| tTestLnormAltPower {EnvStats} | R Documentation |
Power of a One- or Two-Sample t-Test Assuming Lognormal Data
Description
Compute the power of a one- or two-sample t-test, given the sample size, ratio of means, coefficient of variation, and significance level, assuming lognormal data.
Usage
tTestLnormAltPower(n.or.n1, n2 = n.or.n1, ratio.of.means = 1, cv = 1, alpha = 0.05,
sample.type = ifelse(!missing(n2), "two.sample", "one.sample"),
alternative = "two.sided", approx = FALSE)
Arguments
n.or.n1 |
numeric vector of sample sizes. When |
n2 |
numeric vector of sample sizes for group 2. The default value is the value of
|
ratio.of.means |
numeric vector specifying the ratio of the first mean to the second mean.
When |
cv |
numeric vector of positive value(s) specifying the coefficient of
variation. When |
alpha |
numeric vector of numbers between 0 and 1 indicating the Type I error level
associated with the hypothesis test. The default value is |
sample.type |
character string indicating whether to compute power based on a one-sample or
two-sample hypothesis test. When |
alternative |
character string indicating the kind of alternative hypothesis. The possible values
are |
approx |
logical scalar indicating whether to compute the power based on an approximation to
the non-central t-distribution. The default value is |
Details
If the arguments n.or.n1, n2, ratio.of.means, cv, and
alpha are not all the same length, they are replicated to be the same length
as the length of the longest argument.
One-Sample Case (sample.type="one.sample")
Let \underline{x} = x_1, x_2, \ldots, x_n denote a vector of n
observations from a lognormal distribution with mean
\theta and coefficient of variation \tau, and consider the null hypothesis:
H_0: \theta = \theta_0 \;\;\;\;\;\; (1)
The three possible alternative hypotheses are the upper one-sided alternative
(alternative="greater"):
H_a: \theta > \theta_0 \;\;\;\;\;\; (2)
the lower one-sided alternative (alternative="less")
H_a: \theta < \theta_0 \;\;\;\;\;\; (3)
and the two-sided alternative (alternative="two.sided")
H_a: \theta \ne \theta_0 \;\;\;\;\;\; (4)
To test the null hypothesis (1) versus any of the three alternatives (2)-(4), one might be tempted to use Student's t-test based on the log-transformed observations. Unlike the two-sample case with equal coefficients of variation (see below), in the one-sample case Student's t-test applied to the log-transformed observations will not test the correct hypothesis, as now explained.
Let
y_i = log(x_i), \;\; i = 1, 2, \ldots, n \;\;\;\;\;\; (5)
Then \underline{y} = y_1, y_2, \ldots, y_n denote n observations from a
normal distribution with mean \mu and standard deviation \sigma, where
\mu = log(\frac{\theta}{\sqrt{\tau^2 + 1}}) \;\;\;\;\;\; (6)
\sigma = [log(\tau^2 + 1)]^{1/2} \;\;\;\;\;\; (7)
\theta = exp[\mu + (\sigma^2/2)] \;\;\;\;\;\; (8)
\tau = [exp(\sigma^2) - 1]^{1/2} \;\;\;\;\;\; (9)
(see the help file for LognormalAlt). Hence, by Equations (6) and (8) above,
the Student's t-test on the log-transformed data would involve a test of hypothesis
on both the parameters \theta and \tau, not just on \theta.
To test the null hypothesis (1) above versus any of the alternatives (2)-(4), you
can use the function elnormAlt to compute a confidence interval for
\theta, and use the relationship between confidence intervals and hypothesis
tests. To test the null hypothesis (1) above versus the upper one-sided alternative
(2), you can also use
Chen's modified t-test for skewed distributions.
Although you can't use Student's t-test based on the log-transformed observations to
test a hypothesis about \theta, you can use the t-distribution to estimate the
power of a test about \theta that is based on confidence intervals or
Chen's modified t-test, if you are willing to assume the population coefficient of
variation \tau stays constant for all possible values of \theta you are
interested in, and you are willing to postulate possible values for \tau.
First, let's re-write the hypotheses (1)-(4) as follows. The null hypothesis (1) is equivalent to:
H_0: \frac{\theta}{\theta_0} = 1 \;\;\;\;\;\; (10)
The three possible alternative hypotheses are the upper one-sided alternative
(alternative="greater")
H_a: \frac{\theta}{\theta_0} > 1 \;\;\;\;\;\; (11)
the lower one-sided alternative (alternative="less")
H_a: \frac{\theta}{\theta_0} < 1 \;\;\;\;\;\; (12)
and the two-sided alternative (alternative="two.sided")
H_a: \frac{\theta}{\theta_0} \ne 1 \;\;\;\;\;\; (13)
For a constant coefficient of variation \tau, the standard deviation of the
log-transformed observations \sigma is also constant (see Equation (7) above).
Hence, by Equation (8), the ratio of the true mean to the hypothesized mean can be
written as:
R = \frac{\theta}{\theta_0} = \frac{exp[\mu + (\sigma^2/2)]}{exp[\mu_0 + (\sigma^2/2)]} = \frac{e^\mu}{e^\mu_0} = e^{\mu - \mu_0} \;\;\;\;\;\; (14)
which only involves the difference
\mu - \mu_0 \;\;\;\;\;\; (15)
Thus, for given values of R and \tau, the power of the test of the null
hypothesis (10) against any of the alternatives (11)-(13) can be computed based on
the power of a one-sample t-test with
\frac{\delta}{\sigma} = \frac{log(R)}{\sqrt{log(\tau^2 + 1)}} \;\;\;\;\;\; (16)
(see the help file for tTestPower). Note that for the function
tTestLnormAltPower, R corresponds to the argument ratio.of.means,
and \tau corresponds to the argument cv.
Two-Sample Case (sample.type="two.sample")
Let \underline{x}_1 = x_{11}, x_{12}, \ldots, x_{1n_1} denote a vector of
n_1 observations from a lognormal distribution with mean
\theta_1 and coefficient of variaiton \tau, and let
\underline{x}_2 = x_{21}, x_{22}, \ldots, x_{2n_2} denote a vector of
n_2 observations from a lognormal distribution with mean \theta_2 and
coefficient of variation \tau, and consider the null hypothesis:
H_0: \theta_1 = \theta_2 \;\;\;\;\;\; (17)
The three possible alternative hypotheses are the upper one-sided alternative
(alternative="greater"):
H_a: \theta_1 > \theta_2 \;\;\;\;\;\; (18)
the lower one-sided alternative (alternative="less")
H_a: \theta_1 < \theta_2 \;\;\;\;\;\; (19)
and the two-sided alternative (alternative="two.sided")
H_a: \theta_1 \ne \theta_2 \;\;\;\;\;\; (20)
Because we are assuming the coefficient of variation \tau is the same for
both populations, the test of the null hypothesis (17) versus any of the three
alternatives (18)-(20) can be based on the Student t-statistic using the
log-transformed observations.
To show this, first, let's re-write the hypotheses (17)-(20) as follows. The null hypothesis (17) is equivalent to:
H_0: \frac{\theta_1}{\theta_2} = 1 \;\;\;\;\;\; (21)
The three possible alternative hypotheses are the upper one-sided alternative
(alternative="greater")
H_a: \frac{\theta_1}{\theta_2} > 1 \;\;\;\;\;\; (22)
the lower one-sided alternative (alternative="less")
H_a: \frac{\theta_1}{\theta_2} < 1 \;\;\;\;\;\; (23)
and the two-sided alternative (alternative="two.sided")
H_a: \frac{\theta_1}{\theta_2} \ne 1 \;\;\;\;\;\; (24)
If coefficient of variation \tau is the same for both populations, then the
standard deviation of the log-transformed observations \sigma is also the
same for both populations (see Equation (7) above). Hence, by Equation (8), the
ratio of the means can be written as:
R = \frac{\theta_1}{\theta_2} = \frac{exp[\mu_1 + (\sigma^2/2)]}{exp[\mu_2 + (\sigma^2/2)]} = \frac{e^\mu_1}{e^\mu_2} = e^{\mu_1 - \mu_2} \;\;\;\;\;\; (25)
which only involves the difference
\mu_1 - \mu_2 \;\;\;\;\;\; (26)
Thus, for given values of R and \tau, the power of the test of the null
hypothesis (21) against any of the alternatives (22)-(24) can be computed based on
the power of a two-sample t-test with
\frac{\delta}{\sigma} = \frac{log(R)}{\sqrt{log(\tau^2 + 1)}} \;\;\;\;\;\; (27)
(see the help file for tTestPower). Note that for the function
tTestLnormAltPower, R corresponds to the argument ratio.of.means,
and \tau corresponds to the argument cv.
Value
a numeric vector powers.
Note
The normal distribution and
lognormal distribution are probably the two most
frequently used distributions to model environmental data. Often, you need to
determine whether a population mean is significantly different from a specified
standard (e.g., an MCL or ACL, USEPA, 1989b, Section 6), or whether two different
means are significantly different from each other (e.g., USEPA 2009, Chapter 16).
When you have lognormally-distributed data, you have to be careful about making
statements regarding inference for the mean. For the two-sample case with
assumed equal coefficients of variation, you can perform the
Student's t-test on the log-transformed observations.
For the one-sample case, you can perform a hypothesis test by constructing a
confidence interval for the mean using elnormAlt, or use
Chen's t-test modified for skewed data.
In the course of designing a sampling program, an environmental scientist may wish
to determine the relationship between sample size, significance level, power, and
scaled difference if one of the objectives of the sampling program is to determine
whether a mean differs from a specified level or two means differ from each other.
The functions tTestLnormAltPower, tTestLnormAltN,
tTestLnormAltRatioOfMeans, and plotTTestLnormAltDesign
can be used to investigate these relationships for the case of
lognormally-distributed observations.
Author(s)
Steven P. Millard (EnvStats@ProbStatInfo.com)
References
van Belle, G., and D.C. Martin. (1993). Sample Size as a Function of Coefficient of Variation and Ratio of Means. The American Statistician 47(3), 165–167.
Also see the list of references in the help file for tTestPower.
See Also
tTestLnormAltN, tTestLnormAltRatioOfMeans,
plotTTestLnormAltDesign, LognormalAlt,
t.test, Hypothesis Tests.
Examples
# Look at how the power of the one-sample test increases with increasing
# sample size:
seq(5, 30, by = 5)
#[1] 5 10 15 20 25 30
power <- tTestLnormAltPower(n.or.n1 = seq(5, 30, by = 5),
ratio.of.means = 1.5, cv = 1)
round(power, 2)
#[1] 0.14 0.28 0.42 0.54 0.65 0.73
#----------
# Repeat the last example, but use the approximation to the power instead of the
# exact power. Note how the approximation underestimates the true power for
# the smaller sample sizes:
power <- tTestLnormAltPower(n.or.n1 = seq(5, 30, by = 5),
ratio.of.means = 1.5, cv = 1, approx = TRUE)
round(power, 2)
#[1] 0.09 0.25 0.40 0.53 0.64 0.73
#==========
# Look at how the power of the two-sample t-test increases with increasing
# ratio of means:
power <- tTestLnormAltPower(n.or.n1 = 20, sample.type = "two",
ratio.of.means = c(1.1, 1.5, 2), cv = 1)
round(power, 2)
#[1] 0.06 0.32 0.73
#----------
# Look at how the power of the two-sample t-test increases with increasing
# values of Type I error:
power <- tTestLnormAltPower(30, sample.type = "two", ratio.of.means = 1.5,
cv = 1, alpha = c(0.001, 0.01, 0.05, 0.1))
round(power, 2)
#[1] 0.07 0.23 0.46 0.59
#==========
# The guidance document Soil Screening Guidance: Technical Background Document
# (USEPA, 1996c, Part 4) discusses sampling design and sample size calculations
# for studies to determine whether the soil at a potentially contaminated site
# needs to be investigated for possible remedial action. Let 'theta' denote the
# average concentration of the chemical of concern. The guidance document
# establishes the following goals for the decision rule (USEPA, 1996c, p.87):
#
# Pr[Decide Don't Investigate | theta > 2 * SSL] = 0.05
#
# Pr[Decide to Investigate | theta <= (SSL/2)] = 0.2
#
# where SSL denotes the pre-established soil screening level.
#
# These goals translate into a Type I error of 0.2 for the null hypothesis
#
# H0: [theta / (SSL/2)] <= 1
#
# and a power of 95% for the specific alternative hypothesis
#
# Ha: [theta / (SSL/2)] = 4
#
# Assuming a lognormal distribution with a coefficient of variation of 2,
# determine the power associated with various sample sizes for this one-sample test.
# Based on these calculations, you need to take at least 6 soil samples to
# satisfy the requirements for the Type I and Type II errors.
power <- tTestLnormAltPower(n.or.n1 = 2:8, ratio.of.means = 4, cv = 2,
alpha = 0.2, alternative = "greater")
names(power) <- paste("N=", 2:8, sep = "")
round(power, 2)
# N=2 N=3 N=4 N=5 N=6 N=7 N=8
#0.65 0.80 0.88 0.93 0.96 0.97 0.98
#----------
# Repeat the last example, but use the approximate power calculation instead of
# the exact one. Using the approximate power calculation, you need at least
# 7 soil samples instead of 6 (because the approximation underestimates the power).
power <- tTestLnormAltPower(n.or.n1 = 2:8, ratio.of.means = 4, cv = 2,
alpha = 0.2, alternative = "greater", approx = TRUE)
names(power) <- paste("N=", 2:8, sep = "")
round(power, 2)
# N=2 N=3 N=4 N=5 N=6 N=7 N=8
#0.55 0.75 0.84 0.90 0.93 0.95 0.97
#==========
# Clean up
#---------
rm(power)