Step 1

Let, p: 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 of the value of p is too large.

The maximum limit of p is 0.10 for accepting the washer less faucet.

Let X: Number among the n faucets that leak before the test procedure concludes. It was decided that 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 most0.10.

 
Step 2

X: 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 hypothesis will be as given below.

Null hypothesis:

\[{H_0}:p = 0.10\]

Probability of randomly selected faucet of this type will develop a leak within 2 years under normal use is 0.10.

Alternative hypothesis:

\[{H_0}:p \ne 0.10\]

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 us 0.10 but $X \ge c$.

Thus,

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

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) = \sum\limits_x {b\left( {x;n,p} \right)} $

Locate one sample size n = 10. Locate p =0.1 corresponding to n = 10. The find value corresponding to x = 0.

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

Thus,

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

It 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) = \sum\limits_{y = 0}^x {b\left( {y;n,p} \right)} $

Then,

\[\begin{array}{c} \alpha = 1 - P\left( {X \le 2} \right)\\ = 1 - B\left( {2;10,0.1} \right) \end{array}\] \[B\left( {2;10,0.1} \right) = 0.930\] \[\begin{array}{c} \alpha = 1 - B\left( {2;10,0.1} \right)\\ = 1 - 0.930\\ = 0.07 \end{array}\]

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, then 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.

\[X \sim Bin\left( {10,0.3} \right)\]

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) = \sum\limits_x {b\left( {x;n,p} \right)} $

Procedure for finding binomial probabilities value:

Find the value with sample size n = 10 and p = 0.3 corresponding to n = 10. Find the value corresponding to x = 2.

Hence,

\[B\left( {2;10,0.3} \right) = 0.383\]

Which is quite large.

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, $X \sim Bin\left( {20,0.1} \right)$ .

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) = \sum\limits_x {b\left( {x;n,p} \right)} $

Find the value with n = 20 and p = 0.1 and value corresponding to x = 0 in binomial table.

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

Which is very large 0.10.

Let c = 3

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

Which is very large than 0.10.

Let c = 4

\[\begin{array}{c} \alpha = 1 - P\left( {X \le 3} \right)\\ = 1 - B\left( {3;20,0.1} \right)\\ = 1 - 0.567\\ = 0.133 \end{array}\]

Which is larger than 0.10.

\[\begin{array}{c} \alpha = 1 - P\left( {X \le 4} \right)\\ = 1 - B\left( {4;20,0.1} \right)\\ = 1 - 0.957\\ = 0.043 \end{array}\]

Which is less than 0.10.

Type II error:

\[\begin{array}{c} P\left( {Type\;II\;error} \right) = \beta \\ = P\left( {X \le c - 1,when\;p = 0.30} \right) \end{array}\] \[\begin{array}{c} \beta = B\left( {4;20,0.3} \right)\\ = 0.238 \end{array}\]

Which is quite large.

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,

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

Then,

\[\begin{array}{c} \alpha = 1 - B\left( {0;25,0.2} \right)\\ = 1 - 0.001\\ = 0.999 \end{array}\]

Which is larger than 0.10.

Let c = 3.

\[\begin{array}{c} \alpha = 1 - B\left( {2;25,0.2} \right)\\ = 1 - 0.537\\ = 0.463 \end{array}\]

Which is larger than 0.10.

\[\begin{array}{c} \alpha = 1 - B\left( {3;25,0.2} \right)\\ = 1 - 0.764\\ = 0.236 \end{array}\]

Which is larger than 0.10.

Let c = 5.

\[\begin{array}{c} \alpha = 1 - P\left( {X \le 4} \right)\\ = 1 - B\left( {4;25,0.1} \right)\\ = 1 - 0.902\\ = 0.098 \end{array}\]

Which is less than 0.10.

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

Type II error:

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

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

For c = 5.

\[\begin{array}{c} \beta = P\left( {X \le 4} \right)\\ = B\left( {4;25,0.3} \right)\\ = 0.090 \end{array}\]

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

\[\begin{array}{c} P\left( {Type\;I\;error} \right) = \alpha \\ = 1 - P\left( {X \le 5,when\;p = 0.10} \right)\\ = 1 - B\left( {4;25,0.1} \right)\\ = 1 - 0.902 \end{array}\]

Solve it.

\[P\left( {Type\;I\;error} \right) = 0.098\]

Type II error:

\[\begin{array}{c} P\left( {Type\;II\;error} \right) = \beta \\ = P\left( {X \le 4,When\;p = 0.30} \right)\\ = B\left( {4;25,0.3} \right)\\ = 0.090 \end{array}\]

Hence, for n = 25 the probability of Type I and Type II errors are 0.0198 and 0.090 respectively.