aovPower: Compute the Power of a One-Way Fixed-Effects Analysis of...
In EnvStats: Package for Environmental Statistics, Including US EPA Guidance
aovPower R Documentation
Compute the Power of a One-Way Fixed-Effects Analysis of Variance
Description
Compute the power of a one-way fixed-effects analysis of variance,
given the sample sizes, population means, population standard deviation, and
significance level.
Usage
aovPower(n.vec, mu.vec = rep(0, length(n.vec)), sigma = 1, alpha = 0.05)
Arguments
n.vec
numeric vector of sample sizes for each group. The i^{th} element of n.vec
denotes the sample size for group i. The length of n.vec must be at least 2,
and all elements of n.vec must be greater than or equal to 2. Missing (NA),
undefined (NaN), and infinite (Inf, -Inf) values are not allowed.
mu.vec
numeric vector of population means. The length of mu.vec must be the
same as the length of n.vec. The default value is a vector of zeros.
Missing (NA), undefined (NaN), and infinite (Inf, -Inf)
values are not allowed.
sigma
numeric scalar specifying the population standard deviation (\sigma)
for each group. The default value is sigma=1.
alpha
numeric scalar between 0 and 1 indicating the Type I error level associated
with the hypothesis test. The default value is alpha=0.05.
Details
Consider k normally distributed populations with common standard deviation
\sigma. Let \mu_i denote the mean of the i'th group
(i = 1, 2, \ldots, k), and let
\underline{x}_i = x_{i1}, x_{i2}, \ldots, x_{in_i} denote a vector of
n_i observations from the i'th group.
The statistical method of analysis of variance (ANOVA) tests the null hypothesis:
H_0: \mu_1 = \mu_2 = \cdots = \mu_k \;\;\;\;\;\; (1)
against the alternative hypothesis that at least one of the means is different from
the rest by using the F-statistic given by:
F = \frac{[\sum_{i=1}^k n_i(\bar{x}_{i.} - \bar{x}_{..})^2]/(k-1)}{[\sum_{i=1}^k \sum_{j=1}^{n_i} (x_{ij} - \bar{x}_{i.})^2]/(N-k)} \;\;\;\;\;\; (2)
where
\bar{x}_{i.} = \frac{1}{n_i} \sum_{j=1}^{n_i} x_{ij} \;\;\;\;\;\; (3)
\bar{x}_{..} = \frac{1}{N} \sum_{i=1}^k n_i\bar{x}_{i.} = \frac{1}{N} \sum_{i=1}^k \sum_{j=1}^{n_i} x_{ij} \;\;\;\;\;\; (4)
N = \sum_{i=1}^k n_i \;\;\;\;\;\; (5)
Under the null hypothesis (1), the F-statistic in (2) follows an
F-distribution with k-1 and N-k degrees of freedom.
Analysis of variance rejects the null hypothesis (1) at significance level
\alpha when
F > F_{k-1, N-k}(1 - \alpha) \;\;\;\;\;\; (6)
where F_{\nu_1, \nu_2}(p) denotes the p'th quantile of the
F-distribution with \nu_1 and \nu_2 degrees of freedom
(Zar, 2010, Chapter 10; Berthouex and Brown, 2002, Chapter 24;
Helsel and Hirsh, 1992, pp. 164–169).
The power of this test, denoted by 1-\beta, where \beta denotes the
probability of a Type II error, is given by:
1 - \beta = Pr[F_{k-1, N-k, \Delta} > F_{k-1, N-k}(1 - \alpha)] \;\;\;\;\;\; (7)
where
\Delta = \frac{\sum_{i=1}^k n_i(\mu_i - \bar{\mu}_.)^2}{\sigma^2} \;\;\;\;\;\; (8)
\bar{\mu}_. = \frac{1}{k} \sum_{i=1}^k \mu _i \;\;\;\;\;\; (9)
and F_{\nu_1, \nu_2, \Delta} denotes a
non-central F random variable with \nu_1 and
\nu_2 degrees of freedom and non-centrality parameter \Delta.
Equation (7) can be re-written as:
1 - \beta = 1 - H[F_{k-1, N-k}(1 - \alpha), k-1, N-k, \Delta] \;\;\;\;\;\; (10)
where H(x, \nu_1, \nu_2, \Delta) denotes the cumulative distribution function
of this random variable evaluated at x (Scheffe, 1959, pp.38–39, 62–65).
The power of the one-way fixed-effects ANOVA depends on the
sample sizes for each of the k groups, the value of the
population means for each of the k groups, the population
standard deviation \sigma, and the significance level
\alpha.
Value
a numeric scalar indicating the power of the one-way fixed-effects ANOVA for
the given sample sizes, population means, population standard deviation, and
significance level.
Note
The normal and lognormal distribution are probably the two most
frequently used distributions to model environmental data.
Sometimes it is necessary to compare several means to determine
whether any are significantly different from each other
(e.g., USEPA, 2009, p.6-38). In this case, assuming
normally distributed data, you perform a one-way parametric
analysis of variance.
In the course of designing a sampling program, an environmental
scientist may wish to determine the relationship between sample
size, Type I error level, power, and differences in means if
one of the objectives of the sampling program is to determine
whether a particular mean differs from a group of means. The
functions aovPower, aovN, and
plotAovDesign can be used to investigate these
relationships for the case of normally-distributed observations.
Author(s)
Steven P. Millard (EnvStats@ProbStatInfo.com)
References
Berthouex, P.M., and L.C. Brown. (2002).
Statistics for Environmental Engineers. Second Edition.
Lewis Publishers, Boca Raton, FL.
Helsel, D.R., and R.M. Hirsch. (1992).
Statistical Methods in Water Resources Research.
Elsevier, New York, NY, Chapter 7.
Johnson, N. L., S. Kotz, and N. Balakrishnan. (1995).
Continuous Univariate Distributions, Volume 2.
Second Edition. John Wiley and Sons, New York,
Chapters 27, 29, 30.
Millard, S.P., and Neerchal, N.K. (2001). Environmental Statistics with S-PLUS.
CRC Press, Boca Raton, Florida.
Scheffe, H. (1959). The Analysis of Variance.
John Wiley and Sons, New York, 477pp.
USEPA. (2009). Statistical Analysis of Groundwater Monitoring Data at RCRA Facilities, Unified Guidance.
EPA 530/R-09-007, March 2009. Office of Resource Conservation and Recovery Program Implementation and Information Division.
U.S. Environmental Protection Agency, Washington, D.C. p.6-38.
Zar, J.H. (2010). Biostatistical Analysis.
Fifth Edition. Prentice-Hall, Upper Saddle River, NJ,
Chapter 10.
See Also
aovN, plotAovDesign,
Normal, aov.
Examples
# 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)
EnvStats documentation built on Sept. 11, 2024, 6:03 p.m.
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| aovPower | R Documentation |
Compute the Power of a One-Way Fixed-Effects Analysis of Variance
Description
Compute the power of a one-way fixed-effects analysis of variance, given the sample sizes, population means, population standard deviation, and significance level.
Usage
aovPower(n.vec, mu.vec = rep(0, length(n.vec)), sigma = 1, alpha = 0.05)
Arguments
n.vec |
numeric vector of sample sizes for each group. The |
mu.vec |
numeric vector of population means. The length of |
sigma |
numeric scalar specifying the population standard deviation ( |
alpha |
numeric scalar between 0 and 1 indicating the Type I error level associated
with the hypothesis test. The default value is |
Details
Consider k normally distributed populations with common standard deviation
\sigma. Let \mu_i denote the mean of the i'th group
(i = 1, 2, \ldots, k), and let
\underline{x}_i = x_{i1}, x_{i2}, \ldots, x_{in_i} denote a vector of
n_i observations from the i'th group.
The statistical method of analysis of variance (ANOVA) tests the null hypothesis:
H_0: \mu_1 = \mu_2 = \cdots = \mu_k \;\;\;\;\;\; (1)
against the alternative hypothesis that at least one of the means is different from the rest by using the F-statistic given by:
F = \frac{[\sum_{i=1}^k n_i(\bar{x}_{i.} - \bar{x}_{..})^2]/(k-1)}{[\sum_{i=1}^k \sum_{j=1}^{n_i} (x_{ij} - \bar{x}_{i.})^2]/(N-k)} \;\;\;\;\;\; (2)
where
\bar{x}_{i.} = \frac{1}{n_i} \sum_{j=1}^{n_i} x_{ij} \;\;\;\;\;\; (3)
\bar{x}_{..} = \frac{1}{N} \sum_{i=1}^k n_i\bar{x}_{i.} = \frac{1}{N} \sum_{i=1}^k \sum_{j=1}^{n_i} x_{ij} \;\;\;\;\;\; (4)
N = \sum_{i=1}^k n_i \;\;\;\;\;\; (5)
Under the null hypothesis (1), the F-statistic in (2) follows an
F-distribution with k-1 and N-k degrees of freedom.
Analysis of variance rejects the null hypothesis (1) at significance level
\alpha when
F > F_{k-1, N-k}(1 - \alpha) \;\;\;\;\;\; (6)
where F_{\nu_1, \nu_2}(p) denotes the p'th quantile of the
F-distribution with \nu_1 and \nu_2 degrees of freedom
(Zar, 2010, Chapter 10; Berthouex and Brown, 2002, Chapter 24;
Helsel and Hirsh, 1992, pp. 164–169).
The power of this test, denoted by 1-\beta, where \beta denotes the
probability of a Type II error, is given by:
1 - \beta = Pr[F_{k-1, N-k, \Delta} > F_{k-1, N-k}(1 - \alpha)] \;\;\;\;\;\; (7)
where
\Delta = \frac{\sum_{i=1}^k n_i(\mu_i - \bar{\mu}_.)^2}{\sigma^2} \;\;\;\;\;\; (8)
\bar{\mu}_. = \frac{1}{k} \sum_{i=1}^k \mu _i \;\;\;\;\;\; (9)
and F_{\nu_1, \nu_2, \Delta} denotes a
non-central F random variable with \nu_1 and
\nu_2 degrees of freedom and non-centrality parameter \Delta.
Equation (7) can be re-written as:
1 - \beta = 1 - H[F_{k-1, N-k}(1 - \alpha), k-1, N-k, \Delta] \;\;\;\;\;\; (10)
where H(x, \nu_1, \nu_2, \Delta) denotes the cumulative distribution function
of this random variable evaluated at x (Scheffe, 1959, pp.38–39, 62–65).
The power of the one-way fixed-effects ANOVA depends on the
sample sizes for each of the k groups, the value of the
population means for each of the k groups, the population
standard deviation \sigma, and the significance level
\alpha.
Value
a numeric scalar indicating the power of the one-way fixed-effects ANOVA for the given sample sizes, population means, population standard deviation, and significance level.
Note
The normal and lognormal distribution are probably the two most frequently used distributions to model environmental data. Sometimes it is necessary to compare several means to determine whether any are significantly different from each other (e.g., USEPA, 2009, p.6-38). In this case, assuming normally distributed data, you perform a one-way parametric analysis of variance.
In the course of designing a sampling program, an environmental
scientist may wish to determine the relationship between sample
size, Type I error level, power, and differences in means if
one of the objectives of the sampling program is to determine
whether a particular mean differs from a group of means. The
functions aovPower, aovN, and
plotAovDesign can be used to investigate these
relationships for the case of normally-distributed observations.
Author(s)
Steven P. Millard (EnvStats@ProbStatInfo.com)
References
Berthouex, P.M., and L.C. Brown. (2002). Statistics for Environmental Engineers. Second Edition. Lewis Publishers, Boca Raton, FL.
Helsel, D.R., and R.M. Hirsch. (1992). Statistical Methods in Water Resources Research. Elsevier, New York, NY, Chapter 7.
Johnson, N. L., S. Kotz, and N. Balakrishnan. (1995). Continuous Univariate Distributions, Volume 2. Second Edition. John Wiley and Sons, New York, Chapters 27, 29, 30.
Millard, S.P., and Neerchal, N.K. (2001). Environmental Statistics with S-PLUS. CRC Press, Boca Raton, Florida.
Scheffe, H. (1959). The Analysis of Variance. John Wiley and Sons, New York, 477pp.
USEPA. (2009). Statistical Analysis of Groundwater Monitoring Data at RCRA Facilities, Unified Guidance. EPA 530/R-09-007, March 2009. Office of Resource Conservation and Recovery Program Implementation and Information Division. U.S. Environmental Protection Agency, Washington, D.C. p.6-38.
Zar, J.H. (2010). Biostatistical Analysis. Fifth Edition. Prentice-Hall, Upper Saddle River, NJ, Chapter 10.
See Also
aovN, plotAovDesign,
Normal, aov.
Examples
# 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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