Example:
Suppose we sample 500 times and let x 1
1
, x 1
2
,  , x 1
30
 x
1
2
, x
2
2
,  ,
2 x
30
 x x
2
1
    x
1
500 , x 500
2
,  , x 500

30 x
500 be 500 simple random samples of 30 managers. Let X be the random variable representing the average salary of a random sample of 30 managers.
Then, x
1
, x
2
,  , x
500
are 500 possible values of X . Note there are .


2500
30

 possible x for random variable X .
Properties of
X
:
Let X be the random variable representing the average of a random sample. Then,
1.
E ( X )
   the population mean
2.
For finite population with population size N and the sample size n, then the standard deviation of X is
Var ( X )

E ( X
 
)
2  
X

N
N
 n

1
 n
.
For infinite population (the infinite population size),

Var ( X )
  
X n
.
Note: as n
N

0 .
05 , then
N
N
 n

1

1 . Thus,

X

N
N
 n

1
 n

 n
. That is, the standard deviation of X for the finite population is approximately equal to the one for the infinite population.
1

Sampling Distribution of
X
:
As n

30 or the population is normally distributed, then
X

N (

,

2
X
)

2
, where

2
X

. Thus, for some constants    c
 d
 
, n
P ( c

X
 d )

P ( c
 

P ( c


X



X
Z


  d
 
)

P ( c


X
 d


X

),

X


X

 d


X

) where Z is the standard normal random variable.
Example:
What is the probability of the difference between the sample mean and the population mean will be less or equal to 500 as the sample size n

30 ?
[solution]

X

51800 ,

X

 n

4000
30

730 .
30 . Thus,

N ( 51800 , ( 730 .
30 )
2
)

X

51800
730 .
30

Z
 the standard normal random variable

P ( X

51800

500 )

P (

500

X

51800

500 )


P (
P (

500
730 .
30

0 .
68


Z
X

51800

730 .
30
0 .
68 )


500
730
0 .
5036
.
30
)

There is 50.36% chance that the difference between the sample mean and the population mean is not more than 500.
Sample Size and Sampling Distribution:
Since
 
X
 n
, increasing the sample size will decrease the standard error!! Thus, the larger the sample size is, the larger P ( c

X
 d ) is
2

(since the interval


 c


X

, d


X
 


is larger than the one with smaller sample size)!!
Example (continue):
What is the probability of the difference between the sample mean and the population mean will be less or equal to 500 as the sample size n

100 ?
[solution]
 
X
51800 ,

X

 n

4000

400

730 .
30 ( n

30 ) . Thus,
100

N ( 51800 , ( 400 )
2
)

X

51800
400

Z
 the standard normal random variable

P ( X

51800

500 )

P ( 51300

X

52300 )


51300

51800
P (
P (

1 .
25
400

Z

1 .

25 )
X


51800
400
0 .
7888


52300

51800
0 .
5036
400
( n

30 )
)

As the sample size increases to 100, there is 78.88% chance that the difference between the sample mean and the population mean is not more than 500. That is, the larger sample size will provide a higher probability that the value of the sample mean will be within a specific distance of the population mean.
Example:
The average lifetime of a light bulb is 3000 hours with a standard deviation of 696 hours. A simple random sample of 36 bulbs is taken.
(a) What are the expected value, standard deviation, and shape of the sampling distribution of x ?
(b) What is the probability that the average life time in the sample will be between
2670.56 and 2809.76 hours?
(c) What is the probability that the average life time in the sample will be equal to or greater than 3219.24 hours?
(d) What is the probability that the average life time in the sample will be equal to or less than 3180.96 hours?
(e) How large of a sample needs to be taken to provide a 0.01 probability that the average life time in the sample will be equal to or greater than 3219.24 hours
3

[solution:]
(a)
E
 
  
3000 ,

X

 n

696
36

116 , bell shape like normal distribution.
(b)
P

2670 .
56

X

2809 .
76




P

P


2670 .
56
2 .
84


116
Z
3000



1 .
64
X

 



X
0 .
0482
X

3000

116
2809 .
76

3000

116
(c)

P

3219 .
24
P

1 .
89


X


P


3219 .
24

116
Z


0 .
0294
3000

X


X


X

3000
116


( d)

P

X
P

Z


3180 .
96


P


X
1 .
56


0 .
9406


X


X

3000

116
3180 .
96

3000
116


(e)
0 .
01

P

3219 .
24

X


P


3219 .
24

X

3000


P


219 .
24

X

Z



219 .
24

X
 z
0 .
01

2 .
33
 
X

 n
X


X


696
 n
219 .
24
2 .
33

 n
 n


696

2 .
33
219 .
24

55
2

54 .
71
X


3000


X
4

Example :
MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 25 is taken from a population of 500.
(a) What is the probability that the average aptitude test score in the sample will be between 70.14 and 82.14?
(b) What is the probability that the average aptitude test score in the sample will be equal to or greater than 82.68?
(c) Find a value, C, such that P( X 
C) = .015.
[solution:]

(a) Since

X
 n
P

70 .
14


15
X

3 ,
25

82 .
14



P

70 .
14
P


1 .
62
3


75
Z


2 .
X

 

X

3
38
X


0 .
9387
75

82 .
14

75

3
(b)
P


82 .
68
P

2 .
56


X


Z


P


82 .
68
3

75
0 .
0052

X


X


X

3
75


( c)
P

X

C


P


X


X


X

75
3

C

3
75



P

C

75
3

2 .
17

C

75

2 .
17

3

81 .
51
Z

C

75
3

0 .
015
Example:
A population has a standard deviation of 16.
(a) As the sample size be within n

100 , what is the probability that a sample mean will

2 of the population mean (the sampling error

2 )?
(b) Suppose the probability that a sample mean will be within

2 of the population mean is 0.9. What is the sample size n ?
[solution:]
5

(a)

X

 n

16
100

1 .
6

X

N


,

2
X
 

, 1 .
6
2


X


X


P

X
  
2



P


2
2 P

0


Z
X
  

1 .
25


2


P

2
1 .
6

X


X

2

0 .
3944

0 .
7888

2
1 .
6

X

1 .
6


Z

N

P


1 .
25

Z

1 .
25

(b)
X


X


Z

N

P

X
  
2


P
X


X



2
X

P

 Z


2
X



P




X
2

Z


2
X



2 P

0

Z


2
X


0 .
9

P

 0

Z


2
X
 n


8

1 .
645

2



0 .
45


2
X


2 n

173 .
1856
 n

174

2
16 n

1 .
645
 n

8

1 .
645
Properties of
P
:
Let P be the random variable representing the proportion of a random sample (the sample proportion of a random sample is one possible value of P ). Then,
1.
E ( P )
 p
 the population proportion
2.
For finite population with population size N and the sample size n, then the standard deviation of P is
Var ( P )

E ( P
 p )
2  
P

N
N
 n

1 p ( 1 n
 p )
.
For infinite population (the infinite population size),
Var ( P )
 
P
 p ( 1
 n p )
.
6

Note:
As n
N

0 .
05 , then
N
N
 n

1

1 . Thus, 
P

N
N
 n

1 p ( 1
 p )
 n p ( 1
 p )
. n
That is, the standard deviation of P for the finite population is approximately equal to the one for the infinite population.
Sampling Distribution of
P
:
As np

5 and n ( 1
 p )

5 ,
P

N ( p ,

2
P
)
, where

P
2  p ( 1
 n p )
. Thus, for some constants 0
 c
 d

1
P ( c

P
 d )

P ( c
 p

P
 p
 d
 p )

P ( c


P p


P ( c


P p

Z
 d


P p
),
P


P p
 d


P p
) where Z is the standard normal random variable.
Note: Since

P
 p ( 1
 p )
, increasing the sample size will decrease n the standard error!! Thus, the larger the sample size is, the larger
P ( c

P
 d ) is (since the interval


 c


P p
, d


P p



is larger than the one with smaller sample size)!!
Example (continue):
What is the probability of the difference between the sample proportion and the population proportion will be less or equal to 0.05 as the sample size n

30 ?
What is the probability as we increase the sample size to 100?
[solution] p

0 .
6 ,

P

0 .
6 ( 1

0 .
6 )

0 .
0894 . Thus,
30
7

P

N ( 0 .
6 , ( 0 .
0894 )
2
)

P

0 .
6
0 .
0894

Z
 the standard normal random variable

P ( P

0 .
6

0 .
05 )

P (

0 .
05

P

0 .
6

0 .
05 )


P (
P (

0 .
05
0 .
0894

0 .
56


Z
P

0 .
6
0 .
0894

0 .
56 )


0 .
05
0 .
0894
)
0 .
4246

There is 42.46% chance that the difference between the sample proportion and the population proportion is not more than 0.05 as .
n

30 .
As sample size is increased to 100, then p

0 .
6 ,
P

P

0 .
6 ( 1

0 .
6 )

0 .
0490

0 .
0894 ( n

30 ).
Thus,
100

N ( 0 .
6 , ( 0 .
0490 )
2
)

P

0 .
6
0 .
0490

Z
 the standard normal random variable

P ( P

0 .
6

0 .
05 )

P ( 0 .
55

P

0 .
65 )


0 .
55

0 .
6
P (
P (

0 .
0490
1 .
02

Z


P

0 .
0490
1 .
02 )
0 .
6

0

.
0 .
65
6922


0
0 .
0490
0
.
6
)
.
4246 ( n

30 )

There is 69.22% chance that the difference between the sample proportion and the population proportion is not more than 0.05 n

100 . That is, the larger sample size will provide a higher probability that the value of the sample proportion will be within a specific distance of the population proportion.
Example:
A new soft drink is being market tested. It is estimated that 60% of consumers will like the new drink. A sample of 96 taste-tested the new drink.
(a) Determine the standard error of the proportion
(b)What is the probability that equal to or more than 70.4% of consumers will indicate they like the drink?
(c) What is the probability that equal to or more than 30% of consumers will indicate they do not like the drink?
[solution:]
(a)

P

0 .
6

0 .
4
96

0 .
05
(b)
P

0 .
704

P



P

0 .
704
0 .

05
0 .
6

P

2 .
08


P
Z

 p

P

0 .
6
0 .
05

P

0 .
0188

8

(c) We need to compute the probability that less than 70% of consumers will indicate they like the drink?
P

P

0 .
7



P

P


P p


P
P

Z


0 .
6
0 .
05
2



0 .
7

0 .
6
0 .
05
0 .
9772

Example:
What is the most important factor for business travelers when they are staying in a hotel? According to USA Today, 74% of business travelers state that having a smoke-free room is the most important factor. Assume that the population proportion is p

0 .
74 and that a sample of 200 business travelers will be selected.
(a) What is the probability that the sample proportion will be within

.
04 of the population proportion?
(b) Suppose the probability that a sample proportion will be within

.
02 of the population mean is 0.9. What is the sample size n ?
[solution:]
(a)

P


P


P p
 p
P

1
 p n

0 .
74
0 .
031



Z
0 .
74

0 .
26

200
N

0 .
031

P

N
 p ,

2
P


N

0 .
74 , 0 .
031
2


P

P
 p

0 .
04



P
2 P

0
P


P

Z p


P

0 .
74
0 .
031
1 .
29


2

0 .
04
0 .
031

0 .
4015


P

0 .
8030
Z

1 .
29

(b)
P


P p

Z

N

P

P
 p

0 .
02


P
P


P p

0 .
02

P

P

Z

2 P

 0

Z

0 .
02



P

0 .
9

P

 0

Z

0 .
02



P

0 .
45

0 .
02


P

P
 
0 .
02

P

Z

0 .
02


P

0 .
02

P
 p
0 .
02

1
 p
 n

0 .
74
0 .

1
02

0 .
74
 n

0 .
02 n
0 .
74

0 .
26
 n
 n


1 .
645

164
2
.
5
0 .
74
2
0 .
02

0 .
26

0 .
74

0 .
26

1301 .
598
 n

1302

1 .
645
9

Online Exercise:
Exercise 9.2.1
Exercise 9.2.2
10