Question

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?

Answer 1

Answer:

The  rejection region is

       $F_{cal} < F_{{1-\frac{\alpha}{2}} , df_1 , df_2 }$

or     $F_{cal} > F_{{\frac{\alpha}{2}} , df_1 , df_2 }$

and  

From the value obtained we see that  

$F_{cal} < F_{\frac{\alpha }{n} , df_1 , df_2 }$    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  $n_1 = 12$

   The first sample standard deviation is $s_1 = 0.038 \ mil$

   The second sample size is  $n_2 = 10$

   The first sample standard deviation is $s_2 = 0.042 \ mil$

    The significance level is  $\alpha =0.05$

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2_2$

The alternative hypothesis is   $H_o : \sigma^2_1 \ne \sigma^2_2$

Generally the test statistics is mathematically represented as

        $F_{cal} = \frac{s_1^2 }{s_2^2}$

=>     $F_{cal} = \frac{ 0.038^2 }{0.042^2}$

=>     $F_{cal} = 0.81859$    

Generally the first degree of freedom is  $df_1 = n_1 -1 = 12-1 = 11$

Generally the second degree of freedom is  $df_2 = n_2 -1 = 10 -1 = 9$

From the F-table the critical  value of  $\frac{\alpha}{2}$ at the degrees of freedom of   $df_1 = 11$  and   $df_2 =9$ is  

         $F_{\frac{\alpha }{n} , df_1 , df_2 } = 3.91207$

The  rejection region is

       $F_{cal} < F_{{1-\frac{\alpha}{2}} , df_1 , df_2 }$

or     $F_{cal} > F_{{\frac{\alpha}{2}} , df_1 , df_2 }$

and  

From the value obtained we see that  

$F_{cal} < F_{\frac{\alpha }{n} , df_1 , df_2 }$    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