Answer:
The value is [tex]0.152 \le \frac{\sigma_1^2}{\sigma_2^2} \le 1.835[/tex]
Step-by-step explanation:
From the question we are told that
The number of observation for X is [tex]n_1 = 8[/tex]
The sample mean for X is [tex]\= x _1 = 15.7 5[/tex]
The sample variance for X is [tex]s^2_1 = 46.21[/tex]
The number of observation for Y is [tex]n_1 = 10[/tex]
The sample mean for Y is [tex]\= x _1 = 23.3[/tex]
The sample variance for Y is [tex]s^2_1 = 92.8[/tex]
Generally the degree of freedom for X is
[tex]df_1 = n_1 - 1[/tex]
=> [tex]df_1 = 8 - 1[/tex]
=> [tex]df_1 = 7[/tex]
Generally the degree of freedom for X is
[tex]df_2 = n_2 - 1[/tex]
=> [tex]df_2 = 10 - 1[/tex]
=> [tex]df_2 = 9[/tex]
From the question we are told the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 90 ) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
=> [tex]\frac{\alpha}{2} = \frac{0.10}{2}[/tex]
=> [tex]\frac{\alpha}{2} = 0.05 }[/tex]
Generally the 90% confidence interval for the ratio of variances is mathematically represented as
[tex]\frac{s_1^2}{s_2^2 } * \frac{1}{F_{ 1 - \frac{\alpha }{2} , df_1 , df_2 }} \le \frac{\sigma_1^2}{\sigma_2^2} \le \frac{s_1^2}{s_2^2 } * \frac{1}{F_{ \frac{\alpha }{2} , df_1 , df_2 }}[/tex]
=> [tex]\frac{46.21}{92.68 } * \frac{1}{F_{ 1 - 0.05 , 7 , 9 }} \le \frac{\sigma_1^2}{\sigma_2^2} \le \frac{46.21}{92.68 } * \frac{1}{F_{ 0.05 , 7 , 9 }}[/tex]
=> [tex]\frac{46.21}{92.68 } * 0.304 \le \frac{\sigma_1^2}{\sigma_2^2} \le \frac{46.21}{92.68 } * 3.677[/tex]
=> [tex]0.152 \le \frac{\sigma_1^2}{\sigma_2^2} \le 1.835[/tex]
