Let X and Y equal the concentration in parts per billion of chromium in the blood for healthy persons and for persons with a suspected disease, respectively. Assume that the distributions of X and Y are normal. A sample of 8 observations for X have a sample mean of 15.75 and a sample variance of 46.21. A sample of 10 observations of Y have a sample mean of 23.3 and a sample variance of 92.68. Find a 90% confidence interval for the ratio of variances,

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]

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