sanderlings {moult}

Sanderling Moult Data

Description

This data set gives moult indices for 164 Sanderlings trapped on 11 days.

Usage

data(sanderlings)

Format

A data frame with 164 observations on the following 2 variables.

Day

a numeric vector of day bird was measured, 1 = 1 July

MIndex

a numeric vector of moult indices, 0 = bird has not started moult, 1 = bird has completed moult

Details

This data set gives moult indices for 164 Sanderlings trapped on 11 days in the southwestern Cape, South Africa, between October 1978 and April 1979. Day 1 = 1 July). Moult indices are a transformation of moult scores so that moult index increases linearly with time. See Underhill and Zucchini (1988) for details.

Source

Underhill and Zucchini (1998)

References

Underhill, L. G. and Zucchini, W. (1988) A model for avian primary moult. Ibis 130, 358–372.

Examples

data(sanderlings)

## fit model of type 1 to data
m1 <- moult(MIndex ~ Day, data = sanderlings, type = 1)               
summary(m1)

## model of type 2 (default)
m2 <- moult(MIndex ~ Day, data = sanderlings)                     
summary(m2)

## model of type 3
m3 <- moult(MIndex ~ Day, data = sanderlings, type = 3)              
summary(m3)

## find intercept and slope of mean moult trajectory line
uza <- - coef(m2, "mean") / coef(m2, "duration")    
uzb <- 1 / coef(m2, "duration")

## extract how many birds observed on each of the days
nn <- as.numeric(table(sanderlings$Day))        
## extract days of observations
day <- unique(sanderlings$Day)                                            

## probabilities of moult stages
## Table 6 in Underhill and Zucchini 1988
p1 <- predict(m2, newdata = data.frame(day))       
p1$M * nn

## Table 7 in Underhill and Zucchini 1988
days2 <- seq(70, 310, by = 10)
p2 <- predict(m2, newdata = data.frame(days2))
p2$M * 100                                  

p3 <- predict(m3, newdata = data.frame(day))      
p3

## Comparison with regression models
MInd <- sanderlings$MIndex[sanderlings$MIndex > 0 &
                           sanderlings$MIndex < 1]
MTime <- sanderlings$Day[sanderlings$MIndex > 0 &
                         sanderlings$MIndex < 1]

lm1 <- lm(MTime ~ MInd)                           
lm1.int <- coef(lm1)[1]
lm1.slope <- coef(lm1)[2]

lm2 <- lm(MInd ~ MTime)

## regression of Index on Time
plot(MTime, MInd, pch = 19, cex=0.7)

## regression of Time on Index: gives better estimates 
## for mean start day and duration of moult	    
abline(lm2, col = "blue", lwd = 2) 
abline(-lm1.int / lm1.slope, 1 / lm1.slope, col = "orange", lwd = 2) 
abline(uza, uzb, col = "red", lty = 2, lwd = 2)