Answer to Question #248636 in Statistics and Probability for joe

Question

Borden et al. (A-24) performed experiments on cadaveric knees to test the effectiveness of several meniscal repair techniques. Specimens were loaded into a servohydraulic device and tension-loaded to failure. The biomechanical testing was performed by using a slow loading rate to simulate the stresses that the medial meniscus might be subjected to during early rehabilitation exercises and activities of daily living. One of the measures is the amount of displacement that occurs. Of the 12 specimens receiving the vertical mattress suture and the FasT-FIX method, the displacement values measured in millimeteres are 16.9, 20.2, 20.1, 15.7, 13.9, 14.9, 18.0, 18.5, 9.2, 18.8, 22.8, 17.5. Construct a 90 percent confidence interval for the variance of the displacement in millimeters for a population of subjects receiving these repair techniques.

Answer 1

Since there is a situation where neither mean or deviation is known, it is appropriate to use Fishcher theorem

$H = {\frac {(n-1)*s^{2}} {σ^{2}}}$ , where n - sample size, s - sample standart deviation, σ - unbiased standart deviation

H ~ $Xi^{2}(n-1)$

In the given case: u(sample mean) = 17.2

$s^{2} = {\frac{(16.9 -17.2)^{2}+ ... + (17.5 - 17.2)^{2}} {11}} = 12.4$

H = ${\frac {136} {σ^{2}}}$

H ~ $Xi^{2}(11)$ , $\alpha = 0.1$

$P({\frac{136} {Xi^{2}(11;{\frac{\alpha} 2} )}} < σ^{2} < {\frac{136} {Xi^{2}(11;1-{\frac{\alpha} 2})}}) = 0.9$

$Xi^{2}(11;{\frac{\alpha} 2}) = Xi^{2}(11;0.05) = 19.7$

$Xi^{2}(11;1-{\frac{\alpha} 2}) = Xi^{2}(11;0.95) = 4.57$

$P(6.9 < σ^{2} < 29.8) = 0.9$

The 90% confidence interval for the variance is (6.9; 29.8)