This vignette exemplifies ways to decompose and identify variance components in any statistic obtained by running a specification curve analysis.

1. Run the specification curve analysis

In order to have some data to work with, we run the minimal example contained in the package.

library(specr)
library(dplyr)
# run spec analysis
results <- run_specs(example_data,
                     y = c("y1", "y2"),
                     x = c("x1", "x2"),
                     model = "lm",
                     controls = c("c1", "c2"),
                     subset = list(group1 = unique(example_data$group1),
                                   group2 = unique(example_data$group2)))

Let’s quickly get some ideas about the specification curve by using summarise_specs()

# Overall median
summarise_specs(results)
## # A tibble: 1 x 7
##   median   mad   min   max   q25   q75   obs
##    <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1   3.59  4.56 -2.05  9.58  1.03  7.63   123
# By choices
summarise_specs(results, 
                group = c("y", "controls"))
## # A tibble: 8 x 9
## # Groups:   y [2]
##   y     controls      median   mad   min   max   q25   q75   obs
##   <chr> <chr>          <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 y1    c1             7.72  0.744  3.53  8.96 6.44   8.13   123
## 2 y1    c1 + c2        7.03  1.39   3.49  8.71 6.25   8.05   123
## 3 y1    c2             7.15  1.44   3.94  9.58 6.35   8.37   123
## 4 y1    no covariates  7.97  1.40   4.24  9.28 6.80   8.42   123
## 5 y2    c1             1.11  1.40  -2.05  2.79 0.144  1.83   123
## 6 y2    c1 + c2        0.796 1.11  -1.94  2.83 0.101  1.37   123
## 7 y2    c2             0.690 1.00  -1.61  3.64 0.367  1.82   123
## 8 y2    no covariates  1.31  0.931 -1.80  3.67 0.674  1.92   123

We see that it makes quite a difference whether y1 or y2 is used as independent variable.

2. Estimate a simple multilevel model

We can think of the specification curve analysis as a factorial design in which we investigate the influence of different types of analytical choices on a resulting coefficient or test statistic. One way to investigate these effects is to ask how much variance in the specification curve is explained by what (or which combination of) analytical choices.

In a first step, we therefore have to estimate a simple multilevel model without predictors, yet with specific random effects that represent the analytical choices.

# Package to estimate multilevel models
library(lme4)

# Estimate model
m1 <- lmer(estimate ~ 1 + (1|x) + (1|y) + (1|controls) + (1|subsets), data = results)

# Check model summary
summary(m1)
## Linear mixed model fit by REML ['lmerMod']
## Formula: estimate ~ 1 + (1 | x) + (1 | y) + (1 | controls) + (1 | subsets)
##    Data: results
## 
## REML criterion at convergence: 591.7
## 
## Scaled residuals: 
##      Min       1Q   Median       3Q      Max 
## -2.85509 -0.60698 -0.03593  0.55041  2.83207 
## 
## Random effects:
##  Groups   Name        Variance Std.Dev.
##  subsets  (Intercept)  0.76138 0.8726  
##  controls (Intercept)  0.03861 0.1965  
##  y        (Intercept) 19.72906 4.4417  
##  x        (Intercept)  0.42335 0.6507  
##  Residual              1.00930 1.0046  
## Number of obs: 192, groups:  subsets, 12; controls, 4; y, 2; x, 2
## 
## Fixed effects:
##             Estimate Std. Error t value
## (Intercept)    4.079      3.187    1.28

We have now stored the random effects in the object m. We can already see that most variance is related to the random effect y|intercept. Yet, how much exactly?

3. Estimate intraclass correlations

Specr offers to ways to decompose the variance of the specification curve. First, we can simply compute the intraclass correlations using the function icc_specs().

icc_specs(m1) %>%
  mutate_if(is.numeric, round, 2)
##        grp  vcov  icc percent
## 1  subsets  0.76 0.03    3.47
## 2 controls  0.04 0.00    0.18
## 3        y 19.73 0.90   89.83
## 4        x  0.42 0.02    1.93
## 5 Residual  1.01 0.05    4.60

We can see that 89.83% of the variance in the obtained results is related to choosing different dependent variables. Only small parts of the variance are explained by other analytical choices.

4. Plot variance components

Second, we can alternatively use the function plot_variance() to obtain a visualization. The function calls icc_specs() automatically. We can hence pass the multilevel results object directly. Further customizations via the ggplot2-syntax is possible.

plot_variance(m1) +
  ylim(0, 100)

5. Further investigating variance components

You might ask yourself why we did not include the multilevel model estimation in the icc_specs() function. The reason is straight-forward: In many cases, it might be of interest to decompose the variance in different ways (e.g., including interactions between the analytical choices). In this case, we only need to specify the respective multilevel model (with alternative random effects) and we can again obtain a result table of a plot.

m2 <- lmer(estimate ~ 1 + (1|x) + (1|y) + (1|controls) + (1|subsets) + (1|x:y) + (1|y:controls) + (1|y:subsets), data = results)

# Get table
icc_specs(m2) %>%
  mutate_if(is.numeric, round, 2)
##          grp  vcov  icc percent
## 1  y:subsets  0.76 0.03    3.49
## 2    subsets  0.41 0.02    1.87
## 3 y:controls  0.00 0.00    0.00
## 4        x:y  0.02 0.00    0.09
## 5   controls  0.05 0.00    0.22
## 6          y 19.46 0.90   89.55
## 7          x  0.42 0.02    1.92
## 8   Residual  0.62 0.03    2.87
# Plot results
plot_variance(m2)

We can see that the interaction between analytical choices relating to subsets and dependent variables explains 3.49%.

6. Investigating other statistics

Finally it should be noted that other obtained statistics can be investigated as well (not only the main estimate, e.g., a regression coefficient).

m3 <- lmer(p.value ~ 1 + (1|x) + (1|y) + (1|controls) + (1|subsets), data = results)
## boundary (singular) fit: see ?isSingular
icc_specs(m3) %>%
  mutate_if(is.numeric, round, 2)
##        grp vcov  icc percent
## 1  subsets 0.00 0.04    4.07
## 2 controls 0.00 0.00    0.00
## 3        y 0.02 0.30   29.99
## 4        x 0.00 0.01    0.74
## 5 Residual 0.03 0.65   65.21