How to calculate $p$-value in a clinical trial

Question

I'm trying to verify the $p$-value mentioned in this article:

$8\%$ ($n=\frac{1}{13}$) of mavrilimumab-treated patients progressed to mechanical ventilation by Day $28$, compared to $35\%$ ($n=\frac{9}{26}$) of control-group patients who progressed to mechanical ventilation or died ($\textbf{p=0.077}$).

How can I calculate the $p$-value for the data above?

I used $3$ different calculators and they all gave different results, none of them matches with the article.

Calculator 1: $p=0.12$

Calculator 2: $p = 0.06876$

Calculator 3: $p = 0.154$

Answer 1

Without knowing which test they chose, and whether the test was stratified or adjusted by any covariates, it is unlikely that we can replicate the $p$-value.

For instance, we could use Fisher's exact test, the chi-squared test (which is the same as the two-sample independent proportion test with a pooled standard error), or a two-sample independent proportion test with an unpooled standard error, or a likelihood ratio test, and these are just different choices of statistic, not considering whether continuity correction is used, or if there was some other adjustment for other prognostic factors that are not stated in the press release.

That said, I would not be surprised if there was an error. Small biotech companies don't always perform the analyses correctly. But to really know for sure, one would have to read the study protocol and the statistical analysis plan, neither of which is generally available to the public.

Answer 2

I can't explain the 'why'. However, looking at an abstract for the research, all of the $p$-values are different:

$\big[\frac{0}{13}, \frac{7}{26}\big]$ Globenewswire: $p=0.086$, Scientific Abstracts: $\log$ rank $p=0.046$

$\big[\frac{13}{13}, \frac{17}{26}\big]$ Globenewswire: $p=0.0001$, Scientific Abstracts: $p=0.018$

$\big[\frac{10}{11}, \frac{11}{18}\big]$ Globenewswire: $p=0.0093$, Scientific Abstracts: $p=0.110$