ECON1310 29
After watching all the online videos for
Lecture 6, be sure you:
• read through all the lecture slides.
• recognise that all the content covered in the
lecture 6 slides and videos is assessable.
Test your understanding so far!
A soup company has produced cans of soup for
many years, with an average weight of 375g and
standard deviation of 15g. Previous experience
has also shown that 12% of the cans become
slightly scratched during production.
a) From a sample of 50 cans of soup, find the
probability the sample mean is larger than 379g.
b) Find the probability that if a single can is
chosen, it will weigh more than 379g.
c) Find the probability that the sample proportion
of slightly scratched cans is more than 15% in
a sample of 50 cans of soup.
11
Summary: Transformation Z formulae
X
Z




n
pp
pppp
Z
p )1(
ˆˆ
ˆ 





)
n
σ
(
μ)X(
Z


Summary: Rearranged useful formulae
32
Z
X
ZofsigncorrectuseZX
ZofsigncorrectuseZX
X
Z












)(
)(
Lecture 6 ECON1310
33
Z
n
n
Z
n
Z
n
Z
)X(
X
X
X














Summary: Rearranged useful formulae
)( Zofsigncorrectuse
)( Zofsigncorrectuse
12
34
Z
pp
Zpp
Zpp
n
pp
pppp
Z
p
p
p
p









ˆ
ˆ
ˆ
)1(
ˆˆ
ˆ
ˆ
ˆ
ˆ




Summary: Rearranged useful formulae
)( Zofsigncorrectuse
)( Zofsigncorrectuse
Lecture 6 ECON1310 35
Next lecture…
 Estimation using Confidence Intervals – the
first inferential statistics topic