# Look at how the power of a one-way ANOVA increases
# with increasing sample size:
aovPower(n.vec = rep(5, 3), mu.vec = c(10, 15, 20), sigma = 5)
#[1] 0.7015083
aovPower(n.vec = rep(10, 3), mu.vec = c(10, 15, 20), sigma = 5)
#[1] 0.9732551
#----------------------------------------------------------------
# Look at how the power of a one-way ANOVA increases
# with increasing variability in the population means:
aovPower(n.vec = rep(5,3), mu.vec = c(10, 10, 11), sigma=5)
#[1] 0.05795739
aovPower(n.vec = rep(5, 3), mu.vec = c(10, 10, 15), sigma = 5)
#[1] 0.2831863
aovPower(n.vec = rep(5, 3), mu.vec = c(10, 13, 15), sigma = 5)
#[1] 0.2236093
aovPower(n.vec = rep(5, 3), mu.vec = c(10, 15, 20), sigma = 5)
#[1] 0.7015083
#----------------------------------------------------------------
# Look at how the power of a one-way ANOVA increases
# with increasing values of Type I error:
aovPower(n.vec = rep(10,3), mu.vec = c(10, 12, 14),
sigma = 5, alpha = 0.001)
#[1] 0.02655785
aovPower(n.vec = rep(10,3), mu.vec = c(10, 12, 14),
sigma = 5, alpha = 0.01)
#[1] 0.1223527
aovPower(n.vec = rep(10,3), mu.vec = c(10, 12, 14),
sigma = 5, alpha = 0.05)
#[1] 0.3085313
aovPower(n.vec = rep(10,3), mu.vec = c(10, 12, 14),
sigma = 5, alpha = 0.1)
#[1] 0.4373292
#==========
# The example on pages 5-11 to 5-14 of USEPA (1989b) shows
# log-transformed concentrations of lead (mg/L) at two
# background wells and four compliance wells, where observations
# were taken once per month over four months (the data are
# stored in EPA.89b.loglead.df.) Assume the true mean levels
# at each well are 3.9, 3.9, 4.5, 4.5, 4.5, and 5, respectively.
# Compute the power of a one-way ANOVA to test for mean
# differences between wells. Use alpha=0.05, and assume the
# true standard deviation is equal to the one estimated from
# the data in this example.
# First look at the data
names(EPA.89b.loglead.df)
#[1] "LogLead" "Month" "Well" "Well.type"
dev.new()
stripChart(LogLead ~ Well, data = EPA.89b.loglead.df,
show.ci = FALSE, xlab = "Well Number",
ylab="Log [ Lead (ug/L) ]",
main="Lead Concentrations at Six Wells")
# Note: The assumption of a constant variance across
# all wells is suspect.
# Now perform the ANOVA and get the estimated sd
aov.list <- aov(LogLead ~ Well, data=EPA.89b.loglead.df)
summary(aov.list)
# Df Sum Sq Mean Sq F value Pr(>F)
#Well 5 5.7447 1.14895 3.3469 0.02599 *
#Residuals 18 6.1791 0.34328
#---
#Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 '' 1
# Now call the function aovPower
aovPower(n.vec = rep(4, 6),
mu.vec = c(3.9,3.9,4.5,4.5,4.5,5), sigma=sqrt(0.34))
#[1] 0.5523148
# Clean up
rm(aov.list)
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