Higher Engineering Mathematics

594 STATISTICS AND PROBABILITY

Table 62.1

Level of Significance,α 0.1 0.05 0.01 0.005 0.002

z-value, one-tailed test { −1.28 −1.645 −2.33 −2.58 −2.88

or 1.28 or 1.645 or 2.33 or 2.58 or 2.88

z-value, two-tailed test { −1.645 −1.96 −2.58 −2.81 −3.08

and 1.645 and 1.96 and 2.58 and 2.81 and 3.08

The problem of the machine producing 3% defec-

tive bolts can now be reconsidered from a signi-

ficance testing point of view. A random sample of

200 bolts is drawn, and the manufacturer is interested

in a change in the defect rate in a specified direc-

tion (i.e. an increase), hence the hypotheses tests are

designed accordingly. If the manufacturer is willing

to accept a defect rate of 3%, but wants adjustments

made to the machine if the defect rate exceeds 3%,

then the hypotheses will be:

(i) a null hypothesis such that the defect rate,p,is

equal to 3%,

i.e.H 0 :p= 0 .03, and

(ii) an alternative hypothesis such that the defect

rate is greater than 3%,

i.e.H 1 :p> 0. 03

The first rule of decision is as follows: let the level
of significance,α, be 0.05; this will limit the type I
error, that is, the error due to rejecting the hypothe-
sis when it should be accepted, to 5%, which means
that the results are probably correct. The second
rule of decision is to decide the number of defec-
tive bolts in a sample for which the machine is
stopped and adjustments are made. For a one-tailed
test, a level of significance of 0.05 and the crit-
ical region lying to the right of the mean of the
standardised normal distribution, thez-value from
Table 62.1 is 1.645. If the defect ratepis 0.03%,
the mean of the normal distribution is given by
Np= 200 × 0. 03 =6 and the standard deviation
is

√

(Npq)=

√

(200× 0. 03 × 0 .97)= 2 .41, using

the normal approximation to a binomial distribu-

tion. Since thez-value is

variate−mean

standard deviation

, then

1. 645 =

variate− 6

2. 41

giving a variate value of 9.96.

This variate is the umber of defective bolts in a
sample such that when this number is reached or
exceeded the null hypothesis is rejected. For 95 times

out of 100 this will be the correct thing to do. The

second rule of decision will thus be ‘rejectH 0 if

the number of defective bolts in a random sample is

equal to or exceeds 10, otherwise acceptH′ 0. That is,

the machine is adjusted when the number of defec-

tive bolts in a random sample reaches 10 and this

will be the correct decision for 95% of the time. The

type II error can now be calculated, but there is lit-

tle point, since having fixed the sample number and

the level of significance, there is nothing that can be

done about it.

A two-tailed test is used when it is required to test

for changes in anunspecified direction. For exam-

ple, if the manufacturer of bolts, used in the previous

example, is inspecting the diameter of the bolts, he

will want to know whether the diameters are too large

or too small. Let the nominal diameter of the bolts

be 2 mm. In this case the hypotheses will be:

H 0 :d= 2 .00 mm H 1 :d= 2 .00 mm,

wheredis the mean diameter of the bolts. His first

decision is to set the level of significance, to limit

his type I error. A two-tailed test is used, since

adjustments must be made to the machine if the

diameter does not lie within specified limits. The

method of using such a significance test is given in

Section 62.3.

When determining the magnitude of type I and

type II errors, it is often possible to reduce the

amount of work involved by using a normal or

a Poisson distribution rather than binomial distri-

bution. A summary of the criteria for the use of

these distributions and their form is given below,

for a sample of sizeN, a probability of defective

componentspand a probability of non-defective

componentsq.

Binomial distribution

From Chapter 57, the probability of having 0, 1, 2,

3,...defective components in a random sample of

Ncomponents is given by the successive terms of