Higher Engineering Mathematics

SIGNIFICANCE TESTING 591

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(ii) if the null hypothesis is rejected and the defect
rate rises to 5% or over, stop the machine and
adjust or renew parts as necessary; since the
machine is not then producing bolts, this will
reduce his profit.

These decisions may seem logical at first sight, but by
applying the statistical concepts introduced in previ-
ous chapters it can be shown that the manufacturer
is not necessarily making very sound decisions. This
is shown as follows.
When drawing a random sample of 200 bolts from
the machine with a defect rate of 3%, by the laws of
probability, some samples will contain no defective
bolts, some samples will contain one defective bolt,
and so on.
Abinomial distributioncan be used to determine
the probabilities of getting 0, 1, 2,..., 9 defective
bolts in the sample. Thus the probability of getting 10
or more defective bolts in a sample,even with a 3%
defect rate, is given by: 1−(the sum of probabilities
of getting 0, 1, 2,..., 9 defective bolts). This is an
extremely large calculation, given by:

1 −

(

0. 97200 + 200 × 0. 97199 × 0. 03

+

200 × 199

2

× 0. 97198 × 0. 032 to 10 terms

)

An alternative way of calculating the required prob-
ability is to use thenormal approximationto the
binomial distribution. This may be stated as follows:

‘if the probability of a defective item ispand a non-

defective item isq,then if a sample ofNitems is drawn

at random from a large population, provided bothNpand

Nqare greater than 5, the binomial distribution approxi-

mates to a normal distribution of meanNpand standard

deviation

√

(Npq)’

The defect rate is 3%, thusp= 0 .03. Sinceq= 1 −p,
q= 0 .97. Sample sizeN=200. SinceNpandNq
are greater than 5, a normal approximation to the
binomial distribution can be used.

The mean of the normal distribution,

x=Np= 200 × 0. 03 = 6

The standard deviation of the normal distribution

σ=

√

(Npq)

=

√

[(200)(0.03)(0.97)]= 2. 41

The normal standard variate for 10 bolts is

z=

variate−mean

standard deviation

=

10 − 6

2. 41

= 1. 66

Table 58.1 on page 561 is used to determine the

area between the mean and az-value of 1.66, and

is 0.4515.

The probability of having 10 or more defective

bolts is the total area under the standardised nor-

mal curve minus the area to the left of thez= 1. 66

ordinate, i.e. 1−(0. 5 + 0 .4515), i.e., 1− 0. 9515 =

0. 0485 ≈5%. Thus the probability of getting 10 or

more defective bolts in a sample of 200 bolts,even

though the defect rate is still 3%, is 5%. It follows

that as a result of the manufacturer’s decisions, for

5 times in every 100 the number of defects in the

sample will exceed 10, the alternative hypothesis

will be adopted and the machine will be stopped

(and profit lost) unnecessarily. In general terms:

‘a hypothesis has been rejected when it should have

been accepted’.

When this occurs, it is called atype I error, and, in

this example, the type I error is 5%.

Assume now that the defect rate has risen to 5%,

i.e. the expectancy of a defective bolt is now 10. A

second error resulting from this decisions occurs,

due to the probability of getting less than 10 defec-

tive bolts in a random sample, even though the defect

rate has risen to 5%. Using the normal approxima-

tion to a binomial distribution:N=200,p= 0 .05,

q= 0 .95.NpandNqare greater than 5, hence a nor-

mal approximation to a binomial distribution is a

satisfactory method. The normal distribution has:

mean,x=Np=(200)(0.05)= 10

standard deviation,

σ=

√

(Npq)

=

√

[(200)(0.05)(0.95)]= 3. 08

The normal standard variate for 9 defective bolts,

z=

variate−mean

standard deviation

=

9 − 10

3. 08

=− 0. 32