EBK STATISTICS FOR BUSINESS & ECONOMICS
EBK STATISTICS FOR BUSINESS & ECONOMICS
12th Edition
ISBN: 9780100460461
Author: Anderson
Publisher: YUZU
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Chapter 19, Problem 24SE

a.

To determine

Explain about the given values of p0.

a.

Expert Solution
Check Mark

Answer to Problem 24SE

The fraction defective at which the producer would tolerate a probability of rejecting a good lot of 0.075 is 0.03.

Explanation of Solution

Calculation:

An acceptance sampling plan with n=15 and c=1 is given. The producer’s risk is 0.75.

Binomial probability function for acceptance sampling:

f(x)=n!x!(nx)!px(1p)(nx)

Here

n=the sample sizep=the porportion of defective items in the lotx=the number of defective items in the samplef(x)=the probability of x defective items in the sample

For p0=.01:

The probability of accepting the lot when p0=.01 is calculated as follows:

P(Accepting lot)=f(0)+f(1)=15!0!(150)!(0.01)0(10.01)(150)+15!1!(151)!(0.01)1(10.01)(151)=15!15!(0.99)15+15!14!(0.01)(0.99)14=0.8601+15(0.01)(0.8687)

                            =0.8601+0.1303=0.9904

Thus, the probability of accepting the lot when p0=.01 is 0.9904.

The producer’s risk, α, is obtained as follows:

α=10.9904=0.0096

Thus, the producer’s risk when p0=.01 is 0.0096.

For p0=.02:

The probability of accepting the lot when p0=.02 is calculated as follows:

P(Accepting lot)=f(0)+f(1)=15!0!(150)!(0.02)0(10.02)(150)+15!1!(151)!(0.02)1(10.02)(151)=15!15!(0.98)15+15!14!(0.02)(0.98)14=0.7386+15(0.02)(0.7536)

                            =0.7386+0.2261=0.9647

Thus, the probability of accepting the lot when p0=.02 is 0.9647.

The producer’s risk, α, is obtained as follows:

α=10.9647=0.0353

Thus, the producer’s risk when p0=.02 is 0.0353.

For p0=.03:

The probability of accepting the lot when p0=.03 is calculated as follows:

P(Accepting lot)=f(0)+f(1)=15!0!(150)!(0.03)0(10.03)(150)+15!1!(151)!(0.03)1(10.03)(151)=15!15!(0.97)15+15!14!(0.03)(0.97)14=0.6333+15(0.03)(0.6528)

                            =0.6333+0.2938=0.9271

Thus, the probability of accepting the lot when p0=.03 is 0.9271.

The producer’s risk, α, is obtained below:

α=10.9271=0.0729

Thus, the producer’s risk when p0=.03 is 0.0729.

For p0=.04:

The probability of accepting the lot when p0=.04 is calculated as follows:

P(Accepting lot)=f(0)+f(1)=15!0!(150)!(0.04)0(10.04)(150)+15!1!(151)!(0.04)1(10.04)(151)=15!15!(0.96)15+15!14!(0.04)(0.96)14=0.5421+15(0.04)(0.5647)

                            =0.5421+0.3388=0.8809

Thus, the probability of accepting the lot when p0=.04 is 0.8809.

The producer’s risk, α, is obtained as follows:

α=10.8809=0.1191

Thus, the producer’s risk when p0=.04 is 0.1191.

For p0=.05:

The probability of accepting the lot when p0=.05 is calculated as follows:

P(Accepting lot)=f(0)+f(1)=15!0!(150)!(0.05)0(10.05)(150)+15!1!(151)!(0.05)1(10.05)(151)=15!15!(0.95)15+15!14!(0.05)(0.95)14=0.4633+15(0.05)(0.4877)

                            =0.4633+0.3658=0.8291

Thus, the probability of accepting the lot when p0=.05 is 0.8291.

The producer’s risk, α, is obtained as follows:

α=10.8291=0.1709

Thus, the producer’s risk when p0=.05 is 0.1709.

Here, the producer’s risk obtained using p0=.03 is approximately equal to 0.075.

Therefore, the fraction defective at which the producer would tolerate a probability of rejecting a good lot of 0.075 is 0.03.

b.

To determine

Find the consumer’s risk associated with this plan when p1=0.25.

b.

Expert Solution
Check Mark

Answer to Problem 24SE

The consumer’s risk when p1=.25 is 0.0802.

Explanation of Solution

Calculation:

The probability of accepting the lot when p1=.25 is calculated as follows:

P(Accepting lot)=f(0)+f(1)=15!0!(150)!(0.25)0(10.25)(150)+15!1!(151)!(0.25)1(10.25)(151)=15!15!(0.75)15+15!14!(0.25)(0.75)14=0.0134+15(0.25)(0.0178)

                            =0.0134+0.0668=0.0802

Thus, the probability of accepting the lot when p1=.25 is 0.0802.

The consumer’s risk, β, is β=0.0802.

Thus, the consumer’s risk when p1=.25 is 0.0802.

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Chapter 19 Solutions

EBK STATISTICS FOR BUSINESS & ECONOMICS

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