Chapter 19, Problem 24SE

a.

To determine

Explain about the given values of $p_0$.

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:

$f\left(x\right)=\frac{n!}{x!\left(n-x\right)!}{p}^x{\left(1-p\right)}^{\left(n-x\right)}$

Here

$\begin{array}{l}n=\text{the sample size}\\ p=\text{the porportion of defective items in the lot}\\ x=\text{the number of defective items in the sample}\\ f\left(x\right)=\text{the probability of }x\text{ defective items in the sample}\end{array}$

For $p_0=.01$:

The probability of accepting the lot when $p_0=.01$ is calculated as follows:

$\begin{array}{c}P\left(\text{Accepting lot}\right)=f\left(0\right)+f\left(1\right)\\ =\frac{15!}{0!\left(15-0\right)!}{\left(0.01\right)}^0{\left(1-0.01\right)}^{\left(15-0\right)}+\frac{15!}{1!\left(15-1\right)!}{\left(0.01\right)}^1{\left(1-0.01\right)}^{\left(15-1\right)}\\ =\frac{15!}{15!}{\left(0.99\right)}^{15}+\frac{15!}{14!}\left(0.01\right){\left(0.99\right)}^{14}\\ =0.8601+15\left(0.01\right)\left(0.8687\right)\end{array}$

                            $\begin{array}{c}=0.8601+0.1303\\ =0.9904\end{array}$

Thus, the probability of accepting the lot when $p_0=.01$ is 0.9904.

The producer’s risk, $\alpha$, is obtained as follows:

$\begin{array}{c}\alpha =1-0.9904\\ =0.0096\end{array}$

Thus, the producer’s risk when $p_0=.01$ is 0.0096.

For $p_0=.02$:

The probability of accepting the lot when $p_0=.02$ is calculated as follows:

$\begin{array}{c}P\left(\text{Accepting lot}\right)=f\left(0\right)+f\left(1\right)\\ =\frac{15!}{0!\left(15-0\right)!}{\left(0.02\right)}^0{\left(1-0.02\right)}^{\left(15-0\right)}+\frac{15!}{1!\left(15-1\right)!}{\left(0.02\right)}^1{\left(1-0.02\right)}^{\left(15-1\right)}\\ =\frac{15!}{15!}{\left(0.98\right)}^{15}+\frac{15!}{14!}\left(0.02\right){\left(0.98\right)}^{14}\\ =0.7386+15\left(0.02\right)\left(0.7536\right)\end{array}$

                            $\begin{array}{c}=0.7386+0.2261\\ =0.9647\end{array}$

Thus, the probability of accepting the lot when $p_0=.02$ is 0.9647.

The producer’s risk, $\alpha$, is obtained as follows:

$\begin{array}{c}\alpha =1-0.9647\\ =0.0353\end{array}$

Thus, the producer’s risk when $p_0=.02$ is 0.0353.

For $p_0=.03$:

The probability of accepting the lot when $p_0=.03$ is calculated as follows:

$\begin{array}{c}P\left(\text{Accepting lot}\right)=f\left(0\right)+f\left(1\right)\\ =\frac{15!}{0!\left(15-0\right)!}{\left(0.03\right)}^0{\left(1-0.03\right)}^{\left(15-0\right)}+\frac{15!}{1!\left(15-1\right)!}{\left(0.03\right)}^1{\left(1-0.03\right)}^{\left(15-1\right)}\\ =\frac{15!}{15!}{\left(0.97\right)}^{15}+\frac{15!}{14!}\left(0.03\right){\left(0.97\right)}^{14}\\ =0.6333+15\left(0.03\right)\left(0.6528\right)\end{array}$

                            $\begin{array}{c}=0.6333+0.2938\\ =0.9271\end{array}$

Thus, the probability of accepting the lot when $p_0=.03$ is 0.9271.

The producer’s risk, $\alpha$, is obtained below:

$\begin{array}{c}\alpha =1-0.9271\\ =0.0729\end{array}$

Thus, the producer’s risk when $p_0=.03$ is 0.0729.

For $p_0=.04$:

The probability of accepting the lot when $p_0=.04$ is calculated as follows:

$\begin{array}{c}P\left(\text{Accepting lot}\right)=f\left(0\right)+f\left(1\right)\\ =\frac{15!}{0!\left(15-0\right)!}{\left(0.04\right)}^0{\left(1-0.04\right)}^{\left(15-0\right)}+\frac{15!}{1!\left(15-1\right)!}{\left(0.04\right)}^1{\left(1-0.04\right)}^{\left(15-1\right)}\\ =\frac{15!}{15!}{\left(0.96\right)}^{15}+\frac{15!}{14!}\left(0.04\right){\left(0.96\right)}^{14}\\ =0.5421+15\left(0.04\right)\left(0.5647\right)\end{array}$

                            $\begin{array}{c}=0.5421+0.3388\\ =0.8809\end{array}$

Thus, the probability of accepting the lot when $p_0=.04$ is 0.8809.

The producer’s risk, $\alpha$, is obtained as follows:

$\begin{array}{c}\alpha =1-0.8809\\ =0.1191\end{array}$

Thus, the producer’s risk when $p_0=.04$ is 0.1191.

For $p_0=.05$:

The probability of accepting the lot when $p_0=.05$ is calculated as follows:

$\begin{array}{c}P\left(\text{Accepting lot}\right)=f\left(0\right)+f\left(1\right)\\ =\frac{15!}{0!\left(15-0\right)!}{\left(0.05\right)}^0{\left(1-0.05\right)}^{\left(15-0\right)}+\frac{15!}{1!\left(15-1\right)!}{\left(0.05\right)}^1{\left(1-0.05\right)}^{\left(15-1\right)}\\ =\frac{15!}{15!}{\left(0.95\right)}^{15}+\frac{15!}{14!}\left(0.05\right){\left(0.95\right)}^{14}\\ =0.4633+15\left(0.05\right)\left(0.4877\right)\end{array}$

                            $\begin{array}{c}=0.4633+0.3658\\ =0.8291\end{array}$

Thus, the probability of accepting the lot when $p_0=.05$ is 0.8291.

The producer’s risk, $\alpha$, is obtained as follows:

$\begin{array}{c}\alpha =1-0.8291\\ =0.1709\end{array}$

Thus, the producer’s risk when $p_0=.05$ is 0.1709.

Here, the producer’s risk obtained using $p_0=.03$ 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 $p_1=0.25$.

Answer to Problem 24SE

The consumer’s risk when $p_1=.25$ is 0.0802.

Explanation of Solution

Calculation:

The probability of accepting the lot when $p_1=.25$ is calculated as follows:

$\begin{array}{c}P\left(\text{Accepting lot}\right)=f\left(0\right)+f\left(1\right)\\ =\frac{15!}{0!\left(15-0\right)!}{\left(0.25\right)}^0{\left(1-0.25\right)}^{\left(15-0\right)}+\frac{15!}{1!\left(15-1\right)!}{\left(0.25\right)}^1{\left(1-0.25\right)}^{\left(15-1\right)}\\ =\frac{15!}{15!}{\left(0.75\right)}^{15}+\frac{15!}{14!}\left(0.25\right){\left(0.75\right)}^{14}\\ =0.0134+15\left(0.25\right)\left(0.0178\right)\end{array}$

                            $\begin{array}{c}=0.0134+0.0668\\ =0.0802\end{array}$

Thus, the probability of accepting the lot when $p_1=.25$ is 0.0802.

The consumer’s risk, $\beta$, is $\beta =0.0802$.

Thus, the consumer’s risk when $p_1=.25$ is 0.0802.