Paul O'Brien writes:
> I am combining two studies:
>
> Study RR LCI UCI
> Study 1 0.7 0.1 8.2
> Study 2 0.6 0.1 6.4
>
> With the command:
>
> . meta rr ll ul, ci eform gr(f) print id(study)
>
> However, the confidence intervals listed in the print are
> different from what I have entered:
>
> Meta-analysis (exponential form)
> | Pooled 95% CI Asymptotic No. of
> Method | Est Lower Upper z_value p_value studies
> -------+----------------------------------------------------
> Fixed | 0.645 0.142 2.927 -0.568 0.570 2
> Random | 0.645 0.142 2.927 -0.568 0.570
> Test for heterogeneity: Q= 0.010 on 1 degrees of freedom (p= 0.921)
> Moment-based estimate of between studies variance = 0.000
> | Weights Study 95% CI
> Study | Fixed Random Est Lower Upper
> ----------+----------------------------------------
> Study 1 | 0.79 0.79 0.70 0.08 6.34
> Study 2| 0.89 0.89 0.60 0.08 4.80
>
> What is the problem?
The problem is that your input data do not follow the expected
ratios for log-based confidence intervals (probably because
too few digits were retained). -meta- uses your input CI to
compute the standard error (se), assuming log symmetry, then
later recalculates the proper log-symmetric CI endpoints about
the point estimate using this standard error.
Data that follows log symmetry has the characteristic that the
following are all equal:
rr/ll = (ul-ll)/2 = ul/rr
For your input data I get:
rr/ll = (ul-ll)/2 = ul/rr
study 1 7 9.06 11.71
study 2 6 8 10.67
For the (rounded) recalculated values I get:
study 1 8.75 8.90 9.06
study 2 7.50 7.75 8.00
These values are not exactly equal because the two-digit
representation of the recalculated ll, .08, is not
accurate enough.
For a more accurate value, note that -meta- uses the
following calculation to get the se:
se = ( ln(ul) - ln(ll) ) / 2 / z
(where z is an appropriate Normal value)
For your study 1 data this generates:
se = ( ln(8.2) - ln(.1) ) / 2 / 1.96
= ( 2.1041342 - -2.3025851 ) / 2 / 1.96
= 4.4067192 / 2 / 1.96
= 1.1241631
Later, -meta- spits back the recalculated CI endpoints as:
ll = exp( ln(rr) - z * se )
ul = exp( ln(rr) + z * se )
Or, for study 1:
ll = exp( ln(rr) - z * se )
= exp( ln(.7) - 1.96 * 1.1241631 )
= exp( -.35667494 - 1.96 * 1.1241631 )
= exp( -2.5600346 )
= .07730206 (displayed as .08)
ul = exp( ln(rr) + z * se)
= exp( ln(.7) + 1.96 * 1.1241631 )
= exp( -.35667494 + 1.96 * 1.1241631 )
= exp( 1.8466847 )
= 6.3387699 (displayed as 6.34)
Thus, using the exact ll and ul in the ratio calculations:
rr/ll = (ul-ll)/2 = ul/rr = 9.06
This suggests to me that more digits are required from your
original data in order to properly meta-analyze the data.
Tom
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