Sqqs1013 ch6-a122

Definition:
The probability distribution of the population data.
SQQS1013 Elementary Statistics
SAMPLINGSAMPLING
DISTRIBUTIONSDISTRIBUTIONS
6.1 POPULATION AND SAMPLING DISTRIBUTION
6.1.1 Population Distribution
Suppose there are only five students in an advanced statistics class and the midterm scores
of these five students are:
70 78 80 80 95
Let x denote the score of a student.
Chapter 6: Sampling Distributions 1
Example 1

Definition:
The probability distribution of a sample statistic.
Definition:
The sampling distribution of x is a distribution obtained by
using the means computed from random samples of a
specific size taken from a population.
• Mean for Population
Based on Example 1, to calculate mean for population:
6.80
5
9580807870
=
++++
==
∑
N
x
µ
• Standard Deviation for Population
Based on example 1, to calculate standard deviation for population:
( )
( )
0895.8
5
5
403
32809
2
2
2
=
−
=
−
=
∑
∑
N
N
x
x
σ
6.1.2 Sampling Distribution
 Sample statistic such as median, mode, mean and standard deviation
6.1.2.1 The Sampling Distribution of the Sample Mean

Reconsider the population of midterm scores of five students given in example 1. Let say
we draw all possible samples of three numbers each and compute the mean.
Total number of samples = 5
C3 = 10
)!35(!3
!5
=
−
Suppose we assign the letters A, B, C, D and E to scores of the five students, so that
A = 70, B = 78, C=80, D = 80, E = 95
Then the 10 possible samples of three scores each are
ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE
Example 2
for sample ABC,
Use all the possible
means samples to
form the frequency
distribution table

• Sampling Error
 Sampling error is the difference between the value of a sample statistic and
the value of the corresponding population parameter. In the case of mean,
Sampling error = µ−x
• Mean ( ) for the Sampling Distribution of x
Based on example 2,
6.80
10
8533.8433.8433.7967.8167.8167.76817676
=
+++++++++
=xµ
Sampling distribution
offor samples of three
scores
“The mean of the sampling
distribution of is always
equal to the mean of the
population”
µµ =x
FORMULA
ξ∆Σ λϖ
β
xµ

Standard Deviation ( ) for the Sampling Distribution of
nx
σ
σ =
Where: σ is the standard deviation of the population
n is the sample size
This formula is used when
When N is the population size
1−
−
=
N
nN
nx
σ
σ
Where is the finite population correction factor
This formula is used when 05.0>
N
n
 The spread of the sampling distribution of x is smaller than the spread of the
corresponding population distribution, σσ <x .
 The standard deviation of the sampling distribution of x decreases as the
sample size increase.
 The standard deviation of the sample means is called the standard error of
the mean.
6.1.3 Sampling From a Normally Distributed Population
The criteria for sampling from a normally distributed population are:
• µµ =x
FORMULA
ξ∆Σ λϖ
β
xx
σ
FORMULA
ξ∆Σ λϖ
β
05.0≤
N
n
1−
−
N
nN

•
nx
σ
σ =
• The shape of the sampling distribution of x is normal, whatever the value of n.
o Shape of the sampling distribution
6.1.4 Sampling From a Not Normally Distributed Population

• Most of the time the population from which the samples are selected is not
normally distributed. In such cases, the shape of the sampling distribution of
x is inferred from central limit theorem.
Central limit theorem
 For a large sample size, the sampling distribution of x is approximately
normal, irrespective of the population distribution.
 The sample size is usually considered to be large if 30≥n .
 µµ =x

nx
σ
σ =
• Shape of the sampling distribution
6.1.5 Application of the Sampling Distribution of
The z value for a value of x is calculated as:
x

x
x
z
σ
µ−
=
In a study of the life expectancy of 500 people in a certain geographic region, the mean
age at death was 72 years and the standard deviation was 5.3 years. If a sample of 50
people from this region is selected, find the probability that the mean life expectancy will
be less than 70 years.
Solution:
Assume that the weights of all packages of a certain brand of cookies are normally
distributed with a mean of 32 ounces and a standard deviation of 0.3 ounce. Find the
FORMULA
ξ∆Σ λϖ
β
Example 3
Example 4

probability that the mean weight, x of a random sample of 20 packages of this brand of
cookies will be between 31.8 and 31.9 ounces.
Solution:
Although the sample size is small ( 30<n ), the shape of the sampling distribution of x is
normal because the population is normally distributed.

A Bulletin reported that children between the ages of 2 and 5 watch an average of 25 hours
of television per week. Assume the variable is normally distributed and the standard
deviation is 3 hours. If 20 children between the ages of 2 and 5 are randomly selected, find
the probability that the mean of the number of hours they watch television will be:
a) greater than 26.3 hours.
b) less than 24 hours
c) between 24 and 26.3 hours.
Solution:
Example 5

The average number of pounds of meat that a person consumes a year is 218.4 pounds.
Assume that the standard deviation is 25 pounds and the distribution is approximately
normal.
a) Find the probability that a person selected at random consumes less than 224
pounds per year.
b) If a sample of 40 individuals selected, find the probability that the mean of the
sample will be less than 224 pounds per year.
Solution:
Remember!
Sometimes you have difficulty deciding whether to use:
♦
x
x
z
σ
µ−
= ; should be used to gain information about a
sample mean.
OR
♦
σ
µ−
=
x
z ; used to gain information about an
individual data value obtained from the population.
Example 6

6.2 POPULATION AND SAMPLE PROPORTION
Population proportion Sample proportion
N
X
p =
n
x
p =ˆ
Where; N = total number of element in the population
n = total number of element in the sample
X = number of element in the population that possess a specific
characteristic
x = number of element in the sample that possess a specific
characteristic
Suppose a total of 789 654 families live in a city and 563 282 of them own homes. A sample
of 240 families is selected from the city and 158 of them own homes. Find the proportion of
families who own homes in the population and in the sample.
Solution:
The proportion of all families in this city who own homes is
71.0
789654
563282
===
N
X
p
The sample proportion is
66.0
240
158
===
n
x
p
6.2.1 Sampling Distribution of
FORMULA
ξ∆Σ λϖ
β
Example 7
pˆ
Example 8

Boe Consultant Associates has five employees. Table below gives the names of these
five employees and information concerning their knowledge of statistics.
Where: p = population proportion of employees who
know statistics
6.0
5
3
===
N
X
p
Let say we draw all possible samples of three
employees each and compute the proportion.
Total number of samples
=
5
C3 = 10
)!35(!3
!5
=
−
• Mean of the sample proportion:
pp =ˆµ
FORMULA
ξ∆Σ λϖ
β

• Standard deviation of the sample proportion:
n
pq
p =ˆσ
; if
05.0≤
N
n
1
ˆ
−
−
=
N
nN
n
pq
pσ ; if
05.0>
N
n
where p = population proportion
q = 1 – p
n = sample size
Based on Boe Consultant Associates in Example 8,
• Shape of the sampling distribution of pˆ
 According to the central limit theorem, the sampling distribution of pˆ is
approximately normal for a sufficiently large sample size.
 In the case of proportion, the sample size is considered to be sufficiently large if np
> 5 and nq > 5
FORMULA
ξ∆Σ λϖ
β
Example 9

A binomial distribution has p = 0.3. How large must sample size be such that a normal
distribution can be used to approximate sampling distribution of pˆ .
Solution:
6.2.2 Application of the Sampling Distribution of
z value for a value of pˆ
p
pp
z
ˆ
ˆ
σ
−
=
The Dartmouth Distribution Warehouse makes deliveries of a large number of products to its
customers. It is known that 85% of all the orders it receives from its customers are delivered
on time.
a) Find the probability that the proportion of orders in a random sample of 100 are
delivered on time:
i. less than 0.87
ii. between 0.81 and 0.88
Example 10
pˆ
FORMULA
ξ∆Σ λϖ
β
Example 11

b) Find the probability that the proportion of orders in a random sample of 100 are not
delivered on time greater than 0.1.
Solution:
The machine that is used to make CDs is known to produce 6% defective CDs. The quality
control inspector selects a sample of 100 CDs every week and inspects them for being good
or defective. If 8% or more of the CDs in the sample are defective, the process is stopped
and the machine is readjusted. What is the probability that based on a sample of 100 CDs
the process will be stopped to readjust the mashine?
Solution:
Example 12

6.3 OTHER TYPES OF SAMPLING DISTRIBUTIONS
6.3.1 Sampling Distribution for the Difference Between Means With Two
Independent Population
When independent random sample of n1 and n2 observations have been
selected from populations with means 1µ and 2µ and variance
2
1σ and
2
2σ

respectively, for the sampling distribution of the difference 1 2( )x x− has the
following properties:
• the mean of 1 2( )x x− is 1 2µ µ−
• the standard error of 1 2( )x x− is
2 2
1 2
1 2
1 2
s s
s s
n n
− = +
• If the sample populations are normally distributed then the sampling
distribution of 1 2( )x x− is exactly normally distributed, regardless of the
sample size.
• If the sample populations are not normally distributed then the sampling
distribution of 1 2( )x x− is approximately normally distributed when n1 and
n2 are both 30 or more, due to the Central Limit Theorem
6.3.2 Sampling Distribution for the Difference Between Proportions With
Two Independent Population
Assume that independence random samples of n1 and n2 observations have been
selected from binomial populations with parameters p1 and p2 respectively. The
sampling distribution of the difference between sample proportions
 
− = + ÷
 
1 2
1 2
1 2
ˆ ˆ( )
x x
p p
n n
has these properties;
• The mean of 1 2
ˆ ˆ( )p p− is p1 - p2
• The standard error is 1 1 2 2
1 2
ˆ ˆ ˆ ˆp q p q
n n
+
EXERCISE
1. Given a population with mean, µ = 400 and standard deviation, σ = 60.
a) If the population is normally distributed, what is the shape for the sampling
distribution of sample mean with random sample size of 16

b) If we do not know the shape of the population in 1(a), Can we answer 1(a)?
Explain.
c) Can we answer 1(a) if we do not know the population distribution but we have
been given random sample with size 36? Explain.
2. A random sample with size, n = 30, is obtained from a normal distribution population
with µ = 13 and σ = 7.
a) What are the mean and the standard deviation for the sampling distribution of
sample mean.
b) What is the shape of the sampling distribution? Explain.
c) Calculate
i) P ( x < 10)
ii) P ( x < 16)
3. Given a population size of 5000 with standard deviation 25, Calculate the standard
error of mean sample for:
a) n = 300
b) n = 100
4. Given X ~ N (5.55, 1.32). If a sample size of 50 is randomly selected, find the sampling
distribution for x . (Hint: Give the name of distribution, mean and variance).Then,
Calculate:
a) P ( 5.25 ≤ x ≤ 5.90)
b) P (5.45 ≤ x ≤ 5.75)
5. 64 units from a population size of 125 is randomly selected with mean 105 and
variance 289, Find:
a) the standard error of the sampling distribution above
b) P( 107.5 < x < 109.0)
6. The serving time for clerk at the bank counter is normally distributed with mean 8
minutes and standard deviation 2 minutes. If 36 customers is randomly selected:
a) Calculate xσ
b) The probability that the mean of serving time of a clerk at the bank counter is
between 7.7 minutes and 8.3 minutes
7. The workers at the walkie-talkie factory received salary at an average of RM3.70 per
hour and the standard deviation is RM0.80. If a sample of 100 is randomly selected,
find the probability the mean of sample is:
a) at least RM 3.50 per hour
b) between RM 3.20 and R3.60 per hour

8. 1,000 packs of pistachio nut have been sent to one of hyper supermarket in Puchong.
The weight of pistachio nut packs is normally distributed with mean 99.3g and standard
deviation is 1.8g.
a) If a random sample with 300 packs of pistachio nut is selected, find the
probability that the mean of the sample will be between 99.2g and 99.5g.
b) Find the probability that mean of sample 300 packs of pistachio nut is between
99.2g and 99.5g with delivery of
i) 2,000 packs
ii) 5,000 packs
c) What is the consequence of the incremental in population size toward the
probability value on the 9(b)?
9. An average age of 1500 staffs Tebrau Co. Limited is 38 years old with standard
deviation 6.2 years old. If the company selects 50 staffs at random,
a) Do we need correction factor in this situation? Justify your answer.
b) Find the probability of average age for the group of this staff is between 35
and 40 years old.
10. A test of string breaking strength that has been produced by Z factory shows that the
strength of string is only 60%. A random sample of 200 pieces of string is selected for
the test.
a) State the shape of sampling distribution
b) Calculate the probability of string strength is at lest 42%
11. Mr. Jay is a teacher at the Henry Garden School. He has conducted a research about
bully case at his school. 61.6% students said that they are ever being a bully victim. A
random sample of 200 students is selected at random. Find the proportion of bully
victim is
a) between 60% and 66%
b) more than 64%
12. The information given below shows the response of 40 college students for the
question, “Do you work during semester break time?” (The answer is Y=Yes or N=No).
N N Y N N Y N Y N Y N N Y N Y Y N N N Y
N Y N N N N Y N N Y Y N N N Y N N Y N N
If the proportion fo population is 0.30,

a) Find the proportion of sampling for the college student who works during
semester break.
b) Calculate the standard error for the proportion in (a).
13. A credit officer at the Tiger Bank believes that 25% from the total credit card users will
not pay their minimum charge of credit card debt at the end of every month. If a sample
100 credit card user is randomly selected:
a) What is the standard error for the proportion of the customer who does not
pay their minimum charge of credit card debt at the end of every month?
b) Find the probability that the proportion of customer in a random sample of 100
do not pay their minimum charge of credit card debt:
i) less than 0.20
ii) more than 0.30
ANSWER FOR EXERCISE
1. a. Normal distribution with μ = 400 and standard deviation, σ = 15
b. Cannot because the sampling distribution might be not normal
c. Can because based on the central limit theorem, the sampling distribution of mean sample will
normally distributed if the selected sample size is at least 30

2. a. 13xµ µ= = ;
7
1.28
30
xσ = =
b. The sampling distribution is normal because the population distribution is normal
c. i. P ( x < 10 ) =
10 13
P
1.28
z
− 
< ÷
 
= P ( z < -2.34)
= 0.0096
iii. P ( x < 16) =
16 13
P
1.28
z
− 
< ÷
 
= P ( z < 2.34)
= 0.9904
3.
a. n = 300
300
0.06( 0.05)
5000
25 5000 300
5000 1300
1.3995
x
n
N
σ
= = >
− 
=  ÷
− 
=
g
b. n = 100
100
0.02( 0.05)
5000
25
100
2.5
x
n
N
σ
= = <
=
=
4. Normal distribution ( n = 50; >30)
5.55xµ µ= = ;
1.32
0.187
50
xσ = =
a. P (5.25 < x < 5.90) =
5.25 5.55 5.90 5.55
P
0.187 0.187
z
− − 
< < ÷
 
= P (-1.60 < z < 1.87)
= P (z < 1.87) – P (z < -1.60)
= 0.9693 – 0.0548
= 0.9145
b. P (5.45 < x < 5.75) =
5.45 5.55 5.75 5.55
P
0.187 0.187
z
− − 
< < ÷
 
= P (-0.53 < z < 1.07)
= P (z < 1.07) – P (z < -0.53)
= 0.8577 – 0.2981
= 0.5596

5. a. 05.0512.0
125
64
>==
N
n
- need correction factor
17
2.125
64
xσ = =
b. P (107.5 < x < 109.0) =
107.5 105 109.0 105
P
2.125 2.125
z
− − 
< < ÷
 
= P (1.17 < z < 1.88)
= P (z < 1.88) – P (z < 1.18)
= 0.9699 – 0.8810
= 0.0889
6. a.
2
0.333
36
xσ = =
b. P ( 7.7 < x < 8.3) =
7.7 8.0 8.3 8.0
P
0.33 0.33
z
− − 
< < ÷
 
= P (-0.91 < z < 0.91)
= P (z < 0.91) – P (z < -0.91)
= 0.8186 – 0.1814
= 0.6372
7. 3.7xµ µ= =
0.8
0.08
100
xσ = =
a. P ( x > 3.5) =
3.5 3.7
P
0.08
z
− 
> ÷
 
= P (z > -2.5)
= 1 – P (z < -2.5)
= 1 – 0.0062
= 0.9938
b. P (3.2 < x < 3.6 ) =
3.2 3.7 3.6 3.7
P
0.08 0.08
z
− − 
< < ÷
 
= P (-6.25 < z < -1.25)
= P (z < -1.25) – P (z < -6.25)
= 0.1056 – 0
= 0.1056
8. 99.3xµ µ= =

05.03.0
1000
300
>==
N
n
1.8 1000 300
0.1539
1000 1300
xσ
−
= =
−
a. P ( 99.2 < x < 99.5) =
99.2 99.3 99.5 99.3
P
0.1539 0.1539
z
− − 
< < ÷
 
= P (-0.65 < z < 1.30)
= P (z < 1.30) – P (z < -0.65)
= 0.9032 – 0.2578
= 0.6454
b. i) N = 2000
05.015.0
2000
300
>==
N
n
1.8 2000 300
0.1696
2000 1300
xσ
−
= =
−
P ( 99.2 < x < 99.5) =
99.2 99.3 99.5 99.3
P
0.1696 0.1696
z
− − 
< < ÷
 
= P (-0.59 < z < 1.18)
= P (z < 1.18) – P (z < -0.59)
= 0.8810 – 0.2776
= 0.6034
ii) N = 5000
05.006.0
5000
300
>==
N
n
1.8 5000 300
0.1783
5000 1300
xσ
−
= =
−
P ( 99.2 < x < 99.5) =
99.2 99.3 99.5 99.3
P
0.178 0.178
z
− − 
< < ÷
 
= P (-0.56 < z < 1.12)
= P (z < 1.12) – P (z < -0.56)
= 0.8686 – 0.2877
= 0.5809
c. If the sample size is constant but the population size is increasing, the probability value
will be reduced.
9. a. 05.00333.0
1500
50
<==
N
n
- correction factor can be ignored
b.

( )
( )
( ) ( )
6.2
38, 0.8768
50
35 38 40 38
35 40
0.8768 0.8768
3.4215 2.2810
2.2810 3.4215
0.9887 0.0003
x x
n
P x P z
P z
P z P z
σ
µ σ= = = =
− − 
< < = < < ÷
 
= − < <
= < − < −
= −
0.9884=
10. a.
2
ˆ ˆ
np = 200(0.60) = 120 , nq = 200(0.40) = 80
0.55(0.45)
0.55, 0.0012
200
p pµ σ= = =
b.
0.40(0.60)
0.0346
200X
pq
n
σ = = =
( )
( )
0.42 0.40
0.42
0.0346
0.58
1 0.7190 0.5 0.2190
0.2810
P p P Z
P Z
− 
> = > ÷
 
= >
= − −
=
or
11 a.
( )
ˆ ˆ
(1 ) 0.616(0.384)
0.616, 0.0344
200
0.60 0.616 0.66 0.616
ˆ(0.60 0.66)
0.0344 0.0344
0.4651 1.2791
0.8997 0.3192
p p
n
P p P Z
P Z
π π
µ σ
−
= = = =
− − 
< < = < < ÷
 
= − < <
= −
0.5805=
b.
( )
( )
0.64 0.616
ˆ( 0.64)
0.0344
0.6977
1 0.6977
1 0.7580
0.2420
P p P Z
P Z
P Z
− 
> = > ÷
 
= >
= − ≤
= −
=
12. a.
14
ˆ 0.35
40
p = =
b.

ˆ
0.3(0.7)
0.0725
40
p
pq
n
σ = = =
13. a. ˆ
0.25(0.75)
0.043
100
pσ = =
b. i. P ( ˆp < 0.20) =
0.20 0.25
P
0.043
z
− 
< ÷
 
= P (z < -1.16)
= 0.1230
ii. P ( ˆp > 0.30) =
0.30 0.25
P
0.043
z
− 
> ÷
 
= P (z > 1.16)
= 1 – P (z < 1.16)
= 1 – 0.8770
= 0.1230

Sqqs1013 ch6-a122

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Sqqs1013 ch6-a122