A plating company has two silver plating systems with variances σ12 and σ22. You, as the manager, desired to compare the variability in the silver plating done by System-1 with that done by System-2. An independent random sample of size n1= 12 of the System-1 yields s1 = 0.038 mil and sample of size n2= 10 of System-2 yields s2 = 0.042 mil. We need to decide whether σ12= σ22 with α = 0.05. What is the rejection region?

The  rejection region is

or    

and  

From the value obtained we see that  

   Hence

    The decision rule

Fail to reject the null hypothesis

The conclusion

This no sufficient evidence to conclude that there is a difference between the two variance

Step-by-step explanation:

From the question we are told that

   The first sample size is  

   The first sample standard deviation is

   The second sample size is  

   The first sample standard deviation is

    The significance level is  

The null hypothesis is

The alternative hypothesis is  

Generally the test statistics is mathematically represented as

=>    

=>        

Generally the first degree of freedom is  

Generally the second degree of freedom is  

From the F-table the critical  value of   at the degrees of freedom of    and   is  

The  rejection region is

or    

and  

From the value obtained we see that  

   Hence

    The decision rule

Fail to reject the null hypothesis

The conclusion

This no sufficient evidence to conclude that there is a difference between the two variance

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation:

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step-by-step explanation: