Download Weighted Comparison of Means

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Statistics Notes: Weighted Comparison of Means
Author(s): J. Martin Bland and Sally M. Kerry
Reviewed work(s):
Source: BMJ: British Medical Journal, Vol. 316, No. 7125 (Jan. 10, 1998), p. 129
Published by: BMJ Publishing Group
Stable URL: http://www.jstor.org/stable/25176703 .
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Clinical review
Statistics Notes
Weighted
J Martin
Bland,
Sally M Kerry
In a recent
Statistics
two
t test Here
sample
were
data
we
to a
weighted
it is done. The
referred
describe
how
to intervention
two
the
compare
two
1 Number
of requests conforming
and control groups
intervention
of percentages
each observation
on
the result
As
equal
impact
fewer
than others,
contributed
requests
a
to have
practices
difference.
can
We
lesser
do
on
effect
No of requests
+ 7x100
20x100
2
(practice)
some
4_31_90_28_82
5_ 20_90_20_80
6_24_88_19_79
the
of
estimate
the
the practices
7_7_86_9_78
by
8_6_83_25_76
in each
percentage
...+18x56
% Conforming,
84
9_30_83
+
+
group
No of requests
"_7_100_37_89
practices
wish
these
we
this by
weighting
+16x94
20 + 7+16
Control
3_16_94_38_
we
group,
add the observations
and divide
simply
together
by the
number
of observations.
the
To
calculate
weighted
we
each
observation
the
mean,
add,
multiply
by
weight,
sum of the
then divide
by the
weights:
To
in the
100
7 100
the
by
for each practice
% Conforming
20 1
the number
of requests.
calculate
the mean
to guidelines
Intervention group
Practices
groups.2
sets
tmethod,
sample
or control
an
has
Table
from
of requests
percentage
general
were
for x ray examinations
which
judged
had
(table
1), where
general
practitioners
appropriate
been
randomised
usual
we
Note1
the
practitioners
If we
of means
comparison
34
--=
...+18
105
_120_75
10_66_80_88_73
5
11_
22
80
68
12_43_77_76_68
13_43_74_21_67
U
"
~_23_70_126_66
15
_64_69
_22_64
16_6_67_34__62
=
79.50
17_18_
429
56_10_40
Total_429_702_
If
the
are
weights
unweighted,
is not
79.50,
times
unweighted
and
square
a correction
the
of
number
sum
weighted
+
(20 + 7+16
=-=
To
the weighted
to get a
get
weights;
of the
weights.)
we
subtract
mean
the
Dividing
gives
16 =
tion,
this
sum
weighted
of
3 751 934/(704/17)
squares
316
of observations
we
calculate
then
squared,
times
first
the
group,
is:
the
root
the
is
+ 2.04x3.31
=
?=6.99/3.31
10JANUARY
1998
the weighted
of the
estimates
in the usual
two
t
sample
error of the difference
between
therefore
= 13.7.
The
2.11. With
32
test
of
degrees
the unweighted
comparison,
the
variance
estimate
pooled
ard error of the difference
the sum
of
the
about
17x79.502
the
=
second
and
standard
the
interval.
the
not
be
assumptions
uniform.
assumptions
reduction
Some
very
meta-analysis
the
is
squared
the
-
from
worthwhile
tions
is 157.81.
is Vl57.81x(l/17
this
(13.1)
The
statistical
of
The
better
in
the
software
size
will
the
do
as
t test,
these
sum
of
=
George's Hospital
Medical School,
London SW17 ORE
J Martin Bland,
professor ofmedical
statistics
Division of General
Practice and
Primary Care,
St George's
Hospital Medical
School, London
SW17 ORE
Sally M Kerry,
lecturer inmedical
statistics
stand
+ 1/17) =
the
weighted
analysis
a
and
produces
of
the confidence
same basic
simply. The
principle
to combine
of
studies
varying
Department of
Public Health
Sciences, St
Correspondence
Professor Bland
is 8.00
calcula
is used
in
size.
devia
group,
600.68-17x72.512
129
16,
1756.34/
variance,
is the
1=
17-
freedom,
is
significance
of freedom
confidence
is
1 to 17 percent
interval
var
In this
the number
of requests
example
must
to some
between
This
lead
practices.
will
meets
to
0.2
difference
95%
age points.
ies greatly
deviation
variance
the mean
by
=
6.99-2.04x3.31
P = 0.04.
and
squares
observations
is 90
use
variance
standard
The
6.99
gives
18x562
term,
the
common
For
by
divide
of
now
can
We
and
interval
the
18)/17
divided
= 90 600.68
the mean
about
BMJ VOLUME
...+
of
groups,
common
vari
the
and
2976.12
two
the
the
correction
by
degrees
the weighted
estimate
and
the square
109.77,
For
Vl 09.77 = 10.48.
within
the means
is V93.00 x (1/17+ 1/17) = 3.31 and the
difference is 79.50 - 72.51 = 6.99. The 95% confidence
squared
of
squares
=
formulas.
+ ...+
the
1219.78/
we
the
sum
is
then
sum we
the
variance
observations,
means
4.3. The
we
estimated
squares
109 200.59 - 107 444.25 = 1756.34.
107 444.25, giving
of
= 93.00.
the
109 200.59
mean
weighted
To get
sums
1756.34+1219.78
2755709
429/17
the
=
= 8.73.
(17-1) 76.24 and the standard deviation V76.24
We find the pooled sum of squares by adding the
To
observations
+ 7xl002+16x942
20xl002
of
Hence
1219.78.
squares.
about
For
squared.
the observations
of
sum
_73.6
ance estimate for the two groups by dividing by the
combined degrees of freedom, 2976.12/(17+17-2)
in a simi
is found
Here
squared.
the
observations
of
group
= 72.51.
the number
term,
mean
sum
the
to the
conforming
the second
weighted
sum
of
add
in the
significant)
more
make
who
deviation
a
need
mean
weighted
not
Mean(SD)
_81.6(11.9)
usual,
mean,
weighted
mean
(but
is 51050/704
we
Firstiy,
an
the
weighted
subtract
unweighted
slight
standard
weighted
we
subtract
the
a
general
practitioners
a lower
proportion
which
this. For
explains
theweighted mean
mean,
as
is
the
gives
the
to have
guidelines,
lar way.
calculate
that
Note
There
this
for
tendency
referrals
The
same
the
same
the
81.6.
table,
all
mean.
1 Kerry SM, Bland JM.Statistics Notes: analysis of a trial randomised in
clusters. BMJ 1997;316:54.
2 Oakeshott P, Kerry SM,Williams JKRandomised controlled trial of the
effect of the Royal College of Radiologists' guidelines on general practi
tioners' refeiTal for radiographic
examinationJJr J Gen Pract
1994;44:197-200.
BMJ 1998;316:129
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