That was a very nice article, well-researched and thorough; However, I do think I need to point out that you have made an error in your maths. (Interestingly, the effect of the error was to increase the cost of getting a flu shot).
Specifically, take a look at figure one, at the probabilities. You have stated, earlier, that the flu injection has an efficacy of 60%; that is to say, 60% of people who get the flu shot, who would otherwise have got the flu, don’t get the flu.
Given that 90% of people don’t get the flu, whether they have the shot or not, this means that the odds of not getting the flu after getting the shot should be 96%, not 94%.
Similarly, the odds of being hospitalised should be 0.0002, not 0.0003; and the odds of getting the flu but not being hospitalised should be 0.398, not 0.597.
In short, your calculations presumed an efficacy of only 40%. (The fact that it was still shown to be worthwhile, with such a low efficacy, is quite telling).
This reduces the expected cost of getting the injection to $71.40.
That minor point aside, thank you very much for doing the research for this article.
This error has been corrected; thank you for pointing this out!
I actually did a bit more research, and it really seems like flu vaccine efficacy in healthy adults is more like 70% (and sometimes as high as 90%), despite the fact that the average efficacy of the vaccine throughout the population is around 60%. The reason that efficacy in healthy adults is so high, relative to the average efficacy, is that efficacy in the elderly is around 30-40%.
Also, note that about 42% of the US population gets flu shots on any given year. So, if 10% of people on average get the flu, and the vaccine is 60% efficacious throughout the population, then we can write the following equations, defining sick1 as the event in which a person who was vaccinated gets the flu, and sick2 as the event in which a person who was not vaccinated gets the flu:
p(sick1) x 0.42 + p(sick2) x 0.58 = 1 x 0.10
p(sick1) = p(sick2) x 0.60
Solving this system of equations, we get:
p(sick1) = 0.0721
p(sick2) = 0.120 (previous typo: had been written as 0.0120)
The practical implication of this is that the conservative analysis conducted in this report, and shown in Figure 1, assumes that around 5.7% (rather than a more realistic 10 or 12%) of the population will catch the flu in any given year.
So the equations should be (definition of vaccine efficacy from wikipedia)
.6 * p(sick2) = p(sick2) - p(sick1) p(sick1) - .4 p(sick2) = 0
.
i.e. efficacy is the difference be the unvaccinated and vacinated rates of infection divided by the unvaccinated rate. You have to assume there is no selective pressure in terms of who gets the vaccine (they have the same risk pool as the normal population for flu which is surely untrue) to get your assumtion that
This confusion is due to the fact that the system of two equations I wrote in my comment above was originally crammed onto one line, rather than being separated onto two lines. Sorry! This formatting error made the comment hard to read, and has since been corrected. The correct system of equations to solve is:
That was a very nice article, well-researched and thorough; However, I do think I need to point out that you have made an error in your maths. (Interestingly, the effect of the error was to increase the cost of getting a flu shot).
Specifically, take a look at figure one, at the probabilities. You have stated, earlier, that the flu injection has an efficacy of 60%; that is to say, 60% of people who get the flu shot, who would otherwise have got the flu, don’t get the flu.
Given that 90% of people don’t get the flu, whether they have the shot or not, this means that the odds of not getting the flu after getting the shot should be 96%, not 94%.
Similarly, the odds of being hospitalised should be 0.0002, not 0.0003; and the odds of getting the flu but not being hospitalised should be 0.398, not 0.597.
In short, your calculations presumed an efficacy of only 40%. (The fact that it was still shown to be worthwhile, with such a low efficacy, is quite telling).
This reduces the expected cost of getting the injection to $71.40.
That minor point aside, thank you very much for doing the research for this article.
This error has been corrected; thank you for pointing this out!
I actually did a bit more research, and it really seems like flu vaccine efficacy in healthy adults is more like 70% (and sometimes as high as 90%), despite the fact that the average efficacy of the vaccine throughout the population is around 60%. The reason that efficacy in healthy adults is so high, relative to the average efficacy, is that efficacy in the elderly is around 30-40%.
Also, note that about 42% of the US population gets flu shots on any given year. So, if 10% of people on average get the flu, and the vaccine is 60% efficacious throughout the population, then we can write the following equations, defining sick1 as the event in which a person who was vaccinated gets the flu, and sick2 as the event in which a person who was not vaccinated gets the flu:
p(sick1) x 0.42 + p(sick2) x 0.58 = 1 x 0.10
p(sick1) = p(sick2) x 0.60
Solving this system of equations, we get:
p(sick1) = 0.0721
p(sick2) = 0.120 (previous typo: had been written as 0.0120)
The practical implication of this is that the conservative analysis conducted in this report, and shown in Figure 1, assumes that around 5.7% (rather than a more realistic 10 or 12%) of the population will catch the flu in any given year.
...I’m sorry, but I’m rather pedantic when it comes to maths.
I believe that should be p(sick1) = 0.072 and p(sick2) = 0.12, not 0.012.
Which I’m pretty sure is just a typo, and not an actual math error...
So the equations should be (definition of vaccine efficacy from wikipedia)
.6 * p(sick2) = p(sick2) - p(sick1)
p(sick1) - .4 p(sick2) = 0 . i.e. efficacy is the difference be the unvaccinated and vacinated rates of infection divided by the unvaccinated rate. You have to assume there is no selective pressure in terms of who gets the vaccine (they have the same risk pool as the normal population for flu which is surely untrue) to get your assumtion that
.42 p(sick1) + .58p(sick2) = .1 p(sick1) + 1.38p(sick2) = .238
or 1.78 p(sick2) = .238
p(sick2)=.13 (weird I getting a different result) p(sick1) = .05
Did I solve wrong or did you. I do math so I can’t actually manipulate numbers very well but I not seeing the mistake.
This confusion is due to the fact that the system of two equations I wrote in my comment above was originally crammed onto one line, rather than being separated onto two lines. Sorry! This formatting error made the comment hard to read, and has since been corrected. The correct system of equations to solve is:
p(sick1) x 0.42 + p(sick2) x 0.58 = 1 x 0.10
p(sick1) = p(sick2) x 0.60
Instead of:
.42 p(sick1) + .58p(sick2) = .238
.1 p(sick1) + 1.38p(sick2) = .238.