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analyses:normand1999

Normand (1999)

The Data (Part 1)

The article by Normand (1999) is part of the "Tutorial in Biostatistics" series in the journal Statistics in Medicine and provides a nice introduction to meta-analysis, covering fixed- and random-effects models. Two datasets are analyzed in this article. The first is based on data by Hine et al. (1989) regarding the mortality risk due to prophylactic use of lidocaine after a heart attack. The second dataset provides the results from studies on the length of the hospital stay of stroke patients under specialized care and under conventional/routine (non-specialist) care.

We will start with the lidocaine dataset, which is provided in Table I in the article (p. 322). After loading the metafor package with library(metafor), the data can be examined with:

dat.hine1989
  study         source n1i n2i ai ci
1     1  Chopra et al.  39  43  2  1
2     2       Mogensen  44  44  4  4
3     3    Pitt et al. 107 110  6  4
4     4   Darby et al. 103 100  7  5
5     5 Bennett et al. 110 106  7  3
6     6 O'Brien et al. 154 146 11  4

As described under help(dat.hine1989), variables n1i and n2i are the number of patients in the lidocaine and control group, respectively, and ai and ci are the corresponding number of deaths in the two groups. Since these are 2×2 table data, a variety of different outcome measures could be used for the meta-analysis, including the risk difference, the risk ratio (relative risk), and the odds ratio (see Table III). Normand (1999) uses risk differences for the meta-analysis, so we will proceed accordingly.

We can calculate the risk differences and corresponding sampling variances with:

dat <- escalc(measure="RD", n1i=n1i, n2i=n2i, ai=ai, ci=ci, data=dat.hine1989)
dat
  study         source n1i n2i ai ci     yi     vi
1     1  Chopra et al.  39  43  2  1 0.0280 0.0018
2     2       Mogensen  44  44  4  4 0.0000 0.0038
3     3    Pitt et al. 107 110  6  4 0.0197 0.0008
4     4   Darby et al. 103 100  7  5 0.0180 0.0011
5     5 Bennett et al. 110 106  7  3 0.0353 0.0008
6     6 O'Brien et al. 154 146 11  4 0.0440 0.0006

Note that the yi values are the risk differences in terms of proportions. Since Normand (1999) provides the results in terms of percentages, we can make the results directly comparable by multiplying the risk differences by $100$ (and the sampling variances by $100^2$):

dat$yi <- dat$yi * 100
dat$vi <- dat$vi * 100^2

Fixed-Effects Model

Next, a fixed-effects model can be fitted to these data with:

rma(yi, vi, data=dat, method="FE")
Fixed-Effects Model (k = 6)
 
Test for Heterogeneity: 
Q(df = 5) = 0.86, p-val = 0.97
 
Model Results:
 
estimate       se     zval     pval    ci.lb    ci.ub          
    2.94     1.31     2.25     0.02     0.38     5.51        * 
 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

These are the same results as shown in Table VII (p. 349). In particular, patients in the group receiving lidocaine are estimated to have a 2.94 percentage points higher risk of mortality than patients in the control group (with 95% CI: 0.4% to 5.5%).

Random-Effects Model

Next, we can fit a random-effects model to these data with:

rma(yi, vi, data=dat)

When method is not been specified, the rma() function uses REML estimation by default. The results are then:

Random-Effects Model (k = 6; tau^2 estimator: REML)
 
tau^2 (estimated amount of total heterogeneity): 0 (SE = 6.17)
tau (square root of estimated tau^2 value):      0
I^2 (total heterogeneity / total variability):   0.00%
H^2 (total variability / sampling variability):  1.00
 
Test for Heterogeneity: 
Q(df = 5) = 0.86, p-val = 0.97
 
Model Results:
 
estimate       se     zval     pval    ci.lb    ci.ub          
    2.94     1.31     2.25     0.02     0.38     5.51        * 
 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Since the estimate of heterogeneity is 0 (i.e., $\hat{\tau}^2 = 0$), the results are the same as for the fixed-effects model. Finally, the random-effects model using the DerSimonian-Laird estimator for $\tau^2$ can be fitted with:

rma(yi, vi, data=dat, method="DL", digits=2)

Since $\hat{\tau}^2 = 0$ also for the DL estimator, the results are again the same (not shown).

Sidenote: The equation for the REML estimator given in the article (on page 336) is actually just an approximation. The exact equation is given, for example, in Viechtbauer (2005) and this is what the rma() will use when fitting a random-effects model. The differences between the approximate and exact equation is usually small, but there really is no need to use the approximate formula. Also, Normand uses SAS to fit the random-effects model with REML estimation and SAS actually uses the exact formula.

The Data (Part 2)

The second dataset analyzed in the article provides the results from studies where specialist inpatient stroke care was compared to conventional non-specialist care. Table II in the article (p. 324) lists the findings from 9 studies regarding the average length of stay during the acute hospital admission and the corresponding standard deviation. The central hypothesis of interest is whether specialist stroke unit care will result in a shorter length of hospitalization compared to routine management.

The data can be examined with:

dat.normand1999
  study             source n1i m1i sd1i n2i m2i sd2i
1     1          Edinburgh 155  55   47 156  75   64
2     2     Orpington-Mild  31  27    7  32  29    4
3     3 Orpington-Moderate  75  64   17  71 119   29
4     4   Orpington-Severe  18  66   20  18 137   48
5     5      Montreal-Home   8  14    8  13  18   11
6     6  Montreal-Transfer  57  19    7  52  18    4
7     7          Newcastle  34  52   45  33  41   34
8     8               Umea 110  21   16 183  31   27
9     9            Uppsala  60  30   27  52  23   20

As described under help(dat.normand1999), variables n1i, m1i, and sd1i are the sample sizes, means, and standard deviations for the groups receiving specialized care, while n2i, m2i, and sd2i are the corresponding variables for the groups receiving routine care.

For the meta-analysis, Normand uses the (raw or unstandardized) mean difference as the outcome measure. The equation for the sampling variance of a mean difference provided on page 331 in the article assumes that the true variances are the same in the treatment and control groups (i.e., the equation assumes homoscedasticity). If we want to proceed accordingly, we first have to compute the pooled standard deviations of the groups with:

dat.normand1999$sdpi <- with(dat.normand1999, sqrt(((n1i-1)*sd1i^2 + (n2i-1)*sd2i^2)/(n1i+n2i-2)))

And now, we can compute the mean differences and corresponding sampling variances with:1)

dat <- escalc(m1i=m1i, sd1i=sdpi, n1i=n1i, m2i=m2i, sd2i=sdpi, n2i=n2i,
              measure="MD", data=dat.normand1999, digits=2)
dat
  study             source n1i m1i sd1i n2i m2i sd2i      sdpi     yi     vi
1     1          Edinburgh 155  55   47 156  75   64 56.174313 -20.00  40.59
2     2     Orpington-Mild  31  27    7  32  29    4  5.677104  -2.00   2.05
3     3 Orpington-Moderate  75  64   17  71 119   29 23.607908 -55.00  15.28
4     4   Orpington-Severe  18  66   20  18 137   48 36.769553 -71.00 150.22
5     5      Montreal-Home   8  14    8  13  18   11 10.000000  -4.00  20.19
6     6  Montreal-Transfer  57  19    7  52  18    4  5.768104   1.00   1.22
7     7          Newcastle  34  52   45  33  41   34 39.964792  11.00  95.38
8     8               Umea 110  21   16 183  31   27 23.491023 -10.00   8.03
9     9            Uppsala  60  30   27  52  23   20 24.009657   7.00  20.69

These are the same values as shown in Table VIII (p. 351).

Fixed-Effects Model

Next, a fixed-effects model can be fitted to these data with:

rma(yi, vi, data=dat, method="FE", digits=2)
Fixed-Effects Model (k = 9)
 
Test for Heterogeneity: 
Q(df = 8) = 241.06, p-val < .01
 
Model Results:
 
estimate       se     zval     pval    ci.lb    ci.ub          
   -3.49     0.78    -4.47     <.01    -5.03    -1.96      *** 
 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The results indicate that patients receiving specialized care spent, on average, three and a half days less in the hospital than those in the control groups ($\hat{\theta} = -3.5$). However, the results of the studies are clearly heterogeneous ($Q(df=8) = 241.06$, $p < .01$).

Random-Effects Model

A random-effects model using the DerSimonian-Laird estimator can be fitted to these data with:

rma(yi, vi, data=dat, method="DL", digits=2)
Random-Effects Model (k = 9; tau^2 estimator: DL)
 
tau^2 (estimated amount of total heterogeneity): 218.72 (SE = 195.47)
tau (square root of estimated tau^2 value):      14.79
I^2 (total heterogeneity / total variability):   96.68%
H^2 (total variability / sampling variability):  30.13
 
Test for Heterogeneity: 
Q(df = 8) = 241.06, p-val < .01
 
Model Results:
 
estimate       se     zval     pval    ci.lb    ci.ub          
  -14.10     5.28    -2.67     <.01   -24.45    -3.75       ** 
 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

As shown in Table X (p. 352), we obtain an estimate for the amount of heterogeneity equal to $\hat{\tau}^2 = 219$ and an estimate for the average mean difference equal to $\hat{\mu} = -14$ (with 95% CI: $-24$ to $-4$).

The random-effects model using REML estimation can be fitted with:

rma(yi, vi, data=dat, digits=2)
Random-Effects Model (k = 9; tau^2 estimator: REML)
 
tau^2 (estimated amount of total heterogeneity): 685.20 (SE = 360.37)
tau (square root of estimated tau^2 value):      26.18
I^2 (total heterogeneity / total variability):   98.92%
H^2 (total variability / sampling variability):  92.26
 
Test for Heterogeneity: 
Q(df = 8) = 241.06, p-val < .01
 
Model Results:
 
estimate       se     zval     pval    ci.lb    ci.ub          
  -15.12     8.95    -1.69     0.09   -32.67     2.43        . 
 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The estimate of the amount of heterogeneity is now quite a bit larger (i.e., $\hat{\tau}^2 = 685$), but we still obtain a very similar value for the average mean difference (i.e., $\hat{\mu} = -15$). The 95% confidence interval for $\mu$ is now quite a bit wider (i.e., $-33$ to $2$) and in fact includes the value $0$ now (hence, we cannot reject $H_0: \mu = 0$ at $\alpha = .05$). Again, these results match what is reported in Table X (p. 352).2)

References

Hine, L. K., Laird, N., Hewitt, P., & Chalmers, T. C. (1989). Meta-analytic evidence against prophylactic use of lidocaine in acute myocardial infarction. Archives of Internal Medicine, 149(12), 2694–2698.

Normand, S. T. (1999). Meta-analysis: Formulating, evaluating, combining, and reporting. Statistics in Medicine, 18(3), 321–359.

Viechtbauer, W. (2005). Bias and efficiency of meta-analytic variance estimators in the random-effects model. Journal of Educational and Behavioral Statistics, 30(3), 261–293.

1)
If we would not want to assume homoscedasticity, then we could skip the last step and just use dat <- escalc(m1i=m1i, sd1i=sd1i, n1i=n1i, m2i=m2i, sd2i=sd2i, n2i=n2i, measure="MD", data=dat.normand1999, digits=2) to compute the mean differences and corresponding sampling variances.
2)
The estimated standard error of $\hat{\tau}^2$ reported by SAS is $380$, while the corresponding value obtained with the metafor package is $360$. This discrepancy is due to the fact that SAS uses by default the observed Hessian to estimate the standard errors of variance components, while the metafor package uses the expected Hessian. If one uses the option scoring=1000 in SAS (so that Fisher scoring is used), then SAS will also use the expected Hessian and the results agree.
analyses/normand1999.txt · Last modified: 2016/03/12 13:40 by Wolfgang Viechtbauer