Question

7.83 A random sample of n observations is selected from a normal population to test the null hypothesis that ${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathbf{25}$ . Specify the rejection region for each of the following combinations of ${\mathit{H}}_{\mathbf{a}}\mathbf{,}\mathit{\alpha }$and $\mathit{n}$.

a.$H_a:{\sigma }^2\ne 25;\alpha =0.5;n=16$

b. $H_a:{\sigma }^2>25;\alpha =.01;n=23$

c. $H_a:{\sigma }^2>25;\alpha =.10;n=15$

d. $H_a:{\sigma }^2<25;\alpha =.01;n=13$

e. $H_a:{\sigma }^2\ne 25;\alpha =.10;n=7$

f.$H_a:{\sigma }^2<25;\alpha =.05;n=25$

Short answer

a. ${\chi }^2<6.26214$or ${\chi }^2>27.4884$

b. ${\chi }^2>40.2894$

c. ${\chi }^2>21.0642$

d. ${\chi }^2<3.57056$

e. ${\chi }^2<1.63539$or ${\chi }^2>12.5916$

f. ${\chi }^2<13.8484$

Step-by-step solution

Defining the Rejection Region

Rejection Region holds significance in the sense that whenever a numerical value of the test statistic falls in the rejection region, the null hypothesis is rejected.

Solving for part a.

Chi-square distribution depends on $(n-1)$degrees of freedom. With $\alpha =0.05$and , $(n-1)=15$the ${\chi }^2$value for rejection is found in Table IV of Appendix D.

Since this is a two-tailed test, therefore we have ${\chi }^2<6.26214$or ${\chi }^2>27.4884$which are the values for the two tailed rejection region.

Solving for part b.

With$\alpha =0.01,n=23$ , we have$(n-1)=22$ .

Therefore, the rejection value will be ${\chi }^2>40.2894$.

Solving for part c.

With $\alpha =0.10,n=15$, we have $(n-1)=14$.

Therefore, the rejection value will be ${\chi }^2>21.0642$.

Solving for part d.

With $\alpha =0.01,n=13$, we have$(n-1)=12$

Therefore, the rejection value will be${\chi }^2<3.57056$ .

Solving for part e.

With$\alpha =0.10,n=7$ , we have $(n-1)=6$.

Since this is a two-tailed test, therefore the values for rejection will be${\chi }^2<1.63539$ or ${\chi }^2>12.5916$.

Solving for part f.

With $\alpha =0.05,n=25$, we have $(n-1)=24$

Therefore, the rejection value will be ${\chi }^2<13.8484$.