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 to Problem 61AP

The producers risk $\alpha$ is 0.0862.

The consumers risk $\left(\beta \right)$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 $\left(\alpha \right)$ and $\left(\beta \right)$ 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 $X\ge 2$ , $p=p_o$ and $p=p_1$ .

The hypothesis of the test is

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

  $p_1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

  $c_1$ Is the rejection level $\ge 2$

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 $c_1$ the lot can be accepted or else r ejected. Apply the above values to find $\left(\alpha \right)$ producers risk as follows:

  $X\ge c_1$ , reject lot

  $X\ge 2$ reject lot or else accept

Consider $p=p_o=0.05$

  $\alpha =1-P\left\{rejectlot|p=p_o\right\}$

  $\alpha =1-P\left\{x\ge 2|p=0.05,n=10\right\}$

  $\alpha =1-P\left\{x\le 1|p=0.05\right\}$

  $\alpha =1-{\displaystyle \sum _{x=0}^1\left(\begin{array}{l}10\\ x\end{array}\right){\left(0.05\right)}^x{\left(1-0.05\right)}^{10-x}}$

  $\alpha =1-0.9138$

  $\alpha =0.0862$

The producers risk $\alpha$ is 0.0862.

Now, calculate the consumers risk $\left(\beta \right)$as follows:

  $p_1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

  $c_1$ Is the rejection level $\le 2$

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 $c_1$ the lot can be accepted or else r ejected. Apply the above values to find $\left(\beta \right)$ as follows:

  $X\le c_1$ , accept lot

  $X\le 2$ accepted lot or else reject

Consider $p=p_1=0.10$

  $\beta =P\left\{x\le 2|p=0.10\right\}$

  $\beta =1-P\left\{x\le 2|p=0.10,n=10\right\}$

  $\beta =1-{\displaystyle \sum _{x=0}^1\left(\begin{array}{l}10\\ 0\end{array}\right){\left(0.10\right)}^x{\left(1-0.10\right)}^{10-x}}$

  $\beta =0.7361$

The consumers risk $\left(\beta \right)$ is 0.7381

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

The producers risk $\left(\alpha \right)$and consumers risk $\left(\beta \right)$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 $X\ge 2$ , $p=p_o$ and $p=p_1$ .

The hypothesis of the test is

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

  $p_1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

  $c_1$ Is the rejection level $\ge 2$

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 $c_1$ the lot can be accepted or else r ejected. Apply the above values to find $\left(\alpha \right)$ producers risk as follows:

  $X\ge c_1$ , reject lot

  $X\ge 2$ reject lot or else accept

Consider $p=p_o=0.05$

  $\alpha =1-P\left\{rejectlot|p=p_o\right\}$

  $\alpha =1-P\left\{x\ge 2|p=0.05,n=100\right\}$

  $\alpha =\left\{x\le 1|\theta =5\right\}\left(poissondistributionapplication\right)$

  $\alpha =1-{\displaystyle \sum _{x=0}^1\frac{e^{-5}5}{x!}}$

  $\alpha =1-0.0404$

  $\alpha =0.9596$

The producers risk $\left(\alpha \right)$is 0.9596.Now, calculate the consumers risk $\left(\beta \right)$as follows:

  $p_1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

  $c_1$ Is the rejection level $\le 2$

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 $c_1$ the lot can be accepted or else r ejected. Apply the above values to find $\left(\beta \right)$ as follows:

  $X\le c_1$ , accept lot

  $X\le 2$ accepted lot or else reject

Consider $p=p_1=0.10$

  $\beta =P\left\{x\le 2|p=0.10\right\}$

  $\beta =1-P\left\{x\le 2|p=0.10,n=10\right\}$

  $\text{β=P}\left[\text{X£1|θ=10}\right]\left(\text{poisson}\text{distribution}\text{application}\right)$

  $\beta ={\displaystyle \sum _{x=0}^1\frac{e^{-10}{10}^x}{x!}}$

  $\beta =0.0005$

The consumers risk $\left(\beta \right)$ 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

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

  $p_1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

  $c_1$ Is the rejection level $\ge 10$

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 $c_1$ the lot can be accepted or else r ejected. Apply the above values to find $\left(\alpha \right)$ producers risk as follows:

  $X\ge c_1$ , reject lot

  $X\ge 2$ reject lot or else accept

Consider $p=p_o=0.05$

  $\alpha =1-P\left\{rejectlot|p=p_o\right\}$

  $\alpha =1-P\left\{x\ge 10|p=0.05,n=100\right\}$

  $\alpha =1-P\left\{x\le 10|p=0.05\right\}$

  $\alpha =0.0137$

The producers risk $\left(\alpha \right)$ is 0.0137.

Now, calculate the consumers risk$\left(\beta \right)$as follows:

  $p_1$ Is Lot Tolerance Percent Defective (LTPD) = 0.10

  $c_1$ Is the rejection level $\le 10$

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 $c_1$ the lot can be accepted or else r ejected. Apply the above values to find $\left(\beta \right)$ as follows:

  $X\le c_1$ , accept lot

  $X\le 2$ accepted lot or else reject

Consider $p=p_1=0.10$

  $\beta =P\left\{x\le 10|p=0.10\right\}$

  $\beta =P\left\{x\le 10|p=0.10,n=100\right\}$

  $\beta =0.583$

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