Chapter 9.7, Problem 11E

a)

To determine

To find power of the test.

Answer to Problem 11E

Power of the test is 0.2595

Explanation of Solution

Given:

   n = sample size = 100

  $\sigma$ = population standard deviation = 5000

  $\alpha$ = Level of significance = 0.05

Null and Alternative Hypotheses:

  $\begin{array}{l}{H}_0:\mu =50000\\ H_a:\mu >50000\end{array}$

So, ${\mu }_0=50000$

Formula:

Power of the test is

  $1-P\left(Z>-z-\frac{{\mu }_1-{\mu }_0}{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}\right)$

Calculation:

Alternative test is right tailed test with ${\mu }_1$ = 49500

First need to calculate $z_{\alpha }$ for given $\alpha$ = 0.05

Using z-table, $z_{\alpha }$ =1.645

Put all values in the formula,

  $\begin{array}{l}1-P\left(Z>-1.645-\frac{49500-50000}{\raisebox{1ex}{$5000$}\!\left/ \!\raisebox{-1ex}{$\sqrt{100}$}\right.}\right)\\ =1-P(Z>-0.645)\\ =1-(1-P(Z<-0.645))\\ =P(Z<-0.645)\\ =0.2595\end{array}$ ...Using z-table

b)

To determine

To find power of the test.

Answer to Problem 11E

Power of the test is 0.6388.

Explanation of Solution

Given:

  $\alpha$ = Level of significance = 0.05

Formula:

Power of the test is

  $1-P\left(Z>-z-\frac{{\mu }_1-{\mu }_0}{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}\right)$

Calculation:

Alternative test is right tailed test with ${\mu }_1$ = 49000

First need to calculate $z_{\alpha }$ for given $\alpha$ = 0.05

Using z-table, $z_{\alpha }$ =1.645

Put all values in the formula,

  $\begin{array}{l}1-P\left(Z>-1.645-\frac{49000-50000}{\raisebox{1ex}{$5000$}\!\left/ \!\raisebox{-1ex}{$\sqrt{100}$}\right.}\right)\\ =1-P(Z>0.355)\\ =1-(1-P(Z<0.335))\\ =P(Z<0.335)\\ =0.6388\end{array}$ ...Using z-table

c)

To determine

To find power of the test.

Answer to Problem 11E

Power of the test is 0.0918

Explanation of Solution

Given:

  $\alpha$ = Level of significance = 0.01

Formula:

Power of the test is

  $1-P\left(Z>-z-\frac{{\mu }_1-{\mu }_0}{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}\right)$

Calculation:

Alternative test is right tailed test with ${\mu }_1$ = 49500

First need to calculate $z_{\alpha }$ for given $\alpha$ = 0.01

Using z-table, $z_{\alpha }$ =2.33

Put all values in the formula,

  $\begin{array}{l}1-P\left(Z>-2.33-\frac{49500-50000}{\raisebox{1ex}{$5000$}\!\left/ \!\raisebox{-1ex}{$\sqrt{100}$}\right.}\right)\\ =1-P(Z>-1.33)\\ =1-(1-P(Z<-1.33))\\ =P(Z<-1.33)\\ =0.0958\end{array}$ ...Using z-table

d)

To determine

To find power of the test.

Answer to Problem 11E

Power of the test is 0.3748

Explanation of Solution

Given:

  $\alpha$ = Level of significance = 0.01

Formula:

Power of the test is

  $1-P\left(Z>-z-\frac{{\mu }_1-{\mu }_0}{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}\right)$

Calculation:

Alternative test is right tailed test with ${\mu }_1$ = 49000

First need to calculate $z_{\alpha }$ for given $\alpha$ = 0.01

Using z-table, $z_{\alpha }$ =2.33

Put all values in the formula,

  $\begin{array}{l}1-P\left(Z>-2.33-\frac{49000-50000}{\raisebox{1ex}{$5000$}\!\left/ \!\raisebox{-1ex}{$\sqrt{100}$}\right.}\right)\\ =1-P(Z>-0.33)\\ =1-(1-P(Z<-0.33))\\ =P(Z<-0.33)\\ =0.3748\end{array}$ ...Using z-table