Values of alpha and beta are to be determined. Concept Introduction: Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 12, Problem 61AP

Summary Introduction

Interpretation:Values of alpha and beta are to be determined.

Concept Introduction:Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Expert Solution & Answer

Answer to Problem 61AP

The producers risk $α$ is 0.0862.

The consumers risk $(β)$ is 0.583

Explanation of Solution

Given information:

The AQL and LTPD are 5 percent and 10 percent respectively.

A and B supply the product D to company N in the lot sizes of 100 and 1,000 respectively. 10 percent chips are selected at random and the lot is rejected if two or more defective chips are found.

First, calculate the producers risk $(α)$ and $(β)$ of manufacturer A as follows:

Manufacturer A supplies the chips in the lot size of 100

10 percent of chips in a lot of 100 is the sample size, n = 10

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=10}$

$α=1−P{x≤1|p=0.05}$

$α=1−∑x=01(10x)(0.05)x(1−0.05)10−x$

$α=1−0.9138$

$α=0.0862$

The producers risk $α$ is 0.0862.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=1−∑x=01(100)(0.10)x(1−0.10)10−x$

$β=0.7361$

The consumers risk $(β)$ is 0.7381

The consumer risk is more than the producer risk for manufacturer A.

The producers risk $(α)$ and consumers risk $(β)$ for manufacturers A and B is calculated as follows:

Manufacturer A supplies the chips in the lot size of 1,000

10 percent of chips in a lot of 100 is the sample size, n = 100

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=100}$

$α={x≤1|θ=5}(poisson distribution application)$

$α=1−∑x=01e−55x!$

$α=1−0.0404$

$α=0.9596$

The producers risk $(α)$ is 0.9596.Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=P[X£1|θ=10](poisson distribution application)$

$β=∑x=01e−1010xx!$

$β=0.0005$

The consumers risk $(β)$ is0.0005.

The consumer risk is less than the producer risk for manufacturer B.

Reject the lot if more than 10 percent of the chips in a sample are defective.

Therefore, for manufacturer A, c = 1 and for manufacturer B, c = 10.

Manufacturer B is in a more advantageous position than A. The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥10|p=0.05,n=100}$

$α=1−P{x≤10|p=0.05}$

$α=0.0137$

The producers risk $(α)$ is 0.0137.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤10|p=0.10}$

$β=P{x≤10|p=0.10,n=100}$

$β=0.583$

The consumers risk $(β)$ is 0.583 Therefore, the consumer risk is more when the lot has more than 10 percent of the sample as defectives.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Learning Activity 6: Canadian Business This Year Read the following webpage: The Business Development Bank of Canada. (2023). Canadian economic outlook for 2024: Shifting into neutral. https://www.bdc.ca/en/articles- tools/blog/canadian-economic-outlook-for-2024- shifting-into-neutral "Despite persistently high inflation and rising. interest rates, the news was generally better than expected for the Canadian economy in 2023" (BDC Blog 2024). Discussion Question: In your view, what are the most pressing problems for Canadian companies or consumers in 2024? Explain your answer using current examples of companies or consumer concerns.

Travelling and working internationally can lead to a life of adventure and unique career experiences. For businesses, selecting the right candidates to take on foreign assignments can propel, delay, or deny the success of the international ventures. As an international manager, identify key competencies you would look for in choosing expatriates. What might be some of their concerns in taking on overseas assignments? What are some best practices in supporting expats during and after their assignments?

Research proposal: The Investigation of Career Development and Job satisfaction atEskom Rotek Industries 1. Introduction (250 words) 2. Research Context: Background (Research Context: Background (250 words) 3. The Research Problem (The Research Problem) 4. Aim of the study (Maximum 50 words) 5. Research Objectives (3 in total) 6. Research Questions (3 in total) 7. Justification or rationale of the study (150 words) 8. Literature Review (2-3 pages / 1000-1500 words) 9. Research Design and Methodology (1 - 2 page in total, from 9.1 to 9.6) 9.1 Research Philosophy 9.2 Research Methodology 9.3 Target Population and Sampling strategy 9.4 Data Collection 9.5 Data Analysis (100 words) 9.6 Pilot Study 10. Ethical considerations 11. Reference

Answer 2

Question

Chapter 12, Problem 61AP

Summary Introduction

Interpretation:Values of alpha and beta are to be determined.

Concept Introduction:Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Expert Solution & Answer

Answer to Problem 61AP

The producers risk $α$ is 0.0862.

The consumers risk $(β)$ is 0.583

Explanation of Solution

Given information:

The AQL and LTPD are 5 percent and 10 percent respectively.

A and B supply the product D to company N in the lot sizes of 100 and 1,000 respectively. 10 percent chips are selected at random and the lot is rejected if two or more defective chips are found.

First, calculate the producers risk $(α)$ and $(β)$ of manufacturer A as follows:

Manufacturer A supplies the chips in the lot size of 100

10 percent of chips in a lot of 100 is the sample size, n = 10

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=10}$

$α=1−P{x≤1|p=0.05}$

$α=1−∑x=01(10x)(0.05)x(1−0.05)10−x$

$α=1−0.9138$

$α=0.0862$

The producers risk $α$ is 0.0862.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=1−∑x=01(100)(0.10)x(1−0.10)10−x$

$β=0.7361$

The consumers risk $(β)$ is 0.7381

The consumer risk is more than the producer risk for manufacturer A.

The producers risk $(α)$ and consumers risk $(β)$ for manufacturers A and B is calculated as follows:

Manufacturer A supplies the chips in the lot size of 1,000

10 percent of chips in a lot of 100 is the sample size, n = 100

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=100}$

$α={x≤1|θ=5}(poisson distribution application)$

$α=1−∑x=01e−55x!$

$α=1−0.0404$

$α=0.9596$

The producers risk $(α)$ is 0.9596.Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=P[X£1|θ=10](poisson distribution application)$

$β=∑x=01e−1010xx!$

$β=0.0005$

The consumers risk $(β)$ is0.0005.

The consumer risk is less than the producer risk for manufacturer B.

Reject the lot if more than 10 percent of the chips in a sample are defective.

Therefore, for manufacturer A, c = 1 and for manufacturer B, c = 10.

Manufacturer B is in a more advantageous position than A. The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥10|p=0.05,n=100}$

$α=1−P{x≤10|p=0.05}$

$α=0.0137$

The producers risk $(α)$ is 0.0137.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤10|p=0.10}$

$β=P{x≤10|p=0.10,n=100}$

$β=0.583$

The consumers risk $(β)$ is 0.583 Therefore, the consumer risk is more when the lot has more than 10 percent of the sample as defectives.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 12, Problem 61AP

Summary Introduction

Interpretation:Values of alpha and beta are to be determined.

Concept Introduction:Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Expert Solution & Answer

Answer to Problem 61AP

The producers risk $α$ is 0.0862.

The consumers risk $(β)$ is 0.583

Explanation of Solution

Given information:

The AQL and LTPD are 5 percent and 10 percent respectively.

A and B supply the product D to company N in the lot sizes of 100 and 1,000 respectively. 10 percent chips are selected at random and the lot is rejected if two or more defective chips are found.

First, calculate the producers risk $(α)$ and $(β)$ of manufacturer A as follows:

Manufacturer A supplies the chips in the lot size of 100

10 percent of chips in a lot of 100 is the sample size, n = 10

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=10}$

$α=1−P{x≤1|p=0.05}$

$α=1−∑x=01(10x)(0.05)x(1−0.05)10−x$

$α=1−0.9138$

$α=0.0862$

The producers risk $α$ is 0.0862.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=1−∑x=01(100)(0.10)x(1−0.10)10−x$

$β=0.7361$

The consumers risk $(β)$ is 0.7381

The consumer risk is more than the producer risk for manufacturer A.

The producers risk $(α)$ and consumers risk $(β)$ for manufacturers A and B is calculated as follows:

Manufacturer A supplies the chips in the lot size of 1,000

10 percent of chips in a lot of 100 is the sample size, n = 100

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=100}$

$α={x≤1|θ=5}(poisson distribution application)$

$α=1−∑x=01e−55x!$

$α=1−0.0404$

$α=0.9596$

The producers risk $(α)$ is 0.9596.Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=P[X£1|θ=10](poisson distribution application)$

$β=∑x=01e−1010xx!$

$β=0.0005$

The consumers risk $(β)$ is0.0005.

The consumer risk is less than the producer risk for manufacturer B.

Reject the lot if more than 10 percent of the chips in a sample are defective.

Therefore, for manufacturer A, c = 1 and for manufacturer B, c = 10.

Manufacturer B is in a more advantageous position than A. The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥10|p=0.05,n=100}$

$α=1−P{x≤10|p=0.05}$

$α=0.0137$

The producers risk $(α)$ is 0.0137.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤10|p=0.10}$

$β=P{x≤10|p=0.10,n=100}$

$β=0.583$

The consumers risk $(β)$ is 0.583 Therefore, the consumer risk is more when the lot has more than 10 percent of the sample as defectives.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 12, Problem 61AP

Summary Introduction

Interpretation:Values of alpha and beta are to be determined.

Concept Introduction:Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Expert Solution & Answer

Answer to Problem 61AP

The producers risk $α$ is 0.0862.

The consumers risk $(β)$ is 0.583

Explanation of Solution

Given information:

The AQL and LTPD are 5 percent and 10 percent respectively.

A and B supply the product D to company N in the lot sizes of 100 and 1,000 respectively. 10 percent chips are selected at random and the lot is rejected if two or more defective chips are found.

First, calculate the producers risk $(α)$ and $(β)$ of manufacturer A as follows:

Manufacturer A supplies the chips in the lot size of 100

10 percent of chips in a lot of 100 is the sample size, n = 10

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=10}$

$α=1−P{x≤1|p=0.05}$

$α=1−∑x=01(10x)(0.05)x(1−0.05)10−x$

$α=1−0.9138$

$α=0.0862$

The producers risk $α$ is 0.0862.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=1−∑x=01(100)(0.10)x(1−0.10)10−x$

$β=0.7361$

The consumers risk $(β)$ is 0.7381

The consumer risk is more than the producer risk for manufacturer A.

The producers risk $(α)$ and consumers risk $(β)$ for manufacturers A and B is calculated as follows:

Manufacturer A supplies the chips in the lot size of 1,000

10 percent of chips in a lot of 100 is the sample size, n = 100

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=100}$

$α={x≤1|θ=5}(poisson distribution application)$

$α=1−∑x=01e−55x!$

$α=1−0.0404$

$α=0.9596$

The producers risk $(α)$ is 0.9596.Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=P[X£1|θ=10](poisson distribution application)$

$β=∑x=01e−1010xx!$

$β=0.0005$

The consumers risk $(β)$ is0.0005.

The consumer risk is less than the producer risk for manufacturer B.

Reject the lot if more than 10 percent of the chips in a sample are defective.

Therefore, for manufacturer A, c = 1 and for manufacturer B, c = 10.

Manufacturer B is in a more advantageous position than A. The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥10|p=0.05,n=100}$

$α=1−P{x≤10|p=0.05}$

$α=0.0137$

The producers risk $(α)$ is 0.0137.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤10|p=0.10}$

$β=P{x≤10|p=0.10,n=100}$

$β=0.583$

The consumers risk $(β)$ is 0.583 Therefore, the consumer risk is more when the lot has more than 10 percent of the sample as defectives.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12, Problem 61AP

Summary Introduction

Interpretation:Values of alpha and beta are to be determined.

Concept Introduction:Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Expert Solution & Answer

Answer to Problem 61AP

The producers risk $α$ is 0.0862.

The consumers risk $(β)$ is 0.583

Explanation of Solution

Given information:

The AQL and LTPD are 5 percent and 10 percent respectively.

A and B supply the product D to company N in the lot sizes of 100 and 1,000 respectively. 10 percent chips are selected at random and the lot is rejected if two or more defective chips are found.

First, calculate the producers risk $(α)$ and $(β)$ of manufacturer A as follows:

Manufacturer A supplies the chips in the lot size of 100

10 percent of chips in a lot of 100 is the sample size, n = 10

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=10}$

$α=1−P{x≤1|p=0.05}$

$α=1−∑x=01(10x)(0.05)x(1−0.05)10−x$

$α=1−0.9138$

$α=0.0862$

The producers risk $α$ is 0.0862.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=1−∑x=01(100)(0.10)x(1−0.10)10−x$

$β=0.7361$

The consumers risk $(β)$ is 0.7381

The consumer risk is more than the producer risk for manufacturer A.

The producers risk $(α)$ and consumers risk $(β)$ for manufacturers A and B is calculated as follows:

Manufacturer A supplies the chips in the lot size of 1,000

10 percent of chips in a lot of 100 is the sample size, n = 100

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=100}$

$α={x≤1|θ=5}(poisson distribution application)$

$α=1−∑x=01e−55x!$

$α=1−0.0404$

$α=0.9596$

The producers risk $(α)$ is 0.9596.Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=P[X£1|θ=10](poisson distribution application)$

$β=∑x=01e−1010xx!$

$β=0.0005$

The consumers risk $(β)$ is0.0005.

The consumer risk is less than the producer risk for manufacturer B.

Reject the lot if more than 10 percent of the chips in a sample are defective.

Therefore, for manufacturer A, c = 1 and for manufacturer B, c = 10.

Manufacturer B is in a more advantageous position than A. The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥10|p=0.05,n=100}$

$α=1−P{x≤10|p=0.05}$

$α=0.0137$

The producers risk $(α)$ is 0.0137.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤10|p=0.10}$

$β=P{x≤10|p=0.10,n=100}$

$β=0.583$

The consumers risk $(β)$ is 0.583 Therefore, the consumer risk is more when the lot has more than 10 percent of the sample as defectives.

Answer 6

Chapter 12, Problem 61AP

Summary Introduction

Interpretation:Values of alpha and beta are to be determined.

Concept Introduction:Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Expert Solution & Answer

Answer to Problem 61AP

The producers risk $α$ is 0.0862.

The consumers risk $(β)$ is 0.583

Explanation of Solution

Given information:

The AQL and LTPD are 5 percent and 10 percent respectively.

A and B supply the product D to company N in the lot sizes of 100 and 1,000 respectively. 10 percent chips are selected at random and the lot is rejected if two or more defective chips are found.

First, calculate the producers risk $(α)$ and $(β)$ of manufacturer A as follows:

Manufacturer A supplies the chips in the lot size of 100

10 percent of chips in a lot of 100 is the sample size, n = 10

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=10}$

$α=1−P{x≤1|p=0.05}$

$α=1−∑x=01(10x)(0.05)x(1−0.05)10−x$

$α=1−0.9138$

$α=0.0862$

The producers risk $α$ is 0.0862.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=1−∑x=01(100)(0.10)x(1−0.10)10−x$

$β=0.7361$

The consumers risk $(β)$ is 0.7381

The consumer risk is more than the producer risk for manufacturer A.

The producers risk $(α)$ and consumers risk $(β)$ for manufacturers A and B is calculated as follows:

Manufacturer A supplies the chips in the lot size of 1,000

10 percent of chips in a lot of 100 is the sample size, n = 100

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=100}$

$α={x≤1|θ=5}(poisson distribution application)$

$α=1−∑x=01e−55x!$

$α=1−0.0404$

$α=0.9596$

The producers risk $(α)$ is 0.9596.Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=P[X£1|θ=10](poisson distribution application)$

$β=∑x=01e−1010xx!$

$β=0.0005$

The consumers risk $(β)$ is0.0005.

The consumer risk is less than the producer risk for manufacturer B.

Reject the lot if more than 10 percent of the chips in a sample are defective.

Therefore, for manufacturer A, c = 1 and for manufacturer B, c = 10.

Manufacturer B is in a more advantageous position than A. The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥10|p=0.05,n=100}$

$α=1−P{x≤10|p=0.05}$

$α=0.0137$

The producers risk $(α)$ is 0.0137.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤10|p=0.10}$

$β=P{x≤10|p=0.10,n=100}$

$β=0.583$

The consumers risk $(β)$ is 0.583 Therefore, the consumer risk is more when the lot has more than 10 percent of the sample as defectives.

Answer 7

Chapter 12, Problem 61AP

Summary Introduction

Interpretation:Values of alpha and beta are to be determined.

Concept Introduction:Control charts are used to measure the effectiveness of the process determining the average value and control limits of the process. The Upper Control Limit (UCL) is the larger value and the Lower Control limit (LCL) is the smaller value of the sample. These limits gives a range to the data available.

Answer 8

Expert Solution & Answer

Answer to Problem 61AP

The producers risk $α$ is 0.0862.

The consumers risk $(β)$ is 0.583

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The AQL and LTPD are 5 percent and 10 percent respectively.

A and B supply the product D to company N in the lot sizes of 100 and 1,000 respectively. 10 percent chips are selected at random and the lot is rejected if two or more defective chips are found.

First, calculate the producers risk $(α)$ and $(β)$ of manufacturer A as follows:

Manufacturer A supplies the chips in the lot size of 100

10 percent of chips in a lot of 100 is the sample size, n = 10

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=10}$

$α=1−P{x≤1|p=0.05}$

$α=1−∑x=01(10x)(0.05)x(1−0.05)10−x$

$α=1−0.9138$

$α=0.0862$

The producers risk $α$ is 0.0862.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 10

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=1−∑x=01(100)(0.10)x(1−0.10)10−x$

$β=0.7361$

The consumers risk $(β)$ is 0.7381

The consumer risk is more than the producer risk for manufacturer A.

The producers risk $(α)$ and consumers risk $(β)$ for manufacturers A and B is calculated as follows:

Manufacturer A supplies the chips in the lot size of 1,000

10 percent of chips in a lot of 100 is the sample size, n = 100

A lot is rejected if the number of defects $X≥2$ , $p=pο$ and $p=p1$ .

The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥2|p=0.05,n=100}$

$α={x≤1|θ=5}(poisson distribution application)$

$α=1−∑x=01e−55x!$

$α=1−0.0404$

$α=0.9596$

The producers risk $(α)$ is 0.9596.Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤2$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤2|p=0.10}$

$β=1−P{x≤2|p=0.10,n=10}$

$β=P[X£1|θ=10](poisson distribution application)$

$β=∑x=01e−1010xx!$

$β=0.0005$

The consumers risk $(β)$ is0.0005.

The consumer risk is less than the producer risk for manufacturer B.

Reject the lot if more than 10 percent of the chips in a sample are defective.

Therefore, for manufacturer A, c = 1 and for manufacturer B, c = 10.

Manufacturer B is in a more advantageous position than A. The hypothesis of the test is

$pο$ Is Acceptable Quality Level (AQL) = 0.05

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≥10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(α)$ producers risk as follows:

$X≥c1$ , reject lot

$X≥2$ reject lot or else accept

Consider $p=pο=0.05$

$α=1−P{reject lot|p=pο}$

$α=1−P{x≥10|p=0.05,n=100}$

$α=1−P{x≤10|p=0.05}$

$α=0.0137$

The producers risk $(α)$ is 0.0137.

Now, calculate the consumers risk $(β)$ as follows:

$p1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

$c1$ Is the rejection level $≤10$

X is the number of defectives in the sample

n is the sample size of the first sample = 100

If the number of defectives in the sample is less than or equal to $c1$ the lot can be accepted or else r ejected. Apply the above values to find $(β)$ as follows:

$X≤c1$ , accept lot

$X≤2$ accepted lot or else reject

Consider $p=p1=0.10$

$β=P{x≤10|p=0.10}$

$β=P{x≤10|p=0.10,n=100}$

$β=0.583$

The consumers risk $(β)$ is 0.583 Therefore, the consumer risk is more when the lot has more than 10 percent of the sample as defectives.

Answer 12

Ch. 12.1 - Prob. 2P Ch. 12.1 - Prob. 3P Ch. 12.1 - Prob. 4P Ch. 12.1 - Prob. 5P Ch. 12.1 - Prob. 6P Ch. 12.2 - Prob. 7P Ch. 12.2 - Prob. 8P Ch. 12.2 - Prob. 9P Ch. 12.2 - Prob. 10P Ch. 12.2 - Prob. 11P

Videos

Answer to Problem 61AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions