1. The mcci Function

(1) Key Arguments in the mcci Function

To compute the MCCI for difference or equivalence tests, the minimum argument is the estimated effect and its standard errors. The function can take up to two sets of effects and their standard errors. For each set, it could include components of a compound effect (i.e., a mediation effect).

When only one set of argument is specified, the effect itself is estimated difference.

d1: The estimated effect in a study.
se1: The standard error(s) of d1.

When two sets of effects/means are specified, they could be like the following.

d1: The estimated mean (or effect(s)) for group 1 (study 1).
se1: The standard error(s) of d1.
d2: The estimated mean (or effect(s)) for group 2 (study 2).
se2: The standard error(s) of d2.

(2) Plots Provided by the Function

The function also provide a plot of the MCCI by default. Arguments are available to adjust the appearance of the plot. See the function documentation for details.

2. The power.eq.2group Function

This function performs power analysis for equivalence test of two-group means. It can calculate statistical power, required sample size, and the minimum detectable difference between equivalence bounds and the group-mean difference depending on which one and only one of parameters is unspecified in the function.

power: statistical power.
n: sample size.
eq.dis: The minimum distance between the equivalence bounds and the difference in means (effect(s)) .

3. Examples

(1) MCCI Example

 library(anomo)
myci <- mcci(d1 = .1, se1 = .1, d2 = .3, se2 = .1)

## [1] "The 95% MCCI for difference test is c(-0.075,0.479)."
## [1] "The 90% MCCI for equivalence test is c(-0.033,0.438)."

# Note. Effect difference (the black square representing d2 - d1), 90% MCCI 
# (the thick horizontal line) for the test of equivalence, and 95% MCCI 
# (the thin horizontal line) for the test of moderation 
# (or difference in effects).

# Adjust the plot
myci <- mcci(d1 = .1, se1 = .1, d2 = .4, se2 = .1,
             eq.bd = c(-0.2, 0.2), xlim = c(-.2, .7))

## [1] "The 95% MCCI for difference test is c(0.022,0.565)."
## [1] "The 90% MCCI for equivalence test is c(0.063,0.526)."

-MCCI for the difference and equivalence in mediation effects (product of the y~m and m~x paths) in two studies

MyCI.Mediation <- mcci(d1 = c(.60, .40), 
             se1 = c(.019, .025),
             d2 = c(.60, .80),
             se2 = c(.016, .023))

## [1] "The 95% MCCI for difference test is c(0.191,0.289)."
## [1] "The 90% MCCI for equivalence test is c(0.198,0.281)."

(2) Power Analysis Example

Conventional Power Analysis

 # 1. Conventional Power Analyses from Difference Perspectives
 # Calculate the required sample size to achieve certain level of power
 mysample <- power.eq.2group(d = .1, eq.dis = 0.1,  p =.5,
                             r12 = .5, q = 1, power = .8)

## The required sample size (n) is 1237.868.

 mysample$out

## $n
## [1] 1237.868

 # Calculate power provided by a sample size allocation
 mypower <- power.eq.2group(d = 0.1, eq.dis = 0.1, n = 1238, p =.5,
                            r12 = .5, q = 1)

## The statistical power (power) is 0.8000373.

 mypower$out

## $power
## [1] 0.8000373

 # Calculate minimum detectable distance a given sample size allocation can achieve
 myeq.dis <- power.eq.2group(d = .1, n = 1238, p =.5,
                            r12 = .5, q = 1, power = .8)

## The minimum detectable difference between equivalence bounds and mean/effect differnce (eq.dis) is 0.09996871.

 myeq.dis$out

## $eq.dis
## [1] 0.09996871

Power Analysis with Costs

 # 2. Power Analyses Using Optimal Sample Allocation
 # Optimal sample allocation identification
 od <- od.eq.2group(r12 = 0.5, c1 = 1, c1t = 10)

## The optimal proportion of units in treatment (p) is 0.2402531.

 # Required budget and sample size at the optimal allocation
 budget <- power.eq.2group(expr = od, d = 0.1, eq.dis = 0.1,
                            q = 1, power = .8)

## The required budget (m) is 5359.788.
## The required sample size (n) is 1694.914.

 # Required budget and sample size by an balanced design with p = .50
 budget.balanced <- power.eq.2group(expr = od, d = 0.1, eq.dis = 0.1,
                                    q = 1, power = .8,
                                    constraint = list(p = .50))

## The required budget (m) is 6808.272.
## The required sample size (n) is 1237.868.

 # 27% more budget required from the balanced design with p = 0.50.
 (budget.balanced$out$m-budget$out$m)/budget$out$m *100

## [1] 27.02501

Power Curve Under the Same Budget: Statistical Power is Maximized at the Optimal Allocation

pwr <- NULL
p.range <- seq(0.01, 0.99, 0.01)
for(p in p.range){
  pwr <- c(pwr, power.eq.2group(expr = od, constraint = list(p = p),
                                m = budget$out$m, d = 0.1, eq.dis = 0.1,
                                q = 1, verbose = FALSE)$out$power)
}
plot(p.range*100,
     pwr*100,
     type = "l", lty = 1,
     xlim = c(0, 100), ylim = c(0, 100),
     xlab = "Proportion of Units in Treated (%)", ylab = "Power (%)",
     main = "", col = "black")
abline(v=od$out$p*100, lty = 2, col = "black")

anomo

Zuchao Shen