A random sample of n observations is selected from a normal population to test the null hypothesis that σ2=25. Specify the rejection region for each of the following combinations of Ha,αand n.

a.Ha:σ225;α=0.5;n=16

b.Ha:σ2>25;α=.10;n=15

c.Ha:σ2>25;α=.01;n=23

d. Ha:σ2<25;α=.01;n=13

e. Ha:σ225;α=.10;n=7

f. Ha:σ2<25;α=.05;n=25

Short Answer

Expert verified

a. χ2<6.26214or χ2>27.4884

b. χ2>40.2894

c. χ2>21.0642

d. χ2<3.57056

e. χ2<1.63539or χ2>12.5916

f.χ2<13.8484

Step by step solution

01

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.

02

Solving for part a.

Chi-square distribution depends on (n-1)degrees of freedom. With α=0.05and (n-1)=15, the χ2value for rejection is found in Table IV of Appendix D.

Since this is a two-tailed test, therefore we have χ2<6.26214or χ2>27.4884which are the values for the two tailed rejection region.

03

Solving for part b.

With α=0.01,n=23, we have (n-1)=22.

Therefore, the rejection value will be χ2>40.2894.

04

Solving for part c.

With α=0.10,n=15, we have (n-1)=14.

Therefore, the rejection value will be χ2>21.0642.

05

Solving for part d.

With α=0.01,n=13, we have (n-1)=12

Therefore, the rejection value will be χ2<3.57056.

06

Solving for part e.

With α=0.10,n=7, we have (n-1)=6.

Since this is a two-tailed test, therefore the values for rejection will be χ2<1.63539or χ2>12.5916.

07

 Step 7: Solving for part f.

With α=0.05,n=25, we have (n-1)=24.

Therefore, the rejection value will beχ2<13.8484.

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