Higher Engineering Mathematics

578 STATISTICS AND PROBABILITY

mean values will also be small, since it depends on

the distance of the mean values from the distribution

mean. The relationship between the standard devi-

ation of the mean values of a sampling distribution

and the number in each sample can be expressed as

follows:

Theorem 1 ‘If all possible samples of sizeNare
drawn from a finite population,Np, without replace-
ment, and the standard deviation of the mean values
of the sampling distribution of means is determined
then:

σx=

σ

√

N

√(

Np−N

Np− 1

)

whereσxis the standard deviation of the sampling

distribution of means andσis the standard deviation

of the population’.

The standard deviation of a sampling distribution
of mean values is called thestandard error of the
means, thus

standard error of the means,

σx=

σ

√

N

√(

Np−N

Np− 1

)

(1)

Equation (1) is used for a finite population of size

Npand/or for sampling without replacement. The

word ‘error’ in the ‘standard error of the means’

does not mean that a mistake has been made but

rather that there is a degree of uncertainty in pre-

dicting the mean value of a population based on

the mean values of the samples. The formula for

the standard error of the means is true for all val-

ues of the number in the sample,N. WhenNpis

very large compared withNor when the popula-

tion is infinite (this can be considered to be the

case when sampling is done with replacement), the

correction factor

√(

Np−N

Np− 1

)

approaches unit and

equation (1) becomes

σx=

σ

√

N

(2)

Equation (2) is used for an infinite population and/or

for sampling with replacement.

Problem 1. Verify Theorem 1 above for the

set of numbers{3, 4, 5, 6, 7}when the sample

size is 2.

The only possible different samples of size 2 which

can be drawn from this set without replacement are:

(3, 4), (3, 5), (3, 6), (3, 7), (4, 5),

(4, 6), (4, 7), (5, 6), (5, 7) and (6, 7)

The mean values of these samples form the following

sampling distribution of means:

3 .5, 4, 4.5, 5, 4.5, 5, 5.5, 5.5, 6 and 6. 5

The mean of the sampling distributions of means,

μx=

(

3. 5 + 4 + 4. 5 + 5 + 4. 5 + 5

+ 5. 5 + 5. 5 + 6 + 6. 5

)

10

=

50

10

= 5

The standard deviation of the sampling distribution

of means,

σx=

√ √ √ √ √ √ √

⎡

⎢

⎢

⎣

(3. 5 −5)^2 +(4−5)^2 +(4. 5 −5)^2

+(5−5)^2 +···+(6. 5 −5)^2

10

⎤

⎥

⎥

⎦

=

√

7. 5

10

=± 0. 866

Thus,the standard error of the means is 0.866.

The standard deviation of the population,

σ=

√√

√

√

√

√

√

⎡

⎢

⎢

⎣

(3−5)^2 +(4−5)^2 +(5−5)^2

+(6−5)^2 +(7−5)

5

⎤

⎥

⎥

⎦

=

√

2 =± 1. 414

But from Theorem 1:

σx=

σ

√

N

√(

Np−N

Np− 1

)

and substituting forNp,Nandσin equation (1) gives:

σx=

± 1. 414

√

2

√(

5 − 2

5 − 1

)

=

√

3

4

=± 0 .866,

as obtained by considering all samples from the

population. Thus Theorem 1 is verified.

In Problem 1 above, it can be seen that the mean of

the population,

(

3 + 4 + 5 + 6 + 7

5

)