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devaboone | 5 years ago

Here is what the article states: "Therefore, a multivariate logistic regression analysis was performed to adjust the model by possible confounding variables such as hypertension and type 2 diabetes mellitus for the probability of the admission to the Intensive Care Unit in patients with Calcifediol treatment vs Without Calcifediol treatment (odds ratio: 0.03 (95%CI: 0.003-0.25) (Table 3). The dependent variable considered was the need to be treated or not in ICU (dichotomous variable).) CI:-0.30 - 0.03 p:0.08." The statement is worded in a confusing way, but that is a non-significant p value. Of course we should not put too much into "statistical significance" but it is interesting to note.

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kovach|5 years ago

The only reason that p value is a bit high in that second multivariate analysis is because of the uncertainty of how much all the different risk factors like hypertension, T2DM, age >= 60 etc. affect ICU admission numbers.

But even with those variables controlled, the 95% confidence interval is 0.003-0.25, which at worst is a 4-fold reduction in ICU risk.

We should also note that the Calcifediol treatment group had 14 patients ≥ 60 years old, and the non-Calcifediol group had 5. So the study looks even better with that in mind...

harterrt|5 years ago

Unless I'm reading your comment wrong, this p-value (0.03) is actually significant for a 95% confidence test.

Jeriko|5 years ago

0.03 is the odds ratio (a measure of effect size, an odds ratio of 1 would mean the treatment and control arms had the same rate of ICU admittance). The p value is 0.08 from here "CI:-0.30 - 0.03 p:0.08." which captures the likelihood the treatment had an effect.

That said, just looking at p values and applying a cutoff at 0.05 is pretty bad practice that is getting a lot of heat thanks to the replication crisis (does it make sense to behave as though p=0.08 is not true and something at p=0.049 is true? almost certainly not). If you get a value in this range and a huge effect size then it's a really good idea to repeat the experiment with way more data. It's also a common stats error to act as though p>0.05 is the same as knowing something DOES NOT work, all you can say is this specific study wasn't able to show that it does work with 95% confidence.

pmayrgundter|5 years ago

That's part of the confidence interval. The adjusted p-value is 0.08. However, the cut-off of 0.05 is just a convention. https://en.wikipedia.org/wiki/Misuse_of_p-values. I'd think of this as a grey scale, where 0.08 is roughly in the significance range.

The null hypothesis is that it's highly unlikely VitD has an effect, and we should expect to see that substantiated often in tests. How often? 95% of the time. 5% of the time we can expect to see spurious results from our simplistic model of random processes. Upshot, it's a small change to move those numbers to 92% vs. 8%. In this context, it's fine to say "this was a small pilot that directionally shows we should do a much bigger test", which is what they're now doing.