Chapter 8.4, Problem 51E

A manufacturer of plumbing fixtures has developed a new type of washerless faucet. Let p = P(a randomly selected faucet of this type will develop a leak within 2 years under normal use). The manufacturer has decided to proceed with production unless it can be determined that p is too large; the borderline acceptable value of p is specified as .10. The manufacturer decides to subject n of these faucets to accelerated testing (approximating 2 years of normal use). With X = the number among the n faucets that leak before the test concludes, production will commence unless the observed X is too large. It is decided that if p = .10, the probability of not proceeding should be at most .10, whereas if p = .30 the probability of proceeding should be at most .10. Can n = 10 be used? n = 20? n = 25? What are the actual error probabilities for the chosen n?

To determine

Check whether for the given experiment sample size can be used as 10, 20 or 25.

Find the error probabilities for the chosen n.

Answer to Problem 51E

For the given experiment sample $n=25$ can be used.

The probability of Type I and Type II errors are 0.098 and 0.090 for $n=25$.

Explanation of Solution

Given info:

A new type of washer less faucet is developed by a manufacturer. Let p be the probability of randomly selected faucet of this type will develop a leak within 2 years under normal use. It was decided that the manufacturer proceed for the production if the value of p is too large. The maximum limit of p is 0.10 for accepting the washer less faucet. Let X be the number among the n faucets that leak before the test procedure concludes. It was decided that if $p=0.10$, the probability of not proceeding should be at most 0.10. If $p=0.30$, the probability of proceeding should be at most 0.10.

Calculation:

Let X be the number among the n faucets that leak before the test procedure concludes.

In the experiment for deciding that the manufacturer proceed for the production or not the hypotheses will be,

Null hypothesis:

$H_0:p=0.10$

That is, the probability of randomly selected faucet of this type will develop a leak within 2 years under normal use is 0.10.

Alternative hypothesis:

$H_0:p>0.10$

That is, the probability of randomly selected faucet of this type will develop a leak within 2 years under normal use is more 0.10.

Rejection rule:

Let the null hypothesis will be rejected if the minimum of c number among n faucets that leak before the test procedure concludes that is the null hypothesis will be rejected if $X\ge c$

For the sample size of 10 and $p=0.10$:

Hence, $X\sim Bin\left(10,0.1\right)$

For finding the appropriate sample size the Type I error and Type II error should be minimum.

Type I error:

Reject the null hypothesis when actually it is true.

It was decided that if $p=0.10$, the probability of not proceeding should be at most 0.10 that means Type I error will be maximum 0.10.

In this situation the Type I error will arise if actually the probability of randomly selected faucet of this type will develop a leak within 2 years under normal use is 0.10 but $X\ge c$.

Thus,

$\begin{array}{c}P\left(\text{Type I error}\right)=\alpha \\ =P\left(X\ge c,\text{when }p=0.10\right)\\ =1-P\left(X<c,\text{when }p=0.10\right)\\ =1-P\left(X\le c-1,\text{when }p=0.10\right)\end{array}$

Let $c=1$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 0\right)\\ =1-B\left(0;10,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=10$
• Locate $p=0.1$ corresponding to $n=10$
• Find the value corresponding $x=0$

Hence, $B\left(0;10,0.1\right)=0.349$

Thus,

$\begin{array}{c}\alpha =1-B\left(0;10,0.1\right)\\ =1-0.349\\ =0.651\end{array}$

Which is very large than 0.10.

Let $c=2$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 1\right)\\ =1-B\left(1;10,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=10$
• Locate $p=0.1$ corresponding to $n=10$
• Find the value corresponding $x=1$

Hence, $B\left(1;10,0.1\right)=0.736$

Thus,

$\begin{array}{c}\alpha =1-B\left(0;10,0.1\right)\\ =1-0.736\\ =0.264\end{array}$

Which is very large than 0.10.

Let $c=3$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 2\right)\\ =1-B\left(2;10,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=10$
• Locate $p=0.1$ corresponding to $n=10$
• Find the value corresponding $x=2$

Hence, $B\left(2;10,0.1\right)=0.930$

Thus,

$\begin{array}{c}\alpha =1-B\left(2;10,0.1\right)\\ =1-0.930\\ =0.07\end{array}$

Which is less than 0.10.

Hence, Type I error will be less for $c=3$.

Type II error:

Fail to reject the null hypothesis when actually it is false.

It was decided that if $p=0.30$, the probability of proceeding should be at most 0.10 that means Type II error will be maximum 0.10.

In this situation the Type II error will arise if actually the probability of randomly selected faucet of this type will develop a leak within 2 years under normal use is more than 0.10 but $X<c$.

Hence,

$\begin{array}{c}P\left(\text{Type II error}\right)=\beta \\ =P\left(X<c,\text{when }p=0.30\right)\\ =P\left(X\le c-1,\text{when }p=0.30\right)\end{array}$

In this situation ,$X\sim Bin\left(10,0.3\right)$.

For $c=3$,

$\begin{array}{c}\beta =P\left(X\le 2\right)\\ =B\left(2;10,0.3\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=10$
• Locate $p=0.3$ corresponding to $n=10$
• Find the value corresponding $x=2$

Hence, $B\left(2;10,0.3\right)=0.383$

Thus,

$\begin{array}{c}\beta =B\left(2;10,0.3\right)\\ =0.383\end{array}$

Which is quite large value.

Hence, simultaneously the Type I error and Type II error can’t be less for $n=10$.

Thus, it can be concluded that $n=10$ can’t be used.

For sample size of 20 and $p=0.10$:

Hence, $X\sim Bin\left(20,0.1\right)$

Thus,

$\begin{array}{c}P\left(\text{Type I error}\right)=\alpha \\ =1-P\left(X\le c-1,\text{when }p=0.10\right)\end{array}$

Let $c=1$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 0\right)\\ =1-B\left(0;20,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=20$
• Locate $p=0.1$ corresponding to $n=20$
• Find the value corresponding $x=0$

Hence, $B\left(0;20,0.1\right)=0.122$

Thus,

$\begin{array}{c}\alpha =1-B\left(0;20,0.1\right)\\ =1-0.122\\ =0.878\end{array}$

Which is very large than 0.10.

Let $c=3$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 2\right)\\ =1-B\left(2;20,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

1. 1. Locate the sample size $n=20$
2. 2. Locate $p=0.1$ corresponding to $n=20$
3. 3. Find the value corresponding $x=2$

Hence, $B\left(2;20,0.1\right)=0.392$

Thus,

$\begin{array}{c}\alpha =1-B\left(2;20,0.1\right)\\ =1-0.392\\ =0.608\end{array}$

Which is very large than 0.10.

Let $c=4$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 3\right)\\ =1-B\left(3;20,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

1. 1. Locate the sample size $n=20$
2. 2. Locate $p=0.1$ corresponding to $n=20$
3. 3. Find the value corresponding $x=3$

Hence, $B\left(3;20,0.1\right)=0.867$

Thus,

$\begin{array}{c}\alpha =1-B\left(3;20,0.1\right)\\ =1-0.567\\ =0.133\end{array}$

Which is larger than 0.10.

Let $c=5$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 4\right)\\ =1-B\left(4;10,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

1. 1. Locate the sample size $n=20$
2. 2. Locate $p=0.1$ corresponding to $n=20$
3. 3. Find the value corresponding $x=4$

Hence, $B\left(4;20,0.1\right)=0.957$

Thus,

$\begin{array}{c}\alpha =1-B\left(4;20,0.1\right)\\ =1-0.957\\ =0.043\end{array}$

Which is less than 0.10.

Hence, Type I error will be less for $c=5$.

Type II error:

$\begin{array}{c}P\left(\text{Type II error}\right)=\beta \\ =P\left(X\le c-1,\text{when }p=0.30\right)\end{array}$

In this situation ,$X\sim Bin\left(20,0.3\right)$.

For $c=5$,

$\begin{array}{c}\beta =P\left(X\le 4\right)\\ =B\left(4;20,0.3\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

1. 1. Locate the sample size $n=20$
2. 2. Locate $p=0.3$ corresponding to $n=10$
3. 3. Find the value corresponding $x=4$

Hence, $B\left(4;20,0.3\right)=0.238$

Thus,

$\begin{array}{c}\beta =B\left(4;20,0.3\right)\\ =0.238\end{array}$

Which is quite large value.

Hence, simultaneously the Type I error and Type II error can’t be less for $n=20$.

Thus, it can be concluded that $n=20$ can’t be used.

For the sample size of 25 and $p=0.10$:

Hence, $X\sim Bin\left(25,0.1\right)$

Thus,

$\begin{array}{c}P\left(\text{Type I error}\right)=\alpha \\ =1-P\left(X\le c-1,\text{when }p=0.10\right)\end{array}$

Let $c=1$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 0\right)\\ =1-B\left(0;25,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=25$
• Locate $p=0.1$ corresponding to $n=20$
• Find the value corresponding $x=0$

Hence, $B\left(0;25,0.1\right)=0.024$

Thus,

$\begin{array}{c}\alpha =1-B\left(0;25,0.1\right)\\ =1-0.001\\ =0.999\end{array}$

Which is very large than 0.10.

Let $c=3$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 2\right)\\ =1-B\left(2;25,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=25$
• Locate $p=0.1$ corresponding to $n=25$
• Find the value corresponding $x=3$

Hence, $B\left(2;25,0.1\right)=0.537$

Thus,

$\begin{array}{c}\alpha =1-B\left(2;25,0.1\right)\\ =1-0.537\\ =0.463\end{array}$

Which is very large than 0.10.

Let $c=4$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 3\right)\\ =1-B\left(3;25,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=25$
• Locate $p=0.1$ corresponding to $n=25$
• Find the value corresponding $x=3$

Hence, $B\left(3;25,0.1\right)=0.764$

Thus,

$\begin{array}{c}\alpha =1-B\left(3;25,0.1\right)\\ =1-0.764\\ =0.236\end{array}$

Which is larger than 0.10.

Let $c=5$,

Then,

$\begin{array}{c}\alpha =1-P\left(X\le 4\right)\\ =1-B\left(4;25,0.1\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=25$
• Locate $p=0.1$ corresponding to $n=25$
• Find the value corresponding $x=4$

Hence, $B\left(4;25,0.1\right)=0.214$

Thus,

$\begin{array}{c}\alpha =1-B\left(4;25,0.1\right)\\ =1-0.902\\ =0.098\end{array}$

Which is less than 0.10.

Hence, Type I error will be less for $c=5$.

Type II error:

$\begin{array}{c}P\left(\text{Type II error}\right)=\beta \\ =P\left(X\le c-1,\text{when }p=0.30\right)\end{array}$

In this situation ,$X\sim Bin\left(25,0.3\right)$.

For $c=5$,

$\begin{array}{c}\beta =P\left(X\le 4\right)\\ =B\left(4;25,0.3\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=25$
• Locate $p=0.3$ corresponding to $n=25$
• Find the value corresponding $x=4$

Hence, $B\left(4;25,0.3\right)=0.090$

Thus,

$\begin{array}{c}\beta =B\left(4;25,0.3\right)\\ =0.090\end{array}$

Here, $\beta$ is less than 0.10.

Hence, simultaneously the Type I error and Type II error will be less for $n=20$.

Thus, $n=25$ can be used as the sample size of the given experiment.

The error probabilities for $n=25$:

From the above calculation, the null hypothesis will be rejected if $X\ge 5$,

$\begin{array}{c}P\left(\text{Type I error}\right)=\alpha \\ =1-P\left(X\le 5,\text{when }p=0.10\right)\end{array}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=25$
• Locate $p=0.1$ corresponding to $n=25$
• Find the value corresponding $x=5$

Hence, $B\left(4;25,0.1\right)=0.902$

$\begin{array}{c}\alpha =1-B\left(4;25,0.1\right)\\ =1-0.902\\ =0.098\end{array}$

Type II error:

$\begin{array}{c}P\left(\text{Type II error}\right)=\beta \\ =P\left(X\le 4,\text{when }p=0.30\right)\end{array}$

In this situation ,$X\sim Bin\left(25,0.3\right)$.

For $c=5$,

$\begin{array}{c}\beta =P\left(X\le 4\right)\\ =B\left(4;25,0.3\right)\end{array}$

Where, $B\left(x;n.p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for finding the binomial probabilities value:

From table A.1 of Appendix

• Locate the sample size $n=25$
• Locate $p=0.3$ corresponding to $n=25$
• Find the value corresponding $x=4$

Hence, $B\left(4;25,0.3\right)=0.090$

Thus,

$\begin{array}{c}\beta =B\left(4;25,0.3\right)\\ =0.090\end{array}$

Hence, for $n=25$ the probability of Type I and Type II errors are 0.098 and 0.090, respectively.