12
$\begingroup$

I was reading through a paper and I saw a table with a comparison between PPV (Positive Predictive Value) and NPV (Negative Predictive Value). They did some kind of statistical test for them, this is a sketch of the table:

PPV    NPV    p-value
65.9   100    < 0.00001
...

Every rows refers to a particular contingency table.

What kind of hypothesis test did they do? Thanks!

$\endgroup$
17
$\begingroup$

Assuming a cross-classification like the one shown below (here, for a screening instrument)

alt text

we can define four measures of screening accuracy and predictive power:

  • Sensitivity (se), a/(a + c), i.e. the probability of the screen providing a positive result given that disease is present;
  • Specificity (sp), d/(b + d), i.e. the probability of the screen providing a negative result given that disease is absent;
  • Positive predictive value (PPV), a/(a+b), i.e. the probability of patients with positive test results who are correctly diagnosed (as positive);
  • Negative predictive value (NPV), d/(c+d), i.e. the probability of patients with negative test results who are correctly diagnosed (as negative).

Each four measures are simple proportions computed from the observed data. A suitable statistical test would thus be a binomial (exact) test, which should be available in most statistical packages, or many online calculators. The tested hypothesis is whether the observed proportions significantly differ from 0.5 or not. I found, however, more interesting to provide confidence intervals rather than a single significance test, since it gives an information about the precision of measurement. Anyway, for reproducing the results you shown, you need to know the total margins of your two-way table (you only gave the PPV and NPV as %).

As an example, suppose that we observe the following data (the CAGE questionnaire is a screening questionnaire for alcohol):

alt text

then in R the PPV would be computed as follows:

> binom.test(99, 142)

    Exact binomial test

data:  99 and 142 
number of successes = 99, number of trials = 142, p-value = 2.958e-06
alternative hypothesis: true probability of success is not equal to 0.5 
95 percent confidence interval:
 0.6145213 0.7714116 
sample estimates:
probability of success 
             0.6971831 

If you are using SAS, then you can look at the Usage Note 24170: How can I estimate sensitivity, specificity, positive and negative predictive values, false positive and negative probabilities, and the likelihood ratios?.

To compute confidence intervals, the gaussian approximation, $p \pm 1.96 \times \sqrt{p(1-p)/n}$ (1.96 being the quantile of the standard normal distribution at $p=0.975$ or $1-\alpha/2$ with $\alpha=5$%), is used in practice, especially when the proportions are quite small or large (which is often the case here).

For further reference, you can look at

Newcombe, RG. Two-Sided Confidence Intervals for the Single Proportion: Comparison of Seven Methods. Statistics in Medicine, 17, 857-872 (1998).

$\endgroup$
  • $\begingroup$ Thanks. Ok, I read in very beginning of the paper that they used Chi-square test for all categorical variables. The classification table written doesn't refer to a variable in particular, it is the output of a classification task. It isn't very clear! Now I suppose they did a classical test on proportion.. maybe Chi-square.. $\endgroup$ – Simone Jan 17 '11 at 14:53
  • $\begingroup$ I had a look to this question again and I have seen that p-value doesn't refer neither PPV nor NPV, it refers to the entire row. I think the test they had should be associated to the whole contingency-table. $\endgroup$ – Simone Apr 12 '11 at 13:33
  • $\begingroup$ @Simone So, if I understand you correctly, you suggest that the authors provide PPV and NPV values but gave the p-value corresponding to a global association test of the 2x2 table? Is it related to this recent question, stats.stackexchange.com/questions/9464/…? $\endgroup$ – chl Apr 12 '11 at 18:37
  • $\begingroup$ Yes, it would be related to that question if p-value were associated either to PPV or to NPV. And in that case you gave the solution. The test corresponds to the whole 2x2 table, I won't ever know what kind of test it is! $\endgroup$ – Simone Apr 12 '11 at 20:11
1
$\begingroup$

Please see

Kosinski, Andrzej S. A weighted generalized score statistic for comparison of predictive values of diagnostic tests. Statistics in Medicine http://dx.doi.org/10.1002/sim.5587 published online: 22 AUG 2012

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.