(1) Key Arguments in the mcci Function

To compute the MCCI for difference or equivalence tests, the minimum required arguments are the estimate(s) and corresponding standard error(s). The function can take up to five sets of estimates and their standard errors. It could include components of one or two mediation effects if mediation is TRUE.

• d: The estimate(s).
• se: The standard error(s) of d.

When two or more sets of parameters are specified (and mediation is FALSE), the function computes the MCCIs for the difference across these estimates. When mediation is TRUE, the function computes the MCCIs for the estimated mediation effects (in one study) or the difference across these mediation effects (in two studies/groups).

(2) Plots Provided by the Function

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

(1) MCCI Example

 library(anomo)
myci <- mcci(d = c(.1, .15), se = c(.01, .01))

## [1] "The 95% MCCI for difference test is (-0.078, -0.022)"
## [1] "The 90% MCCI for equivalence test is (-0.073, -0.027)"

# Note. Effect difference (the black square representing d1 - d2), 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(d = c(.1, .15), se = c(.01, .01),
             eq.bd = c(-0.2, 0.2), xlim = c(-.2, .7))

## [1] "The 95% MCCI for difference test is (-0.078, -0.022)"
## [1] "The 90% MCCI for equivalence test is (-0.073, -0.027)"

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

MyCI.Mediation <- mcci(d = c(.60, .40, .60, .80), 
             se = c(.019, .025, .016, .023), mediation = TRUE)

## [1] "The 95% MCCI for difference test is (-0.289, -0.19)"
## [1] "The 90% MCCI for equivalence test is (-0.281, -0.198)"

#Note. The order of d is a1, b1, a2, and b2 (e.g., treatment-mediator
#   and mediator-outcome path in group/study 1 and 2, respectively). 
#   se is in the same order for the standard errors.

(2) Power Analysis Example

 # 1. Conventional Power Analyses from Difference Perspectives
 # Calculate the required sample size to achieve certain level of power
 mysample <- power.1.eq(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.1.eq(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.1.eq(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

 # 2. Power Analyses Using Optimal Sample Allocation
 # Optimal sample allocation identification
 od <- od.1.eq(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.1.eq(expr = od, d = 0.1, eq.dis = 0.1, 
                         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.1.eq(expr = od, d = 0.1, eq.dis = 0.1,
                                    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

plot.power.eq(expr = od, d = 0.1, eq.dis = 0.1)