Chapter 18, Problem 18.52E

a.

To determine

The probability of type I error that the test reject ${\text{H}}_0$ when $\mu =0$.

Answer to Problem 18.52E

The probability of type I error is 0.5.

Explanation of Solution

Given info:

The hypothesis is given as:

The null hypothesis is,

${\text{H}}_0:\mu =0$

The research hypothesis is,

${\text{H}}_a:\mu >0$

Simple random sample of size n is 25 and standard deviation is $\sigma =1$.

Calculation:

Type I error:

The probability of rejecting null hypothesis when it is true.

Type I error states that rejecting ${\text{H}}_0$ when it is true and it is denoted by $\alpha$.

$\begin{array}{c}P\left(\overline{x}>\text{0}\text{when}\mu =\text{0}\right)=P\left(\frac{\overline{x}-\mu }{\frac{\text{σ}}{\sqrt{n}}}\text{>}\frac{\text{0}-\mu }{\frac{\text{σ}}{\sqrt{n}}}\right)\text{ when }\mu =0\\ =\text{P}\left(z>\text{0}\right)\\ =0.5-P\left(z<0\right)\\ =0.5-0\end{array}$

Further solve the above equation.

$P\left(\overline{x}>\text{0}\text{when}\mu =\text{0}\right)=0.5$

The 0.5 probability value is obtained by using area under normal curve table.

Thus, the probability of type I error is 0.5.

b.

To determine

The probability of type II error that the test reject ${\text{H}}_0$ when $\mu =0.25$.

Answer to Problem 18.52E

The probability of type II error is 0.1056.

Explanation of Solution

Calculation:

Type II error:

The probability of accepting null hypothesis when it is false.

Type II error states that accepting ${\text{H}}_0$ when it is false and it is denoted by $\beta$.

$\begin{array}{c}P\left(\overline{x}>\text{0}\text{when}\mu =\text{0}\text{.25}\right)\text{=}P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\le \frac{0-\mu }{\frac{\sigma }{\sqrt{n}}}\text{when}\mu =\text{0}\text{.25}\right)\\ =P\left(z\le -1.25\right)\\ =P\left(z>1.25\right)\left(\text{By}\text{symmetric}\right)\\ =0.5-P\left(z<1.25\right)\end{array}$

Further salve the above equation.

$\begin{array}{c}P\left(\overline{x}>\text{0}\text{when}\mu =\text{0}\text{.25}\right)=0.5-0.3944\\ =0.1056\end{array}$

The 0.3944 probability value is obtained by using area under normal curve table.

Thus, the probability of type II error is 0.1056.

c.

To determine

The probability of type II error when $\mu =0.5$.

Answer to Problem 18.52E

The probability of type II error is 0.0202.

Explanation of Solution

Calculation:

Type II error:

The probability of accepting null hypothesis when it is false.

Type II error states that accepting ${\text{H}}_0$ when it is false and it is denoted by $\beta$.

$\begin{array}{c}P\left(\overline{x}>\text{0}\text{when}\mu =\text{0}\text{.5}\right)=\text{P}\left(\frac{\overline{x}-\mu }{\frac{\text{σ}}{\sqrt{n}}}\le \frac{\text{0}-\mu }{\frac{\text{σ}}{\sqrt{n}}}\text{when}\mu =\text{0}\text{.5}\right)\\ =P\left(z\le -2.5\right)\\ =P\left(z>2.5\right)\left(\text{By}\text{symmetric}\right)\\ =0.5-P\left(\text{Z}\text{<}\text{}\text{2}\text{.5}\right)\end{array}$

The 0.4798 probability value is obtained by using area under normal curve table.

Thus,

$\begin{array}{c}P\left(\overline{x}>\text{0}\text{when}\mu =\text{0}\text{.5}\right)=0.5-0.4798\\ =0.0202\end{array}$

Thus, the probability of type II error is 0.0202.