Question

In order to determine that the precision of leveling equipment is appropriate for the intended measuring 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

Confidence interval for the ratio of the two population variances is given by,

Lower bound =    $\frac{S_{1}^{2}/S_{2}^{2}}{F_{\alpha /2}}$ ;  F_(α/2) follows f distribution with d.f₁ = n₁-1 and d.f₂ =n₂-1

Upper bound =  $\frac{S_{1}^{2}/S_{2}^{2}}{F_{1-(\alpha /2)}}$ ; F_1-(α/2) follows f distribution with d.f₁ = n₁-1 and d.f₂ =n₂-1

We are given S₁ = 2.5 and S₂ = 2 ; therefore  S₁² = 6.25 and S₂² = 4

Therefore  S₁² /  S₂² = 6.25/4 = 1.5625

and given d.f₁ = 24 and d.f₂ = 15

We can find F_(α/2) using excel function =FINV( α/2, d.f₁ , d.f₂ )

and  F_1-(α/2) using excel function =FINV( 1- α/2, d.f₁ , d.f₂ )

We are given confidence level = 0.90 ; therefore α = 1 - confidence level = 1 - 0.90 =0.10

So α = 0.10 , α/2 = 0.05 and 1 - (α/2) = 0.95

For F_(α/2)     =FINV( 0.05,24,15 ) = 2.2878

For F_1-(α/2)     =FINV( 0.95,24,15 ) = 0.4745

Lower bound =    $\frac{S_{1}^{2}/S_{2}^{2}}{F_{\alpha /2}}$ = $\frac{1.5625}{2.2878}$

Lower bound = 0.6830

Upper bound =  $\frac{S_{1}^{2}/S_{2}^{2}}{F_{1-(\alpha /2)}}$ = $\frac{1.5625}{0.4745}$

Upper bound = 3.2932

Therefore 90% confidence interval for the ratio of the two population variances associated with the two tests is

( 0.6830 , 3.2932 )