Higher Engineering Mathematics

SIGNIFICANCE TESTING 595

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the expansion of (q+p)N, taken from the left. Thus:

Number of Probability

defective

components

0 qN

1 NqN−^1 p

2

N(N−1)

2!

qN−^2 p^2

3

N(N−1)(N−2)

3!

qN−^3 p^3 ...

Poisson approximation to a binomial distribution

WhenN≥50 andNp<5, the Poisson distribution is
approximately the same as the binomial distribution.
In the Poisson distribution, the expectationλ=Np
and from Chapter 57, the probability of 0, 1, 2, 3,...
defective components in a random sample of N
components is given by the successive terms of

e−λ

(

1 +λ+

λ^2

2!

+

λ^3

3!

+···

)

taken from the left. Thus,

Number of defective 01 2 3

components

Probability e−λ λe−λ

λ^2 e−λ

2!

λ^3 e−λ

3!

Normal approximation to a binomial distribution

When bothNpandNqare greater than 5, the nor-
mal distribution is approximately the same as the
binomial distribution, The normal distribution has a
mean ofNpand a standard deviation of

√

(Npq).

Problem 1. Wood screws are produced by

an automatic machine and it is found over a

period of time that 7% of all the screws pro-

duced are defective. Random samples of 80

screws are drawn periodically from the output of

the machine. If a decision is made that produc-

tion continues until a sample contains more than

7 defective screws, determine the type I error

based on this decision for a defect rate of 7%.

Also determine the magnitude of the type II error

when the defect rate has risen to 10%.

N=80, p= 0 .07, q= 0. 93

Since bothNpandNqare greater than 5, a normal

approximation to the binomial distribution is used.

Mean of the normal distribution,

Np= 80 × 0. 07 = 5. 6

Standard deviation of the normal distribution,

√

(Npq)=

√

(80× 0. 07 × 0 .93)= 2. 28

Atype I erroris the probability of rejecting a

hypothesis when it is correct, hence, the type I error

in this problem is the probability of stopping the

machine, that is, the probability of getting more than

7 defective screws in a sample, even though the

defect rate is still 7%. Thez-value corresponding

to 7 defective screws is given by:

variate−mean

standard deviation

=

7 − 5. 6

2. 28

= 0. 61

Using Table 58.1 of partial areas under the stan-

dardised normal curve given on page 561, the area

between the mean and az-value of 0.61 is 0.2291.

Thus, the probability of more than 7 defective screws

is the area to the right of thezordinate at 0.61, that is,

[total area−(area to the left of mean

+area between mean andz= 0 .61)]

i.e. 1−(0. 5 + 0 .2291). This gives a probability of

0.2709. It is usual to express type I errors as a

percentage, giving

type I error= 27 .1%

Atype II erroris the probability of accepting a

hypothesis when it should be rejected. The type II

error in this problem is the probability of a sample

containing less than 7 defective screws, even though

the defect rate has risen to 10%. The values are now:

N=80, p= 0 .1, q= 0. 9

AsNpandNqare both greater than 5, a normal

approximation to a binomial distribution is used, in

which the meanNpis 80× 0. 1 =8 and the standard

deviation

√

(Npq)=

√

(80× 0. 1 × 0 .9)= 2 .68.

Thez-value for a variate of 7 defective screws is

7 − 8

2. 68

=− 0 .37.

Using Table 58.1 of partial areas given on

page 561, the area between the mean andz=− 0. 37

is 0.1443. Hence, the probability of getting less than

7 defective screws, even though the defect rate is