Download stat 360*regression analysis

STAT 405 - BIOSTATISTICS
Handout 9 – Number Needed to Treat (NNT), Number Needed to Harm
(NNH) and Attributable Risk (AR)
The Absolute Risk Reduction (ARR) is the difference between the event rate in the
experimental group and the event rate in the control group. In other words, this is the
change in absolute risk brought about by an experimental intervention.
The Number Needed to Treat (NNT) is the reciprocal of the ARR.
EXAMPLE: An analgesic agent is given to 100 people, and 70 have their pain relieved
within two hours. In contrast, the administration of a placebo tablet containing no active
drug leads to pain relief in only 20 out of 100 people.
Source: http://www.evidence-based-medicine.co.uk/ebmfiles/WhatisanNNT.pdf
Questions:
1. What is the ARR?
2. What is the NNT?
Interpretation: Two people must be given the analgesic for one of them to
obtain effective pain relief.
EXAMPLE: Consider the use of a thrombolytic agent after myocardial infarction. Say
that 10,000 men have no thrombolytic treatment after a heart attack, and 1,000 die
within six weeks. In contrast, of 10,000 men given a thrombolytic agent, the number
dying within six weeks is reduced to 800.
Source: http://www.evidence-based-medicine.co.uk/ebmfiles/WhatisanNNT.pdf
Questions:
1. What is the ARR?
2. What is the NNT?
1
Interpretation: Fifty people must be given the thrombolytic therapy after a
heart attack to prevent one of them from dying within six weeks who would have
died had they not been given thrombolysis.
These interpretations lead to an alternative definition of NNT:
The Number Needed to Treat (NNT) can also be defined as the number of persons
who must be treated for a given period to achieve an event (treatment) or to prevent an
event (prophylaxis).
CALCULATING NNTs AND CONFIDENCE INTERVALS
The NNT can be calculated from the simple formula:
NNT 
1
1

ARR (Proportio n who benefit from treatment - Proportion who benefit from control)
Questions:
1. What does an NNT of 1 imply?
2. If a treatment works well, do you expect to see a “large” NNT? Explain.
We have already discussed the calculation of NNT for two examples. However, it should
be noted that this point estimate is likely to change if we took another sample.
Therefore, we should also report a confidence interval for NNT.
The most commonly used method for constructing a confidence interval for NNT
involves inverting and exchanging the confidence limits for the ARR.
Confidence Interval for the ARR
𝑝
̂(1
−𝑝
̂)
𝑝
̂(1
−𝑝
̂)
1
1
2
2
𝐴𝑅𝑅 ± 𝑧𝛼 √
+
𝑛1
𝑛2
2
2
This is the same the confidence interval for (𝑝1 − 𝑝2 ) covered in introductory statistics
courses.
Confidence Interval for the NNT
Simply invert and exchange the confidence limits for the ARR.
EXAMPLE: Consider the example in which an analgesic agent is given to 100 people,
and 70 have their pain relieved within two hours. In contrast, the administration of a
placebo tablet containing no active drug leads to pain relief in only 20 out of 100 people.

Find a 95% confidence interval for ARR:

Find a 95% confidence interval for NNT:
Another Method for Finding a Confidence Interval for the NNT
Ralf Bender, an epidemiologist, argues that the method used above to find a confidence
interval for the ARR often leads to unreliable confidence limits. He argues that instead
of using this method, one should use Wilson’s score method for calculating the
confidence interval for ARR. You can read more in his paper titled “Improving the
calculation of confidence intervals for the number needed to treat.”
He has made a SAS program available which calculates the ARR, NNT, and confidence
intervals using both the “unreliable” method and his proposed method. The program is
available in the file Bender_CIs.sas.
To use this program, you need only enter your data as follows:
e1
n1
e2
n2
=
=
=
=
70; /* Number who benefit in experimental group */
100; /* Sample size of experimental group */
20; /* Number who benefit in control group */
100; /* Sample size of control group */
The rest of the program should remain untouched!
3
Output:
General Comments on the NNT and the Confidence Interval:
1. The Wilson method has been shown to have a higher coverage rate than the
asymptotic method.
2. The asymptotic method is most unreliable with very small sample sizes or with a
very low absolute risk reduction (ARR).
4
3. Any estimated NNT should be accompanied by its confidence interval, and it is
good practice to state which calculation method was used for the interval.
4. The only time it is appropriate to compare NNTs is if we have NNTs for different
interventions for the same condition with the same outcome of interest.
Number Needed to Harm (NNH)
For adverse effects, we can calculate a number needed to harm (NNH).
NNH 
1
1

ARR (Prop.adversely affected by treatment- Prop.adverselyaffected by control)
This can be regarded as the number of persons that must be given a treatment in order to
cause harm to one patient that would not have otherwise been harmed.
EXAMPLE: An example of a drug that was removed from the market due to a low NNH
is encainide. A 1991 study in the New England Journal of Medicine found proarrhythmic
effects of encainide and flecainide were more likely in the antiarrhythmic group than in
the placebo group. Data from the study are presented below:
Treatment
Subjects Death/cardiac arrest
placebo
743
26
flecainide or encainide
755
63
a=
b=
c=
d=
Source:
http://smbrower.com/mediawiki/index.php/Understanding_basic_statistics#Number_needed_to_harm
Questions:
1. Find the NNH.
2. Interpret this quantity.
3. Find a 95% confidence interval for the NNH.
5
6
Attributable Risk (AR)
In some cases, a risk factor may have a large RR. However, if the risk factor is
relatively rare, only a small proportion of cases may be attributable to this risk
factor. Conversely, if a risk factor is common, then even a moderate RR may
translate to a large number of cases attributable to the risk factor.
Let
RR = be the relative risk for a disease given the person has the risk factor
d = probability of disease for persons without the risk factor
then
_______ = probability of disease for persons with the risk factor.
Now suppose that
p = probability that a person has the risk factor
then the overall probability of disease is
𝑝𝐷 =
If all persons with the risk factor become risk factor free what proportion of cases or
percent of cases would be eliminated?
This is called the attributable risk (AR) which is usually expressed a percent.
AR =
7
Confidence Interval for AR (Equation 13.12, pg.603)
(100%
𝑒 𝑐1
𝑒 𝑐2
,
100%
)
1 + 𝑒 𝑐1
1 + 𝑒 𝑐2
Here (𝑐1 , 𝑐2 ) is the interval obtained from
𝑦 ± 𝑧1−𝛼 (
2
̂
𝑅𝑅
𝑏
𝑑
+
)√
̂
𝑎𝑛1 𝑐𝑛2
|𝑅𝑅 − 1|
and
𝑦 = ln (
̂
𝐴𝑅
̂
100 − 𝐴𝑅
)
Note: There are a very nice derivations of the standard errors (variances) of the RR,
OR, and AR used in the confidence intervals presented in the text. These derivations use
the delta method which is covered in mathematical statistics (STAT 450/460). You
should definitely examine the derivation of the var(RR), var(OR), and var(AR) in the
book if you have this background!
Example of AR drug example:
An example of a drug that was removed from the market due to a low NNH is encainide.
A 1991 study in the New England Journal of Medicine found proarrhythmic effects of
encainide and flecainide were more likely in the antiarrhythmic group than in the
placebo group. Data from the study are presented below:
Treatment
Subjects Death/cardiac arrest
placebo
743
26
flecainide or encainide
755
63
If all patients taking flecainide or encainide were moved to placebo what is the associated
attributable risk (AR) and the associated CI?
> AR(63,755,26,743,p=.05)
Estimated AR = 6.006
CI for AR = ( 2.804 , 12.4 )
> AR(63,755,26,743,p=.20)
Estimated AR = 20.36
CI for AR = ( 10.35 , 36.15 )
> AR(63,755,26,743,p=.50)
Estimated AR = 38.99
CI for AR = ( 22.39 , 58.6 )
8
Analysis of Cardiac Arrest
Deaths Data in JMP
Comparing Cardiac Arrest Deaths (Drug vs. Placebo)
Contingency Table
Count
Row %
Drug
Placebo
N
Y
755
92.30
743
96.62
1498
63
7.70
26
3.38
89
Fisher's Exact
Test
Left
Right
2-Tail
818
769
1587
Prob Alternative Hypothesis
0.0001* Prob(Death/Cardiac Arrest=Y) is greater for Group=Drug than Placebo
1.0000 Prob(Death/Cardiac Arrest=Y) is greater for Group=Placebo than Drug
0.0002* Prob(Death/Cardiac Arrest=Y) is different across Group
Relative Risk
Description
P(Y|Drug)/P(Y|Placebo)
Relative Risk
2.277929
Lower 95%
1.458103
Upper 95%
3.558707
Odds Ratio
Odds Ratio
0.419364
Lower 95%
0.26264
Upper 95%
0.669609
Two Sample Test for Proportions
Description
P(Y|Drug)-P(Y|Placebo)
Adjusted Wald Test
P(Y|Drug)-P(Y|Placebo) ≥ 0
P(Y|Drug)-P(Y|Placebo) ≤ 0
P(Y|Drug)-P(Y|Placebo) = 0
Proportion
Difference
0.043207
Lower 95%
Upper 95%
0.020547
0.065512
Prob
<.0001*
0.9999
0.0002*
9
In R we can write simple programs for inference for RR, OR, and AR
You can simply copy and paste this code in R and these functions will be available to you forever
as long as your workspace.
Odds Ratio (OR)
OR = function (a, b, c, d, alpha = 0.05)
{
OR <- a * d/(b * c)
logOR <- log(OR)
SElogOR <- sqrt((1/a) + (1/b) + (1/c) + (1/d))
marerr <- qnorm(1 - (alpha/2)) * SElogOR
LCL <- exp(logOR - marerr)
UCL <- exp(logOR + marerr)
cat("\n")
cat(paste("Estimated OR =", OR))
cat("\n")
cat(paste("CI for OR = (", format(LCL, dig = 4), ",", format(UCL,
dig = 4), ")"))
cat("\n")
}
Relative Risk (RR)
RR = function (a, b, c, d, alpha = 0.05)
{
RR <- (a/(a + b))/(c/(c + d))
logRR <- log(RR)
SElogRR <- sqrt((b/(a * (a + b))) + (d/(c * (c + d))))
marerr <- qnorm(1 - (alpha/2)) * SElogRR
LCL <- exp(logRR - marerr)
UCL <- exp(logRR + marerr)
cat("\n")
cat(paste("Estimated RR =", RR))
cat("\n")
cat(paste("CI for RR = (", format(LCL, dig = 4), ",", format(UCL,
dig = 4), ")"))
cat("\n")
}
Attributable Risk (AR)
AR = function(a,b,c,d,p=.5,alpha=.05) {
RR <- (a/(a+b))/(c/(c+d))
logRR <- log(RR)
SElogRR <- sqrt((b/(a*(a+b))) + (d/(c*(c+d))))
za2 = qnorm(1 - (alpha/2))
AR = 100*(RR - 1)*p/((RR - 1)*p + 1)
y = log(AR/(100 - AR))
marerr <- za2*(RR/abs(RR-1))*SElogRR
c1 = y - marerr
c2 = y + marerr
LCL <- 100*exp(c1)/(1+exp(c1))
UCL <- 100*exp(c2)/(1+exp(c2))
cat("\n")
cat(paste("Estimated AR =",format(AR,dig=4)))
cat("\n")
cat(paste("CI for AR = (",format(LCL,dig=4),",",format(UCL,dig=4),")"))
cat("\n\n")
}
10