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02.12.2020 •
Mathematics
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,
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Ответ:
The value is![0.152 \le \frac{\sigma_1^2}{\sigma_2^2} \le 1.835](/tpl/images/0941/5640/39284.png)
Step-by-step explanation:
From the question we are told that
The number of observation for X is![n_1 = 8](/tpl/images/0941/5640/264bd.png)
The sample mean for X is![\= x _1 = 15.7 5](/tpl/images/0941/5640/2f9cd.png)
The sample variance for X is![s^2_1 = 46.21](/tpl/images/0941/5640/6504d.png)
The number of observation for Y is![n_1 = 10](/tpl/images/0941/5640/b754d.png)
The sample mean for Y is![\= x _1 = 23.3](/tpl/images/0941/5640/4c4ac.png)
The sample variance for Y is![s^2_1 = 92.8](/tpl/images/0941/5640/b357a.png)
Generally the degree of freedom for X is
=>![df_1 = 8 - 1](/tpl/images/0941/5640/f6693.png)
=>![df_1 = 7](/tpl/images/0941/5640/5044e.png)
Generally the degree of freedom for X is
=>![df_2 = 10 - 1](/tpl/images/0941/5640/d5a75.png)
=>![df_2 = 9](/tpl/images/0941/5640/2371a.png)
From the question we are told the confidence level is 90% , hence the level of significance is
=>![\alpha = 0.10](/tpl/images/0941/5640/b540d.png)
=>![\frac{\alpha}{2} = \frac{0.10}{2}](/tpl/images/0941/5640/62501.png)
=>![\frac{\alpha}{2} = 0.05 }](/tpl/images/0941/5640/1eb55.png)
Generally the 90% confidence interval for the ratio of variances is mathematically represented as
=>![\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 }}](/tpl/images/0941/5640/efaf5.png)
=>![\frac{46.21}{92.68 } * 0.304 \le \frac{\sigma_1^2}{\sigma_2^2} \le \frac{46.21}{92.68 } * 3.677](/tpl/images/0941/5640/202cf.png)
=>![0.152 \le \frac{\sigma_1^2}{\sigma_2^2} \le 1.835](/tpl/images/0941/5640/39284.png)
Ответ:
h
Step-by-step explanation: