Question

Spirometry is a common test used to assess how well your lungs work, by measuring how much air you can inhale, how much you can exhale, and how quickly you can exhale. It can be used to diagnose chronic obstructive pulmonary disease (COPD) (see, for example, Schneider et al., 2009).

We want to study the effectiveness of spirometry in diagosing COPD. Suppose we gave the test to 400 people with COPD and found that 337 tested positive for COPD. We then tested 600 people without COPD and found that 551 tested negative for COPD.

Produce a contingency table of these results.

Estimate the sensitivity and specificity of the test. Interpret the specificity in the context of the data.

Calculate an approximate 95% confidence interval for the sensitivity of the test. Interpret this confidence interval in the context of the data.

Answer 1

Contingency table :

	Tested Positive	Tested negative	Total
With COPD	337	63	400
Without COPD	49	551	600
Total	386	614	1000

Sensitivity = True Positive / (True Positive + False negative ) = 337 / ( 337 + 63 ) = 0.8425

Specificity = True negative / ( true negative + false postive ) = 551 / ( 551 + 49 ) = 0.9183

We can obtain the confidence interval for the sensitivity of the test by Wilson score method

Here, p = r / n = Sensitivity

r =  the number of true positives

n = total positives

p = 0.8425

r = 337

n = 400

Z_0.975 = 1.96

Putting the values in the formula we get,

Upper limit =

(( 2 * 400 * 0.8425 + Z²_0.975 ) + Z_0.975 * ( Z²_0.975 + 4 * 400 * 0.8425 * ( 1 - 0.8425 ))^0.5) / ( 2*(400 + Z²_0.975 )) = 0.8749

Lower limit =

(( 2 * 400 * 0.8425 + Z²_0.975 ) - Z_0.975 * ( Z²_0.975 + 4 * 400 * 0.8425 * ( 1 - 0.8425 ))^0.5) / ( 2*(400 + Z²_0.975 ) )= 0.8036

95% confidence interval for the sensitivity of the test = ( 0.8036 , 0.8749 )