Total CRQ score formulas: Standard
errors and standard deviations for total
CRQ scores
To calculate the standard deviations and
 m  mXE  mXM  mXD 
SEx  SE  XF
 
4


 m  mXE  mXM  mXD 
Var  XF

4


standard errors for the total CRQ treatment
group mean, and the total CRQ control group
and the standard deviation by multiplying the
mean, and the total CRQ mean difference, we
standard error by the squareroot of the
assumed a correlation of ρ=0.5. Let mEF, mEE,
number of patients in the treatment group, nx
mEM and mED denote the means for the
Fatigue(F), Emotional(E), Mastery(M) and
Dyspnea(D) domains in the experimental
group, let mCF, mCE, mCM and mCD denote the
means for the four domains in the control
 m  mXE  mXM  mXD 
SDx  SD  XF
 
4


 m  mXE  mXM  mXD 
nx  SD  XF

4


group. Let x denote the treatment group
(either E or C). We calculate the total
variance for the average of treatment means
across domains using the formula
Now let mF, mE, mM and mD denote the mean
differences for the four domains. We obtain
the standard error of the average of the mean
differences across domains using the formula
 m  mXE  mXM  mXD 
Var  XF
 
4



1
  Var (mxi )    Cov  mxi , mxj  
16  i {F , E , M , D}
i { F , E , M , D} i  j

SDE2 SDC2
 m  mE  mM  mD 
SE  F



4
nE
nC


MID units formulas: Pooling MID
Where we use that
Cov  xi , x j    Var  xi Var  x j 
 0.5 Var  xi  Var  x j 
standardized mean differences
Assume that a trial reports a MD on some
disease-specific HRQL instrument X, and
assume that the minimally important different
for instrument X, MIDX, has been established.
We obtain the standard error by taking the
squareroot of the variance
The estimated MD is a random variable. If we
standardize this random variable by dividing
it by the MIDX, we get a new random
variable, MD/MIDX. We know from basic
probability theory that because MIDX is a
mi 
MDi
MIDB
and
Var  mi  
Var  MDi 
MIDB2
constant, the variance of MD/MIDX is given
by
 MD
Var 
2
 MIDX
 Var  MD 

MIDX2

By defining the trial weights as wi=Var(mi)-1,
we can use the fixed-effect model inverse
variance method to pool the MIDstandardized mean differences using the
That is, the variance of the mean difference
formula
divided by the square of the MID. Further, the
standard error of MD/MIDX is given by
Var  MD  SE  MD 
 MD 
SE 



2
MIDX2
MIDX
 MIDX 
 k
  k

mˆ    wi  mi  /   wi 
 i 1
  i 1 
Where m̂ denotes the pooled MIDstandardized mean difference. The standard
error of m̂ can be calculated using the formula
Now suppose a meta-analysis included k
trials. The first j trials use disease-specific
instrument A, and the last k-j trials use
disease-specific instrument B. Let MDi denote
the mean difference observed in trial i, let
MIDA denote the minimally important
difference established for instrument A, and
let MIDB denote the minimally important
difference established for instrument B.
Further, let mi denote the MID standardized
effect for trial i.
To pool results across trials using MIDs we
must first estimate the mi and its associated
variance for all trials. For i=1, …, j we have
mi 
MDi
MIDA
and
Var  mi  
and for i=j+1, …, k we have
Var  MDi 
MIDA2
 k

se(mˆ )  1 /   wi 
 i 1 
and confidence intervals can subsequently be
derived. Pooling of MID-standardized mean
differences is naturally extended to the
random-effects model using weights
wi=(Var(mi)+τ2)-1, where τ2 denotes the
between-trial variance.