Question

In order to determine that the precision of an instrument is appropriate for the intended task, two calibration tests were carried out with two samples of measurements by the same instrument but different observers. The results of the tests are as follows:

Test 1:

Computed standard deviation of instrument = ± 2.5 mm;

number of degrees of freedom = 24.

Test 2:

Computed standard deviation of instrument = ± 2.0 mm;

number of degrees of freedom = 15.

Construct the 90% confidence interval for the ratio of the two population variances associated with the two tests.

Answer 1

Solution:-

90% confidence interval for the ratio of the two population variances associated with the two tests is $0.663\leq \frac{\sigma _{1}^{2}}{\sigma _{2}^{2}}\leq 3.360$ .

$\frac{1}{F_{1-\alpha /2}}\cdot \frac{s _{1}^{2}}{s _{2}^{2}}\leq \frac{\sigma _{1}^{2}}{\sigma _{2}^{2}}\leq \frac{s_{1}^{2}}{s _{2}^{2}}\cdot \frac{1}{F_{\alpha /2}}$

$\frac{1}{2.357}\cdot \frac{(2.5)^{2}}{(2.0)^{2}}\leq \frac{\sigma _{1}^{2}}{\sigma _{2}^{2}}\leq \frac{(2.5)^{2}}{(2.0)^{2}}\cdot \frac{1}{0.465}$

$0.663\leq \frac{\sigma _{1}^{2}}{\sigma _{2}^{2}}\leq 3.360$