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Sampling Distributions
 Introduction
 Point Estimation
 Sampling Distribution of x
 Sampling Distribution for the
p̂ Difference between
Two Means
 Sampling Distribution of p̂
 Sampling Distribution for the
p̂ Difference between
Two Proportions
Introduction
 A sampling distribution is a distribution of all of the possible
values of a sample statistic for a given size sample selected
from a population.
 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.
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.
Point Estimation
Point estimation is a form of statistical inference.
In point estimation we use the data from the sample
to compute a value of a sample statistic that serves
as an estimate of a population parameter.
We refer to x as the point estimator of the population
mean .
s is the point estimator of the population standard
deviation .
p̂ is the point estimator of the population proportion p.
Sampling Distribution of the Sample Mean
 The probability distribution of X is called its sampling
distribution. It list the various values that X can assume and
the probability of each value of X . In general, the probability
distribution of a sample statistic is called its sampling
distribution.
 If a population is normal with mean μ and standard deviation
σ, the sampling distribution of
is also normally distributed
with
X  
X 
and
Z-value for the sampling distribution of X
Z
( X  )

n

n
Sample Mean Sampling Distribution: If the Population is
not Normal
 We can apply the 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↑
As the sample
size gets large
enough…
the sampling
distribution becomes
almost normal
regardless of shape
of population
Properties and Shape of the Sampling Distribution of
the Sample Mean, X .
 If n≥30, X is normally distributed, where
 2 
X ~ N  , 
n 

 Note: If the 
2
2
s
unknown then it is estimated by
.
 If n<30 and variance is known. X is normally distributed
 2 
X ~ N  , 
n 

 If n<30 and variance is unknown. t distribution with n-1 degree
of freedom is use
T
X 
2
s
n
~ tn 1
Example
Suppose a population has mean μ = 8 and standard deviation σ = 3.
Suppose 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
x is approximately normal
σ
3
 with mean μ x = 8 and standard deviation σ x 

 0.5
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
Example:
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


10
a) Standard error = standard deviation,  
 2.5
16
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
Sampling Distribution for the Difference between
Two Means
 Suppose we have two populations, X1 and X 2 which are
normally distributed. X1 has mean 1 and variance  21 while X 2
has mean  2 and variance  2 2 . These two distributions can be
written as:
X 1 ~ N  1 , 
2
1

and
X 2 ~ N  2 ,  2 2 
Now we are interested in finding out what is the sampling
distribution of the difference between two sample means, the
distribution of X 1  X 2

12  22 
X1  X 2 ~ N  1  2 ,


n
n

1
2 
Example:
A taxi company purchased two brands of tires, brand A and brand B.
It is known that the mean distance travelled before the tires wear
out is 36300 km for brand A with standard deviation of 200 km while
the mean distance travelled before the tires wear out is 36100 km
for brand A with standard deviation of 300 km. A random sample of
36 tires of brand A and 49 tires of brand B are taken. What is the
probability that the
a) difference between the mean distance travelled before the tires
of brand A and brand B wear out is at most 300 km?
b) mean distance travelled by tires with brand A is larger than the
mean distance travelled by tires with brand B before the tires wear
out?
Solution:
X 1 : the mean distance travelled before the tires of brand A wear out
X 2 : the mean distance travelled before the tires of brand B wear out

2002 3002 
X 1  X 2 ~ N  36300  36100,


36
49 

X 1  X 2 ~ N  200, 2947.846 
a) P | X 1  X 2 | 300   P  300  X 1  X 2  300 
300  200 
 300  200
 P
Z

2947.846 
 2947.846
 P  9.21  Z  1.84   0.9671
b) P  X 1  X 2   P  X 1  X 2  0 
0  200 

 PZ 

2947.846 

 P  Z  3.68  0.9999
Sampling Distribution of the Sample Proportion
 The population and sample proportion are denoted by p and p̂ ,
respectively, are calculated as,
X
p
N
and
x
pˆ 
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
Example
If the true proportion of voters who support Proposition A is pˆ  0.40
what is the probability that a sample of size 200 yields a sample
proportion between 0.40 and 0.45? If p  0.40 and n = 200, what
is P  0.40  pˆ  0.45 ?
σ pˆ 
p(1  p)
0.4(1  0.4)

 0.03464
n
200
0.45  0.40 
 0.40  0.40
ˆ
P(0.40  p  0.45)  P 
Z

0.03464
0.03464


 P(0  Z  1.44)  0.4251
Example:
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 of .
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 is:  pˆ  p  0.87
The standard deviation of p̂ is:  
pˆ
pq
0.87(0.13)

 0.011
n
900
Sampling Distribution for the Difference between
Two Proportions
 Now say we have two binomial populations with proportion of
successes p1 and p2 respectively. Samples of size n1 are taken
from population 1 and samples of size n2 are taken from
population 2. Then p̂1 and p̂2 are the proportions from those
samples.
p1 1  p1  

ˆ
P1 ~ N  p1 ,

n1


p2 1  p2  

ˆ
P2 ~ N  p2 ,

n
2


p1 1  p1  p2 1  p2  

ˆ
ˆ
P1  P2 ~ N  p1  p2 ,


n
n
1
2


Example:
A certain change in a process for manufacture of component parts
was considered. It was found that 75 out of 1500 items from the
existing procedure were found to be defective and 80 of 2000
items from the new procedure were found to be defective. If one
random sample of size 49 items were taken from the existing
procedure and a random sample of 64 items were taken from the
new procedure, what is the probability that
a) the proportion of the defective items from the new procedure
exceeds the proportion of the defective items from the existing
procedure?
b) proportions differ by at most 0.015?
c) the proportion of the defective items from the new procedure
exceeds proportion of the defective items from the existing
procedure by at least 0.02?
Solution:
PˆN :The proportion of defective items from the new procedure
PˆE :The proportion of defective items from the existing procedure
80
 0.04
2000
0.04(0.96) 

PˆN ~ N  0.04,

64


pN 
75
 0.05
1500
0.05(0.95) 

PˆE ~ N  0.05,

49


pE 
0.05(0.95) 0.04(0.96) 

ˆ
ˆ
PN  PE ~ N  0.04  0.05,


49
64


Pˆ  Pˆ ~ N  0.01, 0.0016 
N

E
 
a) P PˆN  PˆE  P PˆN  PˆE  0

0   0.01 

 PZ 

0.0016 

 P  Z  0.25 
 0.4013

 
b) P | PˆN  PˆE | 0.015  P 0.015  PˆN  PˆE  0.015

0.015   0.01 
 0.015   0.01
 P
Z

0.0016
0.0016


 P  0.125  Z  0.625 
 0.2838

 
c) P PˆN  PˆE  0.02  P PˆN  PˆE  0.02

0.02   0.01 

 PZ 

0.0016 

 P  Z  0.75 
 0.2266
End of Chapter 1