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Sampling
Distribution
Estimation
Hypothesis Testing
Introduction
Sampling Distribution of Sample Mean (X )
Sampling Distribution of Sample Proportion (p̂)
Introduction



A sampling distribution - probability distribution of a
sample statistic based on all possible simple random
sample of the same size from the same population.
If we take several sample and find mean of the sample,
therefore the distribution of the sample mean called
sampling distribution of the sample mean, X .
For example, suppose you sample 50 students from
your college regarding their mean GPA. If you obtained
many different samples of 50, you will compute a
different mean for each sample. We are interested in
the distribution of all potential mean GPA we might
calculate for any given sample of 50 students.
Introduction

If we take several sample and find the ratio of the
specific characteristic in the sample, therefore the
distribution of the sample proportion called
sampling distribution of the sample proportion,p̂ .
Introduction
The reason we select a sample is to collect data to
answer a research question about a population.
The sample results provide only estimates of the
values of the population characteristics.
The reason is simply that the sample contains only a
portion of the population.
With proper sampling methods, the sample results
can provide “good” estimates of the population
characteristics.
X



The probability distribution of sample is called
sampling distribution.
It lists the various values that X can assume, and the
probability of each value of X .
If the population is normally distributed with mean
(μ) and standard deviation (σ), the sampling
distribution of the sample mean, X is also normally
distributed with:
i) Mean  x  
iii) Standard deviation /
standard error
ii) Variance
 2x 
2
x 

n
n
where n is the sample size
X

Z-value for the sampling distribution of mean ( X ):
x 
Z

n


If all samples of a particular size are selected from
any population, the sampling distribution of the
sample mean is approximately a normal
distribution.
This approximation improves with larger samples.


If the population is not normally distributed, apply
central limit theorem.
Central Limit Theorem:
◦ Even if the population is not normal, sample
means from the population will be approximately
normal as long as the sample size is large enough
n↑
(n≥30).
the sampling
As the
sample size
gets large
enough…
distribution
becomes almost
normal regardless
of shape of
population
X



If n ≥ 30 (large) , the sampling distribution of the
sample mean is normally distributed;
 2 
X ~ N  ,

n 

2
2

Note: If the
unknown then it is estimated by s .
If n < 30 (small), is known, and the sampling
distribution of the sample mean is normally
distributed if the sample is from the normal
population;
 2 
X ~ N  ,

n


2

If n<30 and is unknown. t distribution with n-1
degree of freedom is use;
2
T
x
2
s
n
~ t n 1
Suppose a population has mean μ = 8 and standard deviation σ = 3.
A random sample of size n = 36 is selected. What is the probability
that the sample mean is between 7.8 and 8.2?
Solution:
•
•
•
Even if the population is not normally distributed, the central limit
theorem can be used (n > 30).
So the sampling distribution of the sample mean is approximately
normal with mean,  x    8
and standard deviation,

3
x 

 0.5
Hence,
n
36


 7.8 - 8
X-μ
8.2 - 8 
P(7.8  X  8.2)  P



3
σ
3


36
n
36 

 P(-0.4  Z  0.4)  0.3108
Z
The amount of time required to change the oil and filter
of any vehicles is normally distributed with a mean of
45 minutes and a standard deviation of 10 minutes. A
random sample of 16 cars is selected.
What is the standard error of the sample mean to be?
 What is the probability of the sample mean between
45 and 52 minutes?
 What is the probability of the sample mean between
39 and 48 minutes?
 Find the two values between the middle 95% of all
sample means.

Solution:
X : the amount of time required to change the oil and filter of any vehicles

X ~ N 45,102

n  16
X : the mean amount of time required to change the oil and filter of any vehicles
 102 
X ~ N  45,

16


a) The standard error,  x 
10
16
 2.5
52  45 
 45  45
b) P  45  X  52   P 
Z

2.5 
 2.5
 P  0  Z  2.8 
 0.4974
48  45 
 39  45
c) P  39  X  48   P 
Z

2.5
2.5


 P  2.4  Z  1.2 
 0.4918  0.3849
 0.8767
P  a  X  b   0.95
d)
b  45 
 a  45
P
Z 
  0.95
2.5
2.5


P  za  Z  zb   0.95
from table:
za  1.96
zb  1.96
a  45
 1.96  a  40.1
2.5
b  45
 1.96  b  49.9
2.5
Z
Exercise:
A certain type of thread is manufactured with a mean tensile
strength is 78.3kg, and a standard deviation is 5.6kg.
Assuming that the strength of this type of thread is
distributed approximately normal, find:
a) The probability that the mean strength of a random
sample of 10 such thread falls between 77kg and 78kg.
b) The probability that the mean strength greater than 79kg.
c) The probability that the mean strength is less than 76kg.
d) The value of X to the right of which 15% of the mean
computed from random samples of size 10 would fall.
p̂


The probability distribution of the sample
proportion (p̂) is called its sampling distribution.
The population and sample proportion are denoted
by p and p̂ respectively, and calculated as:
x
X
pˆ 
p
and
n
N
where
 N = total number of elements in the population;
 X = number of elements in the population that
possess a specific characteristic;
 n = total number of elements in the sample; and
 x = number of elements in the sample that
possess a specific characteristic

For the large values of n (n ≥ 30), the sampling
distribution is very closely normally distributed.
pˆ

 pq 
N  p,

n


Mean and Standard Deviation of Sample Proportion
 P̂  p
 P̂ 
pq
n
If the true proportion of voters who support Proposition A is
p = 0.40, what is the probability that a sample of size 200 yields a
sample proportion between 0.40 and 0.45?
Solution:
(0.4)(0.6) 

pˆ ~ N  0.4,

200 

pˆ ~ N 0.4,0.0012
 0.4  0.4
0.45  0.4 
P(0.40  pˆ  0.45)  P
Z

0.0012 
 0.0012
 P(0  Z  1.44)
 0.4251
Z
The National Survey of Engagement shows about 87% of freshmen
and seniors rate their college experience as “good” or “excellent”.
Assume this result is true for the current population of freshmen and
seniors. Let
p̂ be the proportion of freshmen and seniors in a
random sample of 900 who hold this view. Find the mean and
standard deviation.
Solution:
Let p the proportion of all freshmen and seniors who rate their
college experience as “good” or “excellent”. Then,
p = 0.87 and q = 1 – p = 1 – 0.87 = 0.13
The mean of the sample distribution of p̂
The standard deviation of p̂
is:
 pˆ 
is:
 pˆ  p  0.87
pq
0.87(0.13)

 0.011
n
900