power.iv: power.iv
In ivpack: Instrumental Variable Estimation.

Description Usage Arguments Details Value Author(s) References Examples

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

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

`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.

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.

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).

Dylan Small

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.

### 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)