tTestLnormAltPower: 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 sample.type="one.sample", n.or.n1 denotes $n$, the number of observations in the single sample. When sample.type="two.sample", n.or.n1 denotes $n_1$, the nu

numeric vector of sample sizes for group 2. The default value is the value of n.or.n1. This argument is ignored when sample.type="one.sample". Missing (NA), undefined (NaN), and infinite (

ratio.of.means

numeric vector specifying the ratio of the first mean to the second mean. When sample.type="one.sample", this is the ratio of the population mean to the hypothesized mean. When sample.type="two.sample", this is the rati

numeric vector of positive value(s) specifying the coefficient of variation. When sample.type="one.sample", this is the population coefficient of variation. When sample.type="two.sample", this is the coefficient of

alpha

numeric vector of numbers between 0 and 1 indicating the Type I error level associated with the hypothesis test. The default value is alpha=0.05.

sample.type

character string indicating whether to compute power based on a one-sample or two-sample hypothesis test. When sample.type="one.sample", the computed power is based on a hypothesis test for a single mean. When sample.type=

alternative

character string indicating the kind of alternative hypothesis. The possible values are "two.sided" (the default), "greater", and "less".

approx

logical scalar indicating whether to compute the power based on an approximation to the non-central t-distribution. The default value is FALSE.

Value

a numeric vector powers.

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.

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.

Examples

Run this code

# 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)

Run the code above in your browser using DataLab