Question

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 1

The value is
`0.152 \le (\sigma_1^2)/(\sigma_2^2) \le 1.835`

Explanation:

From the question we are told that

The number of observation for X is
`n_1 = 8`

The sample mean for X is
`\= x _1 = 15.7 5`

The sample variance for X is
`s^2_1 = 46.21`

The number of observation for Y is
`n_1 = 10`

The sample mean for Y is
`\= x _1 = 23.3`

The sample variance for Y is
`s^2_1 = 92.8`

Generally the degree of freedom for X is

`df_1 = n_1 - 1`

=>
`df_1 = 8 - 1`

=>
`df_1 = 7`

Generally the degree of freedom for X is

`df_2 = n_2 - 1`

=>
`df_2 = 10 - 1`

=>
`df_2 = 9`

From the question we are told the confidence level is 90% , hence the level of significance is

`\alpha = (100 - 90 ) \%`

=>
`\alpha = 0.10`

=>
`(\alpha)/(2) = (0.10)/(2)`

=>
`(\alpha)/(2) = 0.05 }`

Generally the 90% confidence interval for the ratio of variances is mathematically represented as

`(s_1^2)/(s_2^2 ) * \frac{1}{F_{ 1 - (\alpha )/(2) , df_1 , df_2 }} \le (\sigma_1^2)/(\sigma_2^2) \le (s_1^2)/(s_2^2 ) * \frac{1}{F_{ (\alpha )/(2) , df_1 , df_2 }}`

=>
`(46.21)/(92.68 ) * (1)/(F_( 1 - 0.05 , 7 , 9 )) \le (\sigma_1^2)/(\sigma_2^2) \le (46.21)/(92.68 ) * (1)/(F_( 0.05 , 7 , 9 ))`

=>
`(46.21)/(92.68 ) * 0.304 \le (\sigma_1^2)/(\sigma_2^2) \le (46.21)/(92.68 ) * 3.677`

=>
`0.152 \le (\sigma_1^2)/(\sigma_2^2) \le 1.835`