# Calculating confidence intervals for impact numbers

*BMC Medical Research Methodology* 2006, **6**:32
doi:10.1186/1471-2288-6-32

- proof of assumption done in the letter posted on 14th of April
- Addition to my letter "confidence interval for NNE
- confidence intervals for NNE

### Addition to my letter "confidence interval for NNE

**Rainer Beier**
(2010-07-23 16:03) Federal institute for drugs and medical devices (BfArM)

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

### confidence intervals for NNE

**Rainer Beier**
(2010-04-14 02:50) BfArM

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

## proof of assumption done in the letter posted on 14th of April

Rainer Beier(2010-07-23 16:03) Federal institute for drugs and medical devicesIn 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 interestsThere is no conflict of interest

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