embryogrowth (version 8.0)

predict.tsd: Estimate sex ratio according to constant incubation temperature

Description

Estimate sex ratio according to constant incubation temperature The data.frame has the temperatures or durations in columns and the quantiles in rows. Note that incubation duration is a very bad proxy for sex ratio. See Georges, A., Limpus, C. J. & Stoutjesdijk, R. 1994. Hatchling sex in the marine turtle _Caretta caretta_ is determined by proportion of development at a temperature, not daily duration of exposure. J. Exp. Zool., 270, 432-444. If replicate.CI is 0 or NULL, point estimate for maximum likelihood is returned.

Usage

# S3 method for tsd
predict(
  object,
  temperatures = NULL,
  durations = NULL,
  SD.temperatures = NULL,
  SD.durations = NULL,
  resultmcmc = NULL,
  chain = 1,
  replicate.CI = 10000,
  probs = c(0.025, 0.5, 0.975),
  ...
)

Arguments

object

A result file generated by tsd

temperatures

A vector of temperatures

durations

A vector of durations

SD.temperatures

SD of temperatures

SD.durations

SD of durations

resultmcmc

A result of tsd_MHmcmc()

chain

What chain to be used is resultmcmc is provided

replicate.CI

Number of replicates to estimate CI

probs

The quantiles to be returned, default=c(0.025, 0.5, 0.975)

...

Not used

Value

A data.frame with informations about sex-ratio

Details

predict.tsd Estimate sex ratio according to constant incubation temperature

See Also

Other Functions for temperature-dependent sex determination: DatabaseTSD.version(), DatabaseTSD, P_TRT(), TSP.list, plot.tsd(), stages, tsd_MHmcmc_p(), tsd_MHmcmc(), tsd()

Examples

# NOT RUN {
library(embryogrowth)
m <- c(10, 14, 7, 4, 3, 0, 0)
f <- c(0, 1, 2, 4, 15, 10, 13)
t <- c(25, 26, 27, 28, 29, 30, 31)
result <- tsd(males=m, females=f, temperatures=t)
plot(result)
predict(result, temperatures=c(25, 31), replicate.CI = 10000)
predict(result, temperatures=c(25, 31), SD.temperatures = c(1, 2), replicate.CI = 10000)
d <- c(72, 70, 65, 63, 62, 60, 59)
result <- tsd(males=m, females=f, durations=d)
predict(result, durations=c(67, 68), replicate.CI = 10000)
# }