The confidence intervals for NNE computed in this article were computed by inverting the lower and upper end confidence intervals for ARI. This confidence intervals are very wide. My suggestion is to compute the confidence intervals like these for the RR using (log(1/ARI^) as an estimate for log NNE. Then you have to compute the confidence interval for this log estimate and after this using the exponentialfunction to compute the upper and lower ends for the confidence intervals for NNE. E.g in table 5 we get a 95% CI as follows: 101-148. This confidence interval is much more closer to 122.37 instead of 69.55-508.69 . The estimate log(1/ARI^) could be handled like a continious function of two estimates.
Competing interests
There are no conflicts of interests
Addition to my letter "confidence interval for NNE
Rainer Beier, Federal institute for drugs and medical devices (BfArM)
23 July 2010
In the following a simulation of the problem explained in my first letter to this article posted on 14th of April: The simulation is done with the statistical program R where k is the difference of the upper ends of my suggestion for calculating confifidence intervals and done in the article and v is the difference of the length of the two kinds of calculating confidence intervals with one million trials on the interval (0,1) as follows:
We see that in 965669 cases the upper end of the CI I suggested is minor than of the well known one and also the distance of the suggested one is also minor than of the well known one. So it has been shown that the suggested confidence interval, interpreting the estimate of NNE as a continious function of confidence intervals is more closer to the true value to be estimated. Also this could be proved directly with mathematical methods
Competing interests
There are no conflicts of interests
proof of assumption done in the letter posted on 14th of April
Rainer Beier, Federal institute for drugs and medical devices
23 July 2010
In addition to my posted letters on 14th of April the proof of the assumption is given as follows:
Let p1= P(E+/K+) and p2=P(E-/K+) be the being the well known probabilties of diseased patients being exposed and not exposed.Then the following 95% CI for the true value log(NNE) = log(1/(p1-p2)) is given as follows:
The estimate of NNE was considered as a continious function of two variables f(p1^,p2^) in IR2
Now we have to compare the upper and lower limits of the confidence interval above with the well known confidence interval calculated by the inverse of the limits due to the confidence interval for the risk difference :
We have to prove the following inequality of the distances of the two kind of confidence intervals as mentioned above:
Now we can write the the inequality to be proven as follows using a further substitiution
x=u/(p1-p2) and we get the following transformed inequality
exp(x) – exp(-x) 1/(1-x) – 1/(1+x)
It is well known that the following two inequalities hold for all not negative real numbers x :
exp(x) 1/(1-x) can be proved using the mean value theorem and shows that the upper end of the suggested confidence interval is lower the the upper end of the well known one.
Also exp(x) – exp(-x) 1/(1-x) – 1/(1+x) can be shown using the mean value theorem as well.
Now x and u have to be resubstituted leads to the assumption above.
confidence intervals for NNE
14 April 2010
The confidence intervals for NNE computed in this article were computed by inverting the lower and upper end confidence intervals for ARI. This confidence intervals are very wide. My suggestion is to compute the confidence intervals like these for the RR using (log(1/ARI^) as an estimate for log NNE. Then you have to compute the confidence interval for this log estimate and after this using the exponentialfunction to compute the upper and lower ends for the confidence intervals for NNE. E.g in table 5 we get a 95% CI as follows: 101-148. This confidence interval is much more closer to 122.37 instead of 69.55-508.69 . The estimate log(1/ARI^) could be handled like a continious function of two estimates.
Competing interests
There are no conflicts of interests
Addition to my letter "confidence interval for NNE
23 July 2010
In the following a simulation of the problem explained in my first letter to this article posted on 14th of April:
The simulation is done with the statistical program R where k is the difference of the upper ends of my suggestion for calculating confifidence intervals and done in the article and v is the difference of the length of the two kinds of calculating confidence intervals with one million trials on the interval (0,1) as follows:
> x=runif(1000000,0,1)
> y=runif(1000000,0,1)
> n=1000
> m=1000
> a=x
> b=y
> f=x*(1-x)/n
> g=y*(1-y)/m
> j=(1/(a-b))*exp(-1.96*(sqrt((f+g)/(a-b)^2)))-1/((a-b)+1.96*sqrt(f+g))
> k=(1/(a-b))*exp(1.96*(sqrt((f+g)/(a-b)^2)))-1/((a-b)-1.96*sqrt(f+g))
> v=k-j
> w=data.frame(j,k,v,sum(v<0))
> sum(k<0)
[1] 965669
> sum(v<0)
[1] 957320
We see that in 965669 cases the upper end of the CI I suggested is minor than of the well known one and also the distance of the suggested one is also minor than of the well known one. So it has been shown that the suggested confidence interval, interpreting the estimate of NNE as a continious function of confidence intervals is more closer to the true value to be estimated. Also this could be proved directly with mathematical methods
Competing interests
There are no conflicts of interests
proof of assumption done in the letter posted on 14th of April
23 July 2010
In addition to my posted letters on 14th of April the proof of the assumption is given as follows:
Let p1= P(E+/K+) and p2=P(E-/K+) be the being the well known probabilties of diseased patients being exposed and not exposed.Then the following 95% CI for the true value log(NNE) = log(1/(p1-p2)) is given as follows:
Log(NNE) 1.96*sqrt(((Var(p^1)+Var(p^2))/(p1-p2)2)
So the 95% confidence interval for NNE is now given as follows:
NNE*exp( 1.96*sqrt(((Var(p^1)+Var(p^2))/(p1-p2)2))
The estimate of NNE was considered as a continious function of two variables f(p1^,p2^) in IR2
Now we have to compare the upper and lower limits of the confidence interval above with the well known confidence interval calculated by the inverse of the limits due to the confidence interval for the risk difference :
We have to prove the following inequality of the distances of the two kind of confidence intervals as mentioned above:
NNE*(exp(1.96*sqrt(((Var(p^1)+Var(p^2))/(p1-p2)2))
- exp(-1.96*sqrt(((Var(p^1)+Var(p^2))/(p1-p2)2)))
1/(p1-p2-1.96*sqrt( Var(p^1)+Var(p^2))) - 1/(p1-p2+1.96*sqrt( Var(p^1)+Var(p^2)))
Proof
Now we have to make a substitution as follows:
Let u = 1.96*sqrt( Var(p^1)+Var(p^2)), so we get
NNE*(exp(u/(p1-p2))) - exp(-(u/(p1-p2)))
1/(p1-p2-u) - 1/(p1-p2+u)
Now we can write the the inequality to be proven as follows using a further substitiution
x=u/(p1-p2) and we get the following transformed inequality
exp(x) – exp(-x) 1/(1-x) – 1/(1+x)
It is well known that the following two inequalities hold for all not negative real numbers x :
exp(x) 1/(1-x) can be proved using the mean value theorem and shows that the upper end of the suggested confidence interval is lower the the upper end of the well known one.
Also exp(x) – exp(-x) 1/(1-x) – 1/(1+x) can be shown using the mean value theorem as well.
Now x and u have to be resubstituted leads to the assumption above.
Competing interests
There is no conflict of interest