wald_test_nb: Wald test for NB ratio of means

Description

Wald test for the ratio of means from two independent negative binomial outcomes.

Usage

wald_test_nb(
  data,
  equal_dispersion = FALSE,
  ci_level = NULL,
  link = "log",
  ratio_null = 1,
  distribution = asymptotic(),
  ...
)

Value

A list with the following elements:

Slot	Subslot	Name	Description
1		`chisq`	$\chi^2$ test statistic for the ratio of means.
2		`df`	Degrees of freedom.
3		`p`	p-value.
4		`ratio`	Estimated ratio of means (group 2 / group 1).
4	1	`estimate`	Point estimate.
4	2	`lower`	Confidence interval lower bound.
4	3	`upper`	Confidence interval upper bound.
5		`mean1`	Estimated mean of group 1.
6		`mean2`	Estimated mean of group 2.
7		`dispersion1`	Estimated dispersion of group 1.
8		`dispersion2`	Estimated dispersion of group 2.
9		`n1`	Sample size of group 1.
10		`n2`	Sample size of group 2.
11		`method`	Method used for the results.
12		`ci_level`	The confidence level.
13		`equal_dispersion`	Whether or not equal dispersions were assumed.
14		`link`	Link function used to transform the ratio of means in the test hypotheses.
15		`ratio_null`	Assumed ratio of means under the null hypothesis.
16		`mle_code`	Integer indicating why the optimization process terminated.
17		`mle_message`	Information from the optimizer.

Arguments

data: (list)
A list whose first element is the vector of negative binomial values from group 1 and the second element is the vector of negative binomial values from group 2. NAs are silently excluded. The default output from sim_nb().
equal_dispersion: (Scalar logical: FALSE)
If TRUE, the Wald test is calculated assuming both groups have the same population dispersion parameter. If FALSE (default), the Wald test is calculated assuming different dispersions.
ci_level: (Scalar numeric: NULL; (0, 1))
If NULL, confidence intervals are set as NA. If in (0, 1), confidence intervals are calculated at the specified level.
link: (Scalar string: "log")
The one-to-one link function for transformation of the ratio in the test hypotheses. Must be one of "log" (default), "sqrt", "squared", or "identity". See 'Details' for additional information.
ratio_null: (Scalar numeric: 1; (0, Inf))
The (pre-transformation) ratio of means assumed under the null hypothesis (group 2 / group 1). Typically ratio_null = 1 (no difference). See 'Details' for additional information.
distribution: (function: asymptotic() or simulated())
The method used to define the distribution of the $\chi^2$ Wald test statistic under the null hypothesis. See 'Details' and asymptotic() or simulated() for additional information.
...: Optional arguments passed to the MLE function mle_nb().

Details

This function is primarily designed for speed in simulation. Missing values are silently excluded.

Suppose $X_1 \sim NB(\mu, \theta_1)$ and $X_2 \sim NB(r\mu, \theta_2)$ where $X_1$ and $X_2$ are independent, $X_1$ is the count outcome for items in group 1, $X_2$ is the count outcome for items in group 2, $\mu$ is the arithmetic mean count in group 1, $r$ is the ratio of arithmetic means for group 2 with respect to group 1, $\theta_1$ is the dispersion parameter of group 1, and $\theta_2$ is the dispersion parameter of group 2.

The hypotheses for the Wald test of $r$ are

$$ \begin{aligned} H_{null} &: f(r) = f(r_{null}) \\ H_{alt} &: f(r) \neq f(r_{null}) \end{aligned} $$

where $f(\cdot)$ is a one-to-one link function with nonzero derivative, $r = \frac{\bar{X}_2}{\bar{X}_1}$ is the population ratio of arithmetic means for group 2 with respect to group 1, and $r_{null}$ is a constant for the assumed null population ratio of means (typically $r_{null} = 1$).

rettiganti_2012;textualdepower found that $f(r) = r^2$ and $f(r) = r$ had greatest power when $r < 1$. However, when $r > 1$, $f(r) = \ln r$, the likelihood ratio test, and the Rao score test have greatest power. Note that $f(r) = \ln r$, LRT, and RST were unbiased tests while the $f(r) = r$ and $f(r) = r^2$ tests were biased when $r > 1$. The $f(r) = \ln r$, LRT, and RST produced acceptable results for any $r$ value. These results depend on the use of asymptotic vs. exact critical values.

The Wald test statistic is

$$ W(f(\hat{r})) = \left( \frac{f \left( \frac{\bar{x}_2}{\bar{x}_1} \right) - f(r_{null})}{f^{\prime}(\hat{r}) \hat{\sigma}_{\hat{r}}} \right)^2 $$

where

$$ \hat{\sigma}^{2}_{\hat{r}} = \frac{\hat{r} \left[ n_1 \hat{\theta}_1 (\hat{r} \hat{\mu} + \hat{\theta}_2) + n_2 \hat{\theta}_2 \hat{r} (\hat{\mu} + \hat{\theta}_1) \right]}{n_1 n_2 \hat{\theta}_1 \hat{\theta}_2 \hat{\mu}} $$

Under $H_{null}$, the Wald test statistic is asymptotically distributed as $\chi^2_1$. The approximate level $\alpha$ test rejects $H_{null}$ if $W(f(\hat{r})) \geq \chi^2_1(1 - \alpha)$. However, the asymptotic critical value is known to underestimate the exact critical value and the nominal significance level may not be achieved for small sample sizes. The level of significance inflation also depends on $f(\cdot)$ and is most severe for $f(r) = r^2$ where only the exact critical value should be used. Argument distribution allows control of the distribution of the $\chi^2_1$ test statistic under the null hypothesis by use of functions asymptotic() and simulated().

Note that standalone use of this function with equal_dispersion = FALSE and distribution = simulated(), e.g.

data |>
  wald_test_nb(
    equal_dispersion = FALSE,
    distribution = simulated()
  )

results in a nonparametric randomization test based on label permutation. This violates the assumption of exchangeability for the randomization test because the labels are not exchangeable when the null hypothesis assumes unequal dispersions. However, used inside power(), e.g.

data |>
  power(
    wald_test_nb(
      equal_dispersion = FALSE,
      distribution = simulated()
    )
  )

results in parametric resampling and no label permutation in performed. Thus, setting equal_dispersion = FALSE and distribution = simulated() is only recommended when wald_test_nb() is used inside of power(). See also, simulated().

References

rettiganti_2012depower

aban_2009depower

Examples

Run this code

#----------------------------------------------------------------------------
# wald_test_nb() examples
#----------------------------------------------------------------------------
library(depower)

set.seed(1234)
sim_nb(
  n1 = 60,
  n2 = 40,
  mean1 = 10,
  ratio = 1.5,
  dispersion1 = 2,
  dispersion2 = 8
) |>
  wald_test_nb()

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