Inferential Statistics
UMAR.KHAYAM
Lecturer in Economics and Biz Statistics
KARDAN Institute of Higher Education
Kabul, Afghanistan
Sampling Theory
Sampling
Sampling is statistical technique which is used in almost every
field in order to collect information and on the basis of this
information inferences about the characteristics of a population
are made. The value of the population characteristics are
summarized by certain numerical descriptive measures, called
parameters. The values of the population parameter which are in
most situations unknown, would have to be estimated and to get
estimates , we resort to sampling. The observations composing a
Sample are used to calculate a corresponding numerical
descriptive measure, called a statistic. Thus we use statistic to
estimate parameters.
Advantages of sampling
The important advantages of sampling over complete enumeration
are briefly stated below:
1) Sampling saves money as it is much cheaper to collect the desired
information from a small sample than from the whole population.
2) Sampling saves a lot of time and energy as the needed data are
collected and processed much faster than census information. And
this is a very important consideration in all types of investigation or
surveys.
3) Sampling makes it possible to obtain more detailed information from
each unit of the sample as collecting data from a few units of the
population (i.e. sample) can be more complete.
Sampling With and Without Replacement
Samples may be selected with replacement or without replacement.
Sampling is said to be with replacement when from a population a
sampling unit is drawn, observed and then returned to the population
before another unit is drawn. The population in this case remains the same
and a sampling unit might be selected more than once. If we sample with
replacement, the number of all possible samples of size n that could be
n
selected is N
If on the other hand, a sampling unit is chosen and not returned to the
population after it has been observed, the sampling is said to be without
replacement. Here the sampling units cannot be selected again for the
sample as the unit drawn are not replaced. If we sample without
replacement, the number of all possible samples of size n that could be
N
selected is (n
)
Example No.1
· Assume that a population consists of 5 students and the
marks obtained by them in a certain statistics class are
20, 15, 12, 16 and 18. Draw all possible random
samples of two students when sampling is performed
(i) with replacement
· (ii) without replacement.
· Calculate the mean marks for each sample.
· Solution on white board.
Sampling Distribution
· A sampling distribution is defined as a
probability distribution of the value of a
statistic such as a mean, a standard
deviation etc, computed from all possible
samples of the same size, which might be
selected with or without replacement
from a population.
Standard Error
· The standard deviation of a sampling
distribution of a sample statistic is called
the standard error (abbreviated to S.E.) of
the statistic. It is denoted by  x and is
given as

x 
n
Exercise 6 page 231
If the standard error of the mean for the
sampling distribution of random samples of
size 36 from a large population is 2, how
large must the size of the sample become if
the standard error is to reduce to 1.2?
Solution on white board
Sampling distribution of the mean
The sampling distribution of the mean is the
probability distribution or the relative frequency
distribution of the means X of all possible random
samples of the same size that could be selected from a
given population. The mean of this distribution is
represented by  x and the standard deviation which
is called standard error of the mean, by  x
Cont;
Where
And
x   x f  x 
 x   x f x    x f x 
2
2
2
Example No. 2
Assume that a population consists of 5 similar containers
having the following weights (kilograms).
9.8, 10.2, 10.4, 10.0, and 9.6.
a) Find the mean µ and the standard deviation σ of the given
population.
b) Draw random samples of 2 containers without replacement and
calculate the mean weight of each sample.
c) Form a sampling distribution of X
d) Find the mean and the standard deviation of the sampling
distribution of X
Solution On White Board
Properties of Sampling Distribution of X
The sampling distribution of X has the following properties.
1)
x
2) In case of sampling with replacement
 
x 

n
In case of sampling without replacement
 N n
x 
n N 1
Example Page 211
Assume that a uniform population consists of 4 values 0,1,2 and 3.
a) Find the mean µ and the standard deviation σ.
b) Draw random samples of size 2 with replacement and calculate
the mean X of each sample.
c) Find the sampling distribution of X
d) Find the mean and the standard deviation of the sampling
distribution of X



e) Verify that  x  
and
x
n
Solution Example Page 211
a)
µ
∑X
=N
=
6
0+1+2+3
= 4
4
Standard Deviation
2
X
X
σ=
0
0
1
1
2
4
3
9
6
14
∑X 2
N
-
= 1.5
2
∑X
N
2
6
= 14
4
4
= √1.25 = 1.1180
Example Page 211 Cont
b) Total number of possible samples
n 2
N= 4 =16
S.No
Sample
Mean (X)
S.No
Sample
Mean( X)
1
(0,0)
0
9
(2,0)
1
2
(0,1)
0.5
10
(2,1)
1.5
3
(0,2)
1
11
(2,2)
2
4
(0,3)
1.5
12
(2,3)
2.5
5
(1,0)
0.5
13
(3,0)
1.5
6
(1,1)
1
14
(3,1)
2
7
(1,2)
1.5
15
(3,2)
2.5
8
(1,3)
2
16
(3,3)
3
Example Page 211 Cont
c) Sampling Distribution of X
X
f( X)
0
1/16
0.5
2/16
1
3/16
1.5
4/16
2
3/16
2.5
2/16
3
1/16
Example Page 211 Cont
d) Mean and the standard deviation of the sampling
distribution of X
2
X
f(X)
X f(X)
(X) f(X)
0
1/16
0
0
0.5
2/16
1/16
0.5/16
1
3/16
3/16
3/16
1.5
4/16
6/16
9/16
2
3/16
6/16
12/16
2.5
2/16
5/16
12.5/16
3
1/16
3/16
9/16
24/16
46/16
Example Page 211 Cont
· Mean of X
x  x f  x 
= 24/16 = 1.5
Standard Error of X
 x  x f x    x f x 
2
2
2
= 46/16 – (24/16)
·
·
= 2.875 – 2.25 = √ 0.625 = 0.79
Example Page 211 Cont
(e) Verification
As x  1.5
and
µ =1.5
Hence  x  
Also   1.1 1 8 0

1.1 1 8 0

n

and  x
2
 o.7 9
Hen ce  x 

n
 0. 7 9
Example Page 214
Given the population 1,1,1,3,4,5,6,6,6 and
7, find the probability that a random
sample of size 36, selected with
replacement, will yield a sample mean
greater than 3.8 but less than 4.5 if the
mean is measured to the nearest tenth.
Solution on White Board
Z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
0.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4861
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.49865
0.49903
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4865
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
0.4991
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4983
0.4987
0.4991
0.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4485
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
0.4991
0.04
0.0159
0.0557
0.0948
0.1331
0.1700
0.2054
0.2380
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
0.4992
0.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2083
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
0.4992
0.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4989
0.4992
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3990
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4758
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4980
0.4985
0.4989
0.4992
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2518
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4430
0.4535
0.4625
0.4690
0.4762
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
0.4993
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3880
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4990
0.4993
Exercise Page 230
Thinking Challenge
· Random sample of size 54 is drawn with
replacement, from a finite population
2,4,and 6. what is the probability that a
sample mean will be greater than 4.1 but
less than 4.4? Assume the means to be
measured to the nearest tenth.
Example Page 217
Given the population 1,1,1,3,4,5,6,6,6 and 7, find
the mean and standard deviation of the sampling
distribution of means for samples of size 4
selected at random without replacement. Between
what two would you expect at least ¾ of the
Sample means to fall?
Solution on White Board
Example page 218
An electrical firm manufactures light bulbs that
have a length of life that is normally distributed ,
with mean equal to 800 hours and a standard
deviation of 40 hours. Find the probability that a
random sample of 16 bulbs will have an average
life of less than 775 hours.
Solution on white board
Exercise 8 Page 231
Thinking Challenge
If all possible samples of size 16 are drawn
from a normal population with mean equal
to 50 and standard deviation equal to 5,
what is the probability that a sample mean
X will fall in the interval from 47.63 to
49.5?
Exercise 10 Page 231
The heights of 1000 students are approximately normally distributed
with a mean of 174.5 centimeters and a standard deviation of 6.9
centimeters. If 200 random samples of size 25 are drawn from this
population and the means recorded to the nearest tenth of a
centimeters,
Determine
a) The mean and standard error of the sampling distribution of X
b) The number of sample means that fall between 172.5 and 175.8
centimeters inclusive.
c) The number of sample means falling below 172.0 centimeters.
Solution on white board
Exercise 12 page 232
Thinking challenge
· If a certain machine makes electrical
resistors having a mean resistance of 40
ohms and a standard deviation of 2 ohms,
what is the probability that a random
sample of 36 of these resistors will have
a combined resistance of more than 1458
ohms?
Sampling distribution of the difference
between means
· Suppose we have two distinct populations with means 1 and 2
2
2
and variances  1 and  2 respectively. Let independent random
samples of sizes be selected from the respective populations, and
the differences X 1  X 2 between the means of all possible pairs of
samples be computed.
· Then, a probability distribution of the differences X 1  X 2
can
be obtained. Such a distribution is called the sampling distribution
of the differences of sample means X 1  X 2 .
Properties of the sampling distribution of
X1  X 2
· The sampling distribution of the
differences has the following properties.
· 1)
x x  1  2
1
2
· 2)  2 x1  x2     .
n1
n1
2
1
2
2
Example page 224
·
·
·
·
·
Draw all possible random samples of size n1 = 2 with replacement
from a finite population consisting of 4, 6, 8.Similarly, draw all
possible random samples of size n = 2 with replacement from another
finite population consisting of 1, 2, 3.
a) Find the possible differences between the sample means of the two
population.
b) Construct the sampling distribution of X1  X 2 and compute
its mean and variance.
c) verify that
x x  1  2 and  2 x1  x2
1
2
12  22

 .
n1 n2
Example page 224 cont;
SOLUTION:
Whenever we are sampling with replacement from a
finite population, the total number of possible samples
n
is N (where N is the population size, and n is the
2
sample size).Hence, in this example, there are (3) = 9
possible samples which can be drawn with replacement
from each population.
Example page 224 cont;
These two sets of samples and their means are given below
From Population 1
Sample Sample
x1
No.
Value
1
4, 4
4
2
4, 6
5
3
4, 8
6
4
6, 4
5
5
6, 6
6
6
6, 8
7
7
8, 4
6
8
8, 6
7
9
8, 8
8
From Population 2
Sample Sample
x2
No.
Value
1
1, 1
1.0
2
1, 2
1.5
3
1, 3
2.0
4
2, 1
1.5
5
2, 2
2.0
6
2, 3
2.5
7
3, 1
2.0
8
3, 2
2.5
9
3, 3
3.0
Example page 224 cont;
a) Since there are 9 samples from the first
population as well as 9 from the second, hence,
there are 81 possible combinations ofx1
andx2 . The 81 possible differencesx1 –x2
are presented in the next table table:
Example page 224 cont;
x2
x1
1.0
1.5
2.0
1.5
2.0
2.5
2.0
2.5
3.0
4
5
6
5
6
7
6
7
8
3.0
2.5
2.0
2.5
2.0
1.5
2.0
1.0
1.0
4.0
3.5
3.0
3.5
3.0
2.5
3.0
2.5
2.0
5.0
4.5
4.0
4.5
4.0
3.5
4.0
3.5
3.0
4.0
3.5
3.0
3.5
3.0
2.5
3.0
2.5
2.0
5.0
4.5
4.0
4.5
4.0
3.5
4.0
3.5
3.0
6.0
5.5
5.0
5.5
5.0
4.5
5.0
4.5
4.0
5.0
4.5
4.0
4.5
4.0
3.5
4.0
3.5
3.0
6.0
5.5
5.0
5.5
5.0
4.5
5.0
4.5
4.0
7.0
6.5
6.0
6.5
6.0
5.5
6.0
5.5
5.0
b)The sampling distribution of
X1  X 2
is as follows
Probability
x1  x 2
Tally
d
f
f x1  x 2 
df (d)
d2 f(d)
 f d 
1.0
|
1
1/81
1/81
1.0/81
1.5
||
2
2/81
3/81
4.5/81
2.0
||||
5
5/81
10/81
20.0/81
2.5
||||
|
6
6/81
15/81
37.5/81
3.0
||||
||||
10
10/81
30/81
90.0/81
3.5
||||
||||
10
10/81
35/81
122.5/81
4.0
||||
|||| |||
13
13/81
52/81
208.0/81
4.5
||||
||||
10
10/81
45/81
202.5/81
5.0
||||
||||
10
10/81
50/81
250.0/81
5.5
||||
|
6
6/81
33/81
181.5/81
6.0
||||
5
5/81
30/81
180.0/81
6.5
||
2
2/81
13/81
84.5/81
7.0
|
1
1/81
7/81
49.0/81
81
1
324/81
1431/81
Total
---
Thus the mean and the variance are
x x
1

2
x1  x2
2
324
  df d  
 4 , and
81

 d f d    df d 
2
2
2
1431  324
53
5


 16   1.67
 
81
3
3
 81 
c) In order to verify the properties of the sampling distribution of
X 1  X 2 we first need to compute the mean and variance of the
first population:
The mean and standard deviation of the first population
are:
468
1 
 6 , and
3
2
2
2

4  6  6  6  8  6
8
2
1 

3
3
And the mean and variance of the second population are:
1 2  3


 2 , and
·
2
3
2
2
2

1  2   2  2   3  2 
 
2
2
3

2
.
3
Verification
Now x1x2  4  6  2  1  2 , and
 12
n1

 22
n2

8 1
2 1
.

.
3 2
3 2
4
1
5



3
3
3
 1 .6 7

 x2  x
1
2
Example 7 Page 227
A sample of size n1= 5 is drawn at random from a
population that is normally distributed with mean µ1= 50
2
σ1
and
= 9 and the sample mean X1 is recorded. A second
random sample of size n2= 4 is selected from a second
population that is also normally distributed, with mean
2
2
µ2= 40 and variance σ = 4 and the sample mean X2 is
recorded. What is the P(X1 – X2 < 8.2)?
Solution on white Board
Exercise 21 Page 233
Thinking Challenge
Draw all possible random samples of size n1 = 2 with
replacement from a finite population consisting of 2, 3 and
7.Similarly, draw all possible random samples of size n2 = 2
with replacement from another finite population consisting of
1, 1 and 3.
a) Find the possible differences between the sample means of
the two population.
b) Construct the sampling distribution of X1  X 2 and
compute its mean and variance.
2
2


2
c) verify that x  x  1  2 and x1  x2  1  2 .
1
2
n1
n2
Example 8 page 228
Thinking Challenge
The television picture tubes of manufacturer A have a
mean life time of 6.5 years and a standard deviation of 0.9
years, while those of manufacturer B have a mean life
time of 0.6 years and a standard deviation of 0.8 years.
What is the probability that a random sample of 36 tubes
from manufacturer A will have a mean life time that is at
least 1 year more than the mean lifetime of a sample of 49
tubes from manufacturer B?
Exercise 23 page 233
A random sample of size 25 is taken from a normal population
having a mean of 80 and a standard deviation of 5. a second
random sample of size 36 is taken from a different normal
population having a mean of 75 and a standard deviation of 3.
find the probability that the sample mean computed from the 25
measurements will exceed the sample mean computed from the
36 measurements by at least 3.4 but less than 5.9.
Solution on white board
Exercise 25 page 233
Thinking Challenge
The mean score for freshmen on an aptitude test,
at a college, is 540, with a standard deviation of 50.
what is the probability that two groups of students
selected at random, consisting of 32 and 50 students,
respectively, will differ in their mean score by
(a) more than 20 points; (b) an amount between 5
and 10 points.