Answer:
The rejection region is
[tex]F_{cal} < F_{{1-\frac{\alpha}{2}} , df_1 , df_2 }[/tex]
or [tex]F_{cal} > F_{{\frac{\alpha}{2}} , df_1 , df_2 }[/tex]
and
From the value obtained we see that
[tex] F_{cal} < F_{\frac{\alpha }{n} , df_1 , df_2 } [/tex] 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 [tex]n_1 = 12[/tex]
The first sample standard deviation is [tex]s_1 = 0.038 \ mil[/tex]
The second sample size is [tex]n_2 = 10[/tex]
The first sample standard deviation is [tex]s_2 = 0.042 \ mil[/tex]
The significance level is [tex]\alpha =0.05[/tex]
The null hypothesis is [tex]H_o : \sigma^2_1 = \sigma^2_2[/tex]
The alternative hypothesis is [tex]H_o : \sigma^2_1 \ne \sigma^2_2[/tex]
Generally the test statistics is mathematically represented as
[tex]F_{cal} = \frac{s_1^2 }{s_2^2}[/tex]
=> [tex]F_{cal} = \frac{ 0.038^2 }{0.042^2}[/tex]
=> [tex]F_{cal} = 0.81859[/tex]
Generally the first degree of freedom is [tex]df_1 = n_1 -1 = 12-1 = 11[/tex]
Generally the second degree of freedom is [tex]df_2 = n_2 -1 = 10 -1 = 9[/tex]
From the F-table the critical value of [tex] \frac{\alpha}{2}[/tex] at the degrees of freedom of [tex]df_1 = 11[/tex] and [tex]df_2 =9 [/tex] is
[tex]F_{\frac{\alpha }{n} , df_1 , df_2 } = 3.91207[/tex]
The rejection region is
[tex]F_{cal} < F_{{1-\frac{\alpha}{2}} , df_1 , df_2 }[/tex]
or [tex]F_{cal} > F_{{\frac{\alpha}{2}} , df_1 , df_2 }[/tex]
and
From the value obtained we see that
[tex] F_{cal} < F_{\frac{\alpha }{n} , df_1 , df_2 } [/tex] 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
