Answer:
The rejection region is
![F_{cal} < F_{{1-\frac{\alpha}{2}} , df_1 , df_2 }](https://tex.z-dn.net/?f=F_%7Bcal%7D%20%3C%20F_%7B%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%20%2C%20df_1%20%2C%20df_2%20%7D)
or ![F_{cal} > F_{{\frac{\alpha}{2}} , df_1 , df_2 }](https://tex.z-dn.net/?f=F_%7Bcal%7D%20%3E%20F_%7B%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%20%2C%20df_1%20%2C%20df_2%20%7D)
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 ![n_1 = 12](https://tex.z-dn.net/?f=n_1%20%3D%20%2012)
The first sample standard deviation is ![s_1 = 0.038 \ mil](https://tex.z-dn.net/?f=s_1%20%3D%20%200.038%20%5C%20%20mil)
The second sample size is ![n_2 = 10](https://tex.z-dn.net/?f=n_2%20%3D%20%2010)
The first sample standard deviation is ![s_2 = 0.042 \ mil](https://tex.z-dn.net/?f=s_2%20%3D%20%200.042%20%5C%20%20mil)
The significance level is ![\alpha =0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D0.05)
The null hypothesis is ![H_o : \sigma^2_1 = \sigma^2_2](https://tex.z-dn.net/?f=H_o%20%3A%20%5Csigma%5E2_1%20%20%3D%20%5Csigma%5E2_2)
The alternative hypothesis is ![H_o : \sigma^2_1 \ne \sigma^2_2](https://tex.z-dn.net/?f=H_o%20%3A%20%5Csigma%5E2_1%20%5Cne%20%20%5Csigma%5E2_2)
Generally the test statistics is mathematically represented as
![F_{cal} = \frac{s_1^2 }{s_2^2}](https://tex.z-dn.net/?f=F_%7Bcal%7D%20%3D%20%20%5Cfrac%7Bs_1%5E2%20%7D%7Bs_2%5E2%7D)
=> ![F_{cal} = \frac{ 0.038^2 }{0.042^2}](https://tex.z-dn.net/?f=F_%7Bcal%7D%20%3D%20%20%5Cfrac%7B%200.038%5E2%20%7D%7B0.042%5E2%7D)
=>
Generally the first degree of freedom is ![df_1 = n_1 -1 = 12-1 = 11](https://tex.z-dn.net/?f=df_1%20%3D%20%20n_1%20-1%20%20%3D%20%2012-1%20%3D%2011)
Generally the second degree of freedom is ![df_2 = n_2 -1 = 10 -1 = 9](https://tex.z-dn.net/?f=df_2%20%3D%20%20n_2%20-1%20%20%3D%20%2010%20-1%20%3D%209)
From the F-table the critical value of
at the degrees of freedom of
and
is
![F_{\frac{\alpha }{n} , df_1 , df_2 } = 3.91207](https://tex.z-dn.net/?f=F_%7B%5Cfrac%7B%5Calpha%20%7D%7Bn%7D%20%2C%20df_1%20%2C%20df_2%20%7D%20%3D%20%203.91207)
The rejection region is
![F_{cal} < F_{{1-\frac{\alpha}{2}} , df_1 , df_2 }](https://tex.z-dn.net/?f=F_%7Bcal%7D%20%3C%20F_%7B%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%20%2C%20df_1%20%2C%20df_2%20%7D)
or ![F_{cal} > F_{{\frac{\alpha}{2}} , df_1 , df_2 }](https://tex.z-dn.net/?f=F_%7Bcal%7D%20%3E%20F_%7B%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%20%2C%20df_1%20%2C%20df_2%20%7D)
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