Chapter 18, Problem 18.52E

(a)

To determine

To find: The probability of a Type I error.

Answer to Problem 18.52E

The probability of a Type I error is 0.5.

Explanation of Solution

Calculation:

The probability of Type I error:

Formula for probability of Type I error is,

$\begin{array}{c}P\left(\text{Type I error}\right)=P\left(\overline{x}>0\right){|}_{\mu =0}\\ =P\left(\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}>\frac{0-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\right){|}_{\mu =0}\\ =P\left(Z>\frac{0-0}{\left(\frac{\sigma }{\sqrt{n}}\right)}\right)\end{array}$

                                 $\begin{array}{c}=P\left(Z>0\right)\\ =1-P\left(Z\le 0\right)\end{array}$

From Table A: Standard Normal Cumulative Proportions, the value of $P\left(Z\le 0\right)$ is 0.5.

Therefore,

$\begin{array}{c}P\left(\text{Type I error}\right)=1-P\left(Z\le 0\right)\\ =1-0.5\\ =0.5\end{array}$

Thus, the probability that Type I error is 0.5.

(b)

To determine

To find: The probability of a Type II error when $\mu =0.5$.

Answer to Problem 18.52E

The probability of a Type II error when $\mu =0.5$ is 0.1587.

Explanation of Solution

Calculation:

Probability of Type II error:

$\begin{array}{c}P\left(\text{Type II error}\right)=P\left(\overline{x}\le 0\right){|}_{\mu =0.5}\\ =P\left(\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\le \frac{0-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\right){|}_{\mu =0.5}\\ =P\left(Z\le \frac{0-0.5}{\left(\frac{2.5}{\sqrt{25}}\right)}\right)\\ =P\left(Z\le -1\right)\end{array}$

From Table A: Standard Normal Cumulative Proportions, the value of $P\left(Z\le -1\right)$ is 0.1587.

Therefore,

$P\left(\text{Type II error}\right)=0.1587$

Thus, the probability of a Type II error when $\mu =0.5$ is 0.1587.

(c)

To determine

To find: The probability of a Type II error when $\mu =1.0$.

Answer to Problem 18.52E

The probability of a Type II error when $\mu =1.0$ is 0.0228.

Explanation of Solution

Calculation:

Probability of Type II error:

$\begin{array}{c}P\left(\text{Type II error}\right)=P\left(\overline{x}\le 0\right){|}_{\mu =1.0}\\ =P\left(\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\le \frac{0-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\right){|}_{\mu =1.0}\\ =P\left(Z\le \frac{0-1.0}{\left(\frac{2.5}{\sqrt{25}}\right)}\right)\\ =P\left(Z\le -2\right)\end{array}$

From Table A: Standard Normal Cumulative Proportions, the value of $P\left(Z\le -2\right)$ is 0.0228.

Therefore,

$P\left(\text{Type II error}\right)=0.0228$

Thus, the probability of a Type II error when $\mu =1.0$ is 0.0228.