The Generic Inverse Variance method

The generic inverse variance method

The

new

method

of

analysis

that

is

available

in

Review

Manager

4.2

(RevMan)

is

the

‘generic

inverse variance method’ (GIVM). This method can be applied to a number of different situations

that are encountered by Cochrane authors and this article aims to address three of these.

The

data that

are

required for

the

GIVM

are

an

estimate for

the

relative

treatment effect

and

its

standard error, for each of the studies that are to be included in the analysis. Each study estimate

of the relative treatment is given a weight that is equal to the inverse of the variance of the effect

estimate (i.e. one divided by the standard error squared).

It should be noted that this method should only be used when it is not possible to enter data in the

usual form of dichotomous, continuous or individual patient data to ensure that the reader is able to

see

the

data

by

treatment

group

whenever

possible.

The

GIVM

should

not

be

used

as

an

alternative to the methods that are currently available in RevMan 4.1 but rather in situations where

it is not feasible to use them.

Described below are three situations where authors may need to use the GIVM:

(1) The outcome is

dichotomous

(i.e. the outcome can only take one of two possibilities) but the

published

paper

reports

only

the

odds

ratio

(OR)

or

relative

risk

(RR)

and

its

standard

error.

Previously in RevMan 4.1, such results could not be included in a quantitative analysis and could

only be described in the text of the review.

(2) The outcome is

continuous

and the published paper reports only the difference between the

means for the two groups and the standard error of this difference. These data could not previously

be included in a quantitative analysis in RevMan 4.1.

(3) The design of the trial is

cross-over

, the outcome is continuous, and the average difference

between periods and its standard error are reported by the published paper. There are a number of

different methods that could be used to analyse continuous data that have been reported in papers

conducting cross- over trials. Elbourne discusses problems that arise in the meta-analysis of cross-

over trials and other issues relating to these type of studies.

Each of these situations will be looked at in detail with examples that authors may come across.

Dichotomous Data

In

this

example

there

are

five

included

studies

that

have

all

reported

data

on

a

dichotomous

outcome.

This

outcome

is

not

a

favourable

outcome

(e.g.

death)

and

therefore

the

treatment

is

more beneficial if there are fewer events.

The information that is provided for each trial is given below:

Study 1

Intervention

Treatment A

Treatment B

Total

Event Present

Yes

16

16

32

No

9

29

38

Total

25

45

70

Study 2

RR 2.1 95% Confidence interval (0.83,5.3)

Study 3

Intervention

Study 4

Intervention

Treatment A

Treatment B

Total

Treatment A

Treatment B

Total

Event Present

Yes

21

44

65

No

15

44

59

Total

36

88

124

Event Present

Yes

6

2

8

No

5

4

9

Total

11

6

17

Study 5

RR 1.67 95% CI (0.74,3.75)

In

the

last

edition

of

RevMan,

studies

1,3

and

4

could

be

included

in

a

meta- analysis,

but

not

studies 2 and results from these latter two studies could only be included in the text of the

review.

Using the GIVM it is now possible to combine these data to obtain one overall pooled estimate for

the RR. To do this the following procedure can be followed once you are in RevMan 4.2:

?

add an outcome as normal (click on the comparison of interest and then click add), you will be

?

asked to ‘Add outcome/category’,

?

click on ‘generic inverse variance’.

?

In the next screen a number of details will need to be entered.

?

Enter the description and group labels as normal but two other details will need to be entered.

?

The

‘Name

for

effect

measure’

should

be

entered

as

‘Relative

Risk’

(as

this

is

the

effect

estimate that we are interested in, in this particular example, this may be the odds ratio or the

risk difference for dichotomous data in other examples).

?

Next

to

‘entered

data’

there

are

two

options

that

can

be

chosen,

either

‘Original

Scale’

or

‘Logarithm’.

Due

to

the

fact

that

the

standard

error

of

the

relative

risk

is

on

the

natural

log

scale we need to highlight ‘Logarithm’

(for more information on logarithms

see

Bland and Altman).

Entering

the

required

results

for

dichotomous

outcomes

is

quite

tricky

and

below

there

are

two

detailed examples based on the data that have already been discussed. We need to enter the log

(from this point on were log is written the natural log is assumed, this can also be written as ln) of

the RR and the standard error of the ln(RR).

For studies 1,3 and 4 we could proceed using the following method (based on study 1):

Construct a 2x2 table entering in all the cells as shown for study 1 below:

Event Present

Yes

No

Total

Treatment A

16 (a)

16 (b)

32 (a+b)

Intervention

Treatment B

9 (c)

29 (d)

38 (c+d)

Total

25 (a+c)

45 (b+d)

70 (a+b+c+d)

Table 1: 2x2 table for study 1.

a

(

c

?

d

)

The RR is calculated as :

a

?

b

?

c

c

?

d

c

(

< br>a

?

b

)

a

(

1

)

Which is equal to 2.11 for study 1, we then take the log of this value to obtain 0.747.

The formula for calculating the SE of the ln(RR) is given as:

SE

(ln(

RR

))

?

1

1

1

1

?

?

?

a

a

?

b

c

c

?

d

(

2

)

The SE of the ln(RR) is equal to 0.341.

If these two values are entered as the effect estimate and standard error, from them RevMan 4.2

will calculate the RR and 95% CI (this is not on the log scale).

For studies 2 and 5 we are unable to do this as we do not have the information to construct a 2x2

table. Therefore we need to use the 95% CI to work backwards and calculate the SE of the ln(RR).

Note that the 95% CI for the ln(RR) is usually calculated as:

(ln(

RR

)

?

[

1

.

96

?

SE

( ln(

RR

))],

ln(

)

?

[

1

.

96

?

SE

(ln(

RR

))])< /p>

(

3

)

These figures are then exponentiated to give the CI of the relative risk.

To calculate the SE of the ln(RR) from the CI we

can either use the upper or lower limit to work

with. In this example the upper limit of the confidence interval will be used, a similar procedure can

be used for the lower limit.

We know that the RR for study 2 is 2.1, the 95% CI for the RR is (0.83,5.3) and ln(RR) is 0.7419. If

we take the log of this CI we obtain the 95% CI for the log of the RR, which is (-0.19,1.67).

If we fill in the information that we know in (3) we obtain:

(ln(

2

.

1

)

?

1< /p>

.

96

?

SE< /p>

(ln(

RR

)),

ln(

2

.

1

)

?

1

.

96

?

SE

(ln(

< br>RR

)))

(

4

)

We know that the upper limit of (4) is equal to ln(5.3), which equals 1.67, therefore

< br>ln(

2

.

1

< br>)

?

(

1

.

96

?

SE

(ln(

RR

)))

?

1

.

67

(

5

)

Rearranging (5)

ln(

2

.

1

)

?

(

1

.

96

?

SE

(ln(

RR

) ))

?

1

.

6 7

1

.

96

?

SE

(ln(

RR

)

?

1

.

67

?

ln(

2

.

1

)

1

.

67

?

ln(

2

.

1

)

SE

(ln(

RR

)

?

1

.

96

we obtain that the SE(ln(RR)) is equal to 0.4723.

We now have the two figures that can be entered into RevMan 4.2 to complete our analysis, ln(RR)

= 0.7419 and SE(ln(RR)) = 0.4723.

The results of the analysis including data from all five studies can be seen overleaf:

Rev

iew:

Comparison:

Outcome:

Study

or sub-category

1

2

3

4

5

A

01 Treatment A v

s Treatment B

01 Outcome X

Relativ

e Risk (f

ixed)

95% CI

Weight

%

20.73

10.79

29.37

18.44

20.66

100.00

Relativ

e Risk (f

ixed)

95% CI

2.11 [1.08, 4.12]

2.10 [0.83, 5.30]

1.27 [0.72, 2.23]

1.35 [0.66, 2.74]

2.11 [1.08, 4.12]

1.67 [1.23, 2.27]

log[Relativ

e Risk] (SE)

0.7467 (0.3408)

0.7419 (0.4723)

0.2390 (0.2863)

0.3001 (0.3613)

0.7467 (0.3414)

Total (95% CI)

Test f

or heterogeneity

: Chi?= 2.44, df

= 4 (P = 0.66), I?= 0%

Test f

or ov

erall ef

f

ect: Z = 3.32 (P = 0.0009)

0.1

0.2

0.5

1

2

5

10

Fav

ours Treatment A

Fav

ours Treatment B

Figure 1: Forest plot for data in example 1

Details

of

the

formulae

for

calculating

the

odds

ratio

and

its

standard

error

can

be

seen

in

the

appendix .

Continuous Data

Similar to the previous example we will look at a hypothetical example which includes five studies,

each of which has reported data on a particular outcome. This outcome is a favourable outcome (a

higher result is better) and there are two treatment groups (A and B).

The information that each study reported can be seen in table 2:

Study

6

7

8

9

10

Treatment A

Mean=16.3

sd=3.1

n=16

Mean=14.6

sd=4.6

n=12

----------------------

-----------------------

Mean=12.9

sd=2.1

n=42

Treatment B

Mean=-0.9

sd=1.8

n=19

Mean=1.9

sd=3.2

n=14

---------------------

--------------------

Mean=2.6

sd=1.6

n=40

Difference (A-B)

---------------

---------------

14.21 95% CI (12.45,15.97)

7.6 95%CI (5.05,10.15)

-----------------

Table 2: Information reported by each study

Table

2

shows

that

studies

6,7

and

10

report

information

that

could

be

traditionally

entered

in

RevMan 4.1, namely the mean, standard deviation and the number of participants in each group.

However the information given for studies 8 and 9 could not have been included in a quantitative-

analysis

due

to

the

fact

that

the

difference

between

the

treatments

and

its

standard

error

have

been reported.

All of these studies can be combined using the GIVM in RevMan 4.2. The analysis should be set

up as before, but the appropriate headings should be changed accordingly and the scale should be

left on the default option which is ‘Original Scale’.

The information that needs to be entered for each study is the difference in means and its standard

error. For studies 6,7 and 10 we need to calculate the mean difference and standard error. This

can be done using the following formulae for calculating the mean difference and its SE.

Mean difference = Mean response of Treatment A

–

Mean response Treatment B