ivpack (version 1.2)

power.iv: power.iv

Description

Computes the power for an instrumental variables analysis to be done using the Anderson-Rubin test.

Usage

power.iv(n, lambda, gamma, var.z, sigmau, sigmav, rho, alpha = 0.05)

Arguments

n

Sample size.

lambda

True causal effect minus null hypothesis causal effect

gamma

Regression coefficient for effect of instrument on treatment.

var.z

Variance of instrument.

sigmau

Standard deviation of potential outcome under control (structural error for y)

sigmav

Standard deviation of error from regressing treatment on instrument

rho

Correlation between u(potential outcome under control ) and v (error from regressing treatment on instrument)

alpha

Significance level of test.

Value

• Power for the proposed study, assuming that the Anderson-Rubin (1949) test will be used. The power formula is derived in Jiang, Small and Zhang (2013).

Details

The structural equations model assumed is: D=gamma0+gamma*z+v, Y=beta0+beta1*D+u. This model can also be obtained by assuming the potential outcomes model Y^(d=0)=beta0+u, Y^d=Y^(d=0)+beta1. See Jiang, Small and Zhang (2013) for details. lambda is equal to the true beta1 minus the null hypothesis beta1.

References

Anderson, T.W. and Rubin, H. (1949), Estimation of the parameters of a single equation in a complete system of stochastic equations, Annals of Mathematical Statistics, 20, 46-63. Jiang, Y., Small, D. and Zhang, N. (2013), Sensitivity analysis and power for instrumental variable studies, Working paper.

Examples

### Power for a study with in which the null hypothesis causal effect is 0,
### the true causal effect is 1, the sample size is 250, the instrument is
### binary with probability .5 (so variance = .25), the standard deviation
### of potential outcome under control is 1, the effect of the instrument
### is to increase the probability of a binary treatment being 1 from .25 to 
### .75.  The function sigmav.func computes the SD of v for a binary insrument, 
### binary treatment. The correlation between u and v is assumed to be .5.  The 
### significance level for the study will be alpha = .05
sigmav.func(prob.d1.given.z1=.75,prob.d1.given.z0=.25,prob.z1=.5)
# The sigmav.func finds sigmav=.4330127
power.iv(n=250, lambda=1, gamma=.5, var.z=.25, sigmau=1, sigmav=.4330127, rho=.5, 
alpha = 0.05)