power.noninf | R Documentation |

Function calculates of the power of the one-sided non-inferiority *t*-test
for normal or log-normal distributed data.

power.noninf(alpha = 0.025, logscale = TRUE, margin, theta0, CV, n, design = "2x2", robust = FALSE)

`alpha` |
Significance level (one-sided). Defaults here to 0.025. |

`logscale` |
Should the data used on log-transformed or on original scale? |

`theta0` |
‘True’ or assumed T/R ratio or difference. |

`margin` |
Non-inferiority margin. |

`CV` |
In case of In case of cross-over studies this is the within-subject CV, in case of a parallel-group design the CV of the total variability. |

`n` |
Number of subjects under study. |

`design` |
Character string describing the study design. |

`robust` |
Defaults to |

The power is calculated exact via non-central *t*-distribution.

**Notes on the underlying hypotheses**

If the supplied margin is < 0 (`logscale=FALSE`

) or < 1 (`logscale=TRUE`

),
then it is assumed higher response values are better. The hypotheses are

` H0: theta0 <= margin vs. H1: theta0 > margin`

where `theta0 = mean(test)-mean(reference)`

if `logscale=FALSE`

or

` H0: log(theta0) <= log(margin) vs. H1: log(theta0) > log(margin)`

where `theta0 = mean(test)/mean(reference)`

if `logscale=TRUE`

.

If the supplied margin is > 0 (`logscale=FALSE`

) or > 1 (`logscale=TRUE`

),
then it is assumed lower response values are better. The hypotheses are

` H0: theta0 >= margin vs. H1: theta0 < margin`

where `theta0 = mean(test)-mean(reference)`

if `logscale=FALSE`

or

` H0: log(theta0) >= log(margin) vs. H1: log(theta0) < log(margin)`

where `theta0 = mean(test)/mean(reference)`

if `logscale=TRUE`

.

This latter case may also be considered as ‘non-superiority’.

Value of power according to the input arguments.

The function does not vectorize if design is a vector.

The function vectorize properly if CV or theta0 are vectors.

Other vector input is not tested yet.

This function does not rely on TOST but may be useful in planning BE studies
if the question is not equivalence but ‘non-superiority’.

Hint: Evaluation of Fluctuation in the EMEA’s Note for Guidance between a modified release
formulation and an immediate release product.

D. Labes

Julious SA. *Sample sizes for clinical trials with Normal data.* Stat Med. 2004;23(12):1921–86. doi: 10.1002/sim.1783

`known.designs, sampleN.noninf`

# using all the defaults: margin=0.8, theta0=0.95, alpha=0.025 # log-transformed, design="2x2" # should give: 0.4916748 power.noninf(CV=0.3, n=24)

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