Ivermectin looks promising

From the Daily Mail:

In leaked slides published ahead of the study’s release next month, the scientists behind the research combined results from 11 trials of the drug involving more than 1,400 patients.

This revealed only eight Covid-19 patients out of 573 who received the drug died, compared to the 44 out of the 510 who received a placebo. 

Daily Mail

This looks very promising — a death rate of 1.4% in patients on Ivermectin vs. a death rate of 8.6% in patients on a placebo.

Let’s run some numbers.

One tool we have is what’s known as the “standard error of the proportion”. If we have some number that is a proportion of some population, say, 1.4%, we can calculate the standard error — the likelihood that the “real” proportion is some other number. (Well, really, the likelihood that, if we did the same measurement on another group the same size, we’d get some other number.)

So the standard error of the proportion is given as s = (pq/n)^(1/2), or the square root of the proportion of the population that meets a criterion, times the proportion that doesn’t, divided by the sample size.

So, if 8 out of 573 Covid-19 patients on Ivermectin died, that’s a proportion of about 1.4%. The standard error is sqrt(0.014 * 0.986 / 573) = 0.00491.

Multiply that standard error by 573, and the standard error works out to 2.81 above or below the 8 test group patients who died. So if we run the same test on a bunch of groups of 573 patients, we’d expect to see between 5.2 and 10.8 deaths about two thirds of the time. Two standard error units, which is usually a publishable result, would encompass 95% of the results in other tests. That range is between 2.4 and 13.6.

The same analysis of the placebo group yields s = sqrt(0.0863 * 0.9137 / 510) = 0.01243. So our wiggle room here is 6.34. Taking the two standard deviation range again, we expect other replications of this test to fall between 31.2 deaths and 56.7 deaths.

Gee, these ranges don’t overlap. So we conclude the proportions are in fact different.

It turns out there’s a modification for when we’re comparing two proportions:

images

So for our treatment and placebo groups, we have

s = sqrt (0.014 * 0.986 / 573 + 0.0863 * 0.9137 / 510) = 0.01337.
So what we want to know is whether the proportions are statistically different from each other. Is 0.014 significantly different from 0.0863?
The difference between the two is 0.0723. Divide by the standard error, and we get 5.41 standard error units. That’s pretty darn significant.

If the numbers reported in the Daily Mail are right, I’d have to say Ivermectin works.

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