asbio (version 1.9-7)

r.dist: Visualize the sampling distribution of Pearson's product moment correlation

Description

Stumbling points for many methods of inference for the true correlation \(\rho\) and for independence are: 1) asymmetry, 2) explicit bounds on \(\rho\), and 3) dependence on sample size, of the sampling distribution of r.

The functions here allow visualization of these characteristics. The algorithm used for the sampling distribution of r is based on the first two steps in an asymptotic series (see Kenney and Keeping 1951).

Usage

r.dist(rho, r, n)
see.r.dist.tck()

Arguments

rho

Population correlation

r

A numeric vector containing possible estimates of \(rho\).

n

Sample size, an integer.

Author

Ken Aho

Details

All distributions are standardized to have an area of one.

References

Kenney, J. F. and E. S. Keeping (1951) Mathematics of Statistics, Pt. 2, 2nd ed. Van Nostrand, Princeton, NJ.

Weisstein, E. W. (2012) Correlation Coefficient--Bivariate Normal Distribution. From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CorrelationCoefficientBivariateNormalDistribution.html

See Also

cor

Examples

# dev.new(height=3.5)
op <- par(mfrow=c(1,2),mar=c (0,0,1.5,3), oma = c(5, 4.2, 0, 0))
vals <- r.dist(0.9, seq(-1, 1, .001), 5)
plot(seq(-1, 1, .001), vals, type = "l",ylab = "", xlab = "")
vals <- r.dist(0.5, seq(-1, 1, .001), 5)
lines(seq(-1, 1, .001), vals, lty = 2)
vals <- r.dist(0.0, seq(-1, 1, .001), 5)
lines(seq(-1, 1, .001), vals, lty = 3)
legend("topleft", lty = c(1, 2, 3), title = expression(paste(italic(n)," = 5")), 
legend = c(expression(paste(rho, " = 0.9")),expression(paste(rho, " = 0.5")),
expression(paste(rho, " = 0"))),bty = "n") 

vals <- r.dist(0.9, seq(-1, 1, .001), 30)
plot(seq(-1, 1, .001), vals, type = "l",xlab= "", ylab= "")
vals <- r.dist(0.5, seq(-1, 1, .001), 30)
lines(seq(-1, 1, .001), vals, lty = 2)
vals <- r.dist(0.0, seq(-1, 1, .001), 30)
lines(seq(-1, 1, .001), vals, lty = 3)
legend("topleft", lty = c(1, 2, 3), title = expression(paste(italic(n)," = 30")), 
legend = c(expression(paste(rho, " = 0.9")),expression(paste(rho, " = 0.5")),
expression(paste(rho, " = 0"))), bty = "n") 
mtext(side = 2, expression(paste(italic(f),"(",italic(r),")")), outer = TRUE, line = 3)
mtext(side = 1, expression(italic(r)), outer = TRUE, line = 3, at = .45)
par(op)