Higher Engineering Mathematics

SIGNIFICANCE TESTING 599

J

For large samples, say, a minimum ofNbeing 30, the

factor

√(

N

N− 1

)

is

√

30

29

which is approximately

equal to 1.017. Thus, for large samplessis very

nearly equal toσˆand the factor

√(

N

N− 1

)

can be

omitted without introducing any appreciable error. In
equations (1) and (2),scan be written forσ, giving:

z=

x−μ

s

√

N

for infinite populations (5)

andz=

x−μ

s

√

N

√(

Np−N

Np− 1

) (6)

for populations of sizeNp

For small samples, the factor

√(

N

N− 1

)

cannot

be disregarded and substitutingσ=s

√(

N

N− 1

)

in

equations (3) and (4) gives:

|t|=

x−μ

s

√(

N

N− 1

)

√

N

=

(x−μ)

√

(N−1)

s

(7)

for infinite populations, and

|t|=

x−μ

s

√√

√√

(

N

N− 1

)

√

N

√(

Np−N

Np− 1

)

=

(x−μ)

√

(N−1)

s

√(

Np−N

Np− 1

) (8)

for populations of sizeNp.

The equations given in this section are parts of
tests which are applied to determine population
means. The way in which some of them are used
is shown in the following worked problems.

Problem 4. Sugar is packed in bags by an auto-

matic machine. The mean mass of the contents

of a bag is 1.000 kg. Random samples of 36

bags are selected throughout the day and the

mean mass of a particular sample is found to

be 1.003 kg. If the manufacturer is willing to

accept a standard deviation on all bags packed

of 0.01 kg and a level of significance of 0.05,

above which values the machine must be stopped

and adjustments made, determine if, as a result

of the sample under test, the machine should be

adjusted.

Population meanμ= 1 .000 kg, sample mean

x= 1 .003 kg, population standard deviation

σ=0.01 kg and sample size,N=36.

A null hypothesis for this problem is that the sam-

ple mean and the mean of the population are equal,

i.e.H 0 :x=μ.

Since the manufacturer is interested in deviations

on both sides of the mean, the alternative hypothesis

is that the sample mean is not equal to the population

mean, i.e.H 1 :x=μ.

The decision rules associated with these hypothe-

ses are:

(i) rejectH 0 if thez-value of the sample mean

is outside of the range of the z-values cor-

responding to a level of significance of 0.05

for a two-tailed test, i.e. stop machine and

adjust, and

(ii) accept H 0 otherwise, i.e. keep the machine

running.

The sample size is over 30 so this is a ‘large sample’

problem and the population can be considered to be

infinite. Because values ofx,μ,σandNare all

known, equation (1) can be used to determine the

z-value of the sample mean,

i.e.z=

x−μ

σ

√

N

=

1. 003 − 1. 000

0. 01

√

36

=±

0. 003

0. 0016

=± 1. 8

Thez-value corresponding to a level of significance

of 0.05 for a two-tailed test is given in Table 62.1

on page 594 and is±1.96. Since thez-value of the

sample is within this range,the null hypothesis is

accepted and the machine should not be adjusted.