Higher Engineering Mathematics

596 STATISTICS AND PROBABILITY

10% is (area to the left of mean−area between mean

and az-value of−0.37), i.e. 0. 5 − 0. 1443 = 0 .3557.

It is usual to express type II errors as a percentage,

giving

type II error= 35 .6%

Problem 2. The sample size in Problem 1 is

reduced to 50. Determine the type I error if the

defect rate remains at 7% and the type II error

when the defect rate rises to 9%. The decision

is now to stop the machine for adjustment if a

sample contains 4 or more defective screws.

N=50, p= 0. 07

WhenN≥50 andNp<5, the Poisson approxima-
tion to a binomial distribution is used. The expecta-
tionλ=Np= 3 .5. The probabilities of 0, 1, 2, 3,...

defective screws are given by e−λ,λe−λ,

λ^2 e−λ

2!

,

λ^3 e−λ

3!

,...Thus,

probability of a sample containing

no defective screws, e−λ=0.0302

probability of a sample containing

1 defective screw, λe−λ=0.1057

probability of a sample containing

2 defective screws,

λ^2 e−λ

2!

=0.1850

probability of a sample containing

3 defective screws,

λ^3 e−λ

3

=0.2158

probability of a sample containing

0, 1, 2, or 3 defective screws is 0.5367

Hence, the probability of a sample containing 4 or

more defective screws is 1− 0. 5367 = 0 .4633. Thus

thetype I error, that is, rejecting the hypothesis

when it should be accepted or stopping the machine

for adjustment when it should continue running,

is46.3%.

When the defect rate has risen to 9%,p= 0 .09 and

Np=λ= 4 .5. SinceN≥50 andNp<5, the Poisson

approximation to a binomial distribution can still be

used. Thus,

probability of a sample containing

no defective screws, e−λ=0.0111

probability of a sample containing

1 defective screw, λe−λ=0.0500

probability of a sample containing

2 defective screws,

λ^2 e−λ

2!

=0.1125

probability of a sample containing

3 defective screws,

λ^3 e−λ

3!

=0.1687

probability of a sample containing

0, 1, 2, or 3 defective screws is 0.3423

That is, the probability of a sample containing less

than 4 defective screws is 0.3423. Thus, thetype

II error, that is, accepting the hypothesis when it

should have been rejected or leaving the machine

running when it should be stopped, is34.2%.

Problem 3. The sample size in Problem 1 is

now reduced to 25. Determine the type I error

if the defect rate remains at 7%, and the type II

error when the defect rate rises to 10%. The deci-

sion is now to stop the machine for adjustment

if a sample contains 3 or more defective screws.

N=25, p= 0 .07, q= 0. 93

The criteria for a normal approximation to a binomial

distribution and for a Poisson approximation to a

binomial distribution are not met, hence the binomial

distribution is applied.

Probability of no defective screws in a sample,

qN= 0. 9325 =0.1630

Probability of 1 defective screw in a sample,

NqN−^1 p= 25 × 0. 9324 × 0. 07 =0.3066

Probability of 2 defective screws in a sample,

N(N−1)

2

qN−^2 p^2

=

25 × 24

2

× 0. 9323 × 0. 072 =0.2770

Probability of 0, 1, or 2 defective screws

in a sample =0.7466

Thus, the probability of a type I error, i.e. stop-

ping the machine even though the defect rate is still

7%, is 1− 0. 7466 = 0 .2534. Hence, thetype I error

is 25.3%.

When the defect rate has risen to 10%:

N=25, p= 0 .1, q= 0. 9