Assignment #7

Questions

8E1. For each of the causal relationships below, name a hypothetical third variable that would lead to an interaction effect:

Bread dough rises because of yeast.
Education leads to higher income.
Gasoline makes a car go.

#The activity of yeast is influenced by temperature, so temperature would lead to an interaction effect with yeast to predict amount of bread dough rising.
#Income are different for people with different majors in their education, so major would lead to an interaction effect with years of education to predict income.
#Gasoline will only make a working car go, so the model year would would lead to an interaction effect with gasoline to predict movement.

8E2. Which of the following explanations invokes an interaction?

Caramelizing onions requires cooking over low heat and making sure the onions do not dry out.
A car will go faster when it has more cylinders or when it has a better fuel injector.
Most people acquire their political beliefs from their parents, unless they get them instead from their friends.
Intelligent animal species tend to be either highly social or have manipulative appendages (hands, tentacles, etc.).

#This explanation invokes an interaction between heat and dryness in predicting onion caramelization. 
#This explanation invokes main effects of number of cylinders and fuel injector quality on car speed but does not invoke an interaction. Adding cylinders and increasing the quality of the fuel injector are independent routes to increasing car speed.
#This explanation invokes an interaction between  type and parents’ beliefs on the one hand and between type and friends’ beliefs on the other hand to predict individuals’ political beliefs.
#This explanation invokes an interaction between sociality and the possession of manipulative appendages in predicting a species’ intelligence.

8E3. For each of the explanations in 8E2, write a linear model that expresses the stated relationship.

#μi = βH*Hi + βD*Di + βHD*Hi*Di
#μi = βC*Ci + βF*Fi
#μi = βTP*Ti*Pi + βTF*Ti*Fi
#μi = βS*Si + βA*Ai + βSA*Si*Ai

8M1. Recall the tulips example from the chapter. Suppose another set of treatments adjusted the temperature in the greenhouse over two levels: cold and hot. The data in the chapter were collected at the cold temperature. You find none of the plants grown under the hot temperature developed any blooms at all, regardless of the water and shade levels. Can you explain this result in terms of interactions between water, shade, and temperature?

#An interaction allows the relationship between a predictor and an outcome to depend on another predictor. Since the question is stating that the relationships between blossoms and water and between blossoms and shade depend upon the value of temperature, interaction effects can be used here. Since there are three predictor variables now (water, shade, and temperature), we would have a single three-way interaction and three two-way interactions (WST, WS, WT, and ST).

8M2. Can you invent a regression equation that would make the bloom size zero, whenever the temperature is hot?

#The algebraic form of the regression equation was shown below:
#μi = α+βW*Wi+βS*Si+βT*Ti+βWST*Wi*Si*Ti+βWS*Wi*Si+βWT*Wi*Ti+βST*Si*Ti
#After invention,the regression equation that would make the bloom size zero, whenever the temperature is hot is: 
#μi = α+βW*Wi+βS*Si−α*Ti–βWS*Wi*Si*Ti+βWS*Wi*Si–βW*Wi*Ti–βS*Si*Ti

8M3. In parts of North America, ravens depend upon wolves for their food. This is because ravens are carnivorous but cannot usually kill or open carcasses of prey. Wolves however can and do kill and tear open animals, and they tolerate ravens co-feeding at their kills. This species relationship is generally described as a “species interaction.” Can you invent a hypothetical set of data on raven population size in which this relationship would manifest as a statistical interaction? Do you think the biological interaction could be linear? Why or why not?

#μi = α+βP*Pi+βW*Wi+βPW*Pi*Wi

library(rethinking)

## Loading required package: rstan

## Loading required package: StanHeaders

## Loading required package: ggplot2

## rstan (Version 2.21.2, GitRev: 2e1f913d3ca3)

## For execution on a local, multicore CPU with excess RAM we recommend calling
## options(mc.cores = parallel::detectCores()).
## To avoid recompilation of unchanged Stan programs, we recommend calling
## rstan_options(auto_write = TRUE)

## Do not specify '-march=native' in 'LOCAL_CPPFLAGS' or a Makevars file

## Loading required package: parallel

## rethinking (Version 2.12)

## 
## Attaching package: 'rethinking'

## The following object is masked from 'package:stats':
## 
##     rstudent

N <- 300 #simulation size
#assume there is a positive but not a perfect correlation between prey and wolf populations
rPW <- 0.6 #correlation of the interaction between prey and wolf
bP <- 0.3 #regression coefficient for prey
bW <- 0.1 #regression coefficient for wolf
bPW <- 0.5 #regression coefficient for prey-by-wolf interaction
#Simulate data
prey <- rnorm(
  n = N, 
  mean = 0, 
  sd = 1
)
wolf <- rnorm(
  n = N, 
  mean = rPW * prey, 
  sd = sqrt(1 - rPW^2)
)
raven <- rnorm(
  n = N, 
  mean = bP*prey + bW*wolf + bPW*prey*wolf, 
  sd = 1
)
d <- data.frame(raven, prey, wolf)
str(d)

## 'data.frame':    300 obs. of  3 variables:
##  $ raven: num  1.66 -1.02 1.93 -0.35 0.67 ...
##  $ prey : num  -2.186 -0.429 -2.065 0.6 1.734 ...
##  $ wolf : num  -1.801 0.513 -2.374 0.405 -0.732 ...

#Test that we were successful by estimating the linear model and seeing if the estimates are similar to what we put into the simulation.
m <- map(
  alist(
    raven ~ dnorm(mu, sigma),
    mu <- a + bP*prey + bW*wolf + bPW*prey*wolf,
    a ~ dnorm(0, 1),
    bW ~ dnorm(0, 1),
    bP ~ dnorm(0, 1),
    bPW ~ dnorm(0, 1),
    sigma ~ dunif(0, 5)
  ),
  data = d,
  start = list(a = 0, bP = 0, bW = 0, bPW = 0, sigma = 1)
)
precis(m)

##            mean         sd         5.5%     94.5%
## a     0.0213411 0.06454654 -0.081816746 0.1244989
## bP    0.2760990 0.07512031  0.156042205 0.3961557
## bW    0.1232532 0.07093762  0.009881172 0.2366252
## bPW   0.4461925 0.04635422  0.372109543 0.5202756
## sigma 1.0101361 0.04123821  0.944229492 1.0760427

#The estimates are quite similar to what we put in, including a sizable interaction effect.

8M4. Repeat the tulips analysis, but this time use priors that constrain the effect of water to be positive and the effect of shade to be negative. Use prior predictive simulation. What do these prior assumptions mean for the interaction prior, if anything?

data(tulips)
d <- tulips
str(d)

## 'data.frame':    27 obs. of  4 variables:
##  $ bed   : Factor w/ 3 levels "a","b","c": 1 1 1 1 1 1 1 1 1 2 ...
##  $ water : int  1 1 1 2 2 2 3 3 3 1 ...
##  $ shade : int  1 2 3 1 2 3 1 2 3 1 ...
##  $ blooms: num  0 0 111 183.5 59.2 ...

d$blooms_std <- d$blooms / max(d$blooms)
d$water_cent <- d$water - mean(d$water)
d$shade_cent <- d$shade - mean(d$shade)

a <- rnorm(1e4, 0.5, 1)
sum(a < 0|a > 1) / length(a)

## [1] 0.6188

a <- rnorm(1e4, 0.5, 0.25)
sum(a < 0|a > 1) / length(a)

## [1] 0.0482

m8.4 <- quap(
    alist(
        blooms_std ~ dnorm(mu , sigma),
        mu <- a + bw*water_cent - bs*shade_cent,
        a ~ dnorm(0.5, 0.25),
        bw ~ dnorm(0, 0.25),
        bs ~ dnorm(0, 0.25),
        sigma ~ dexp(1)
) , data=d )

par(mfrow=c(1,3))
for ( s in -1:1 ) {
    idx <- which( d$shade_cent==s )
    plot( d$water_cent[idx] , d$blooms_std[idx], xlim=c(-1,1), ylim=c(0,1),
        xlab="water", ylab="blooms", pch=16, col=rangi2)
    mu <- link(m8.4, data=data.frame( shade_cent=s , water_cent=-1:1))
    for (i in 1:20) lines( -1:1, mu[i,], col=col.alpha("black",0.3))}

#Both water and shades would affect tulip. At high light levels, water does not have much effect as the tulips don’t need that much light to produce tulips. At lower light levels, water has a bigger effect as the tulips need more water to produce blooms.

8H1. Return to the data(tulips) example in the chapter. Now include the bed variable as a predictor in the interaction model. Don’t interact bed with the other predictors; just include it as a main effect. Note that bed is categorical. So to use it properly, you will need to either construct dummy variables or rather an index variable, as explained in Chapter 5.

d <- tulips
d$shade.c <- d$shade - mean(d$shade)
d$water.c <- d$water - mean(d$water)
# Dummy variables
d$bedb <- d$bed == "b"
d$bedc <- d$bed == "c"
# Index variable
d$bedx <- coerce_index(d$bed)

#dummy variables
m_dummy <- map(
  alist(
    blooms ~ dnorm(mu, sigma),
    mu <- a + bW*water.c + bS*shade.c + bWS*water.c*shade.c + bBb*bedb + bBc*bedc,
    a ~ dnorm(130, 100),
    bW ~ dnorm(0, 100),
    bS ~ dnorm(0, 100),
    bWS ~ dnorm(0, 100),
    bBb ~ dnorm(0, 100),
    bBc ~ dnorm(0, 100),
    sigma ~ dunif(0, 100)
  ),
  data = d,
  start = list(a = mean(d$blooms), bW = 0, bS = 0, bWS = 0, bBb = 0, bBc = 0, sigma = sd(d$blooms))
)
precis(m_dummy)

##            mean        sd      5.5%     94.5%
## a      99.36131 12.757521  78.97233 119.75029
## bW     75.12433  9.199747  60.42136  89.82730
## bS    -41.23103  9.198481 -55.93198 -26.53008
## bWS   -52.15060 11.242951 -70.11901 -34.18219
## bBb    42.41139 18.039255  13.58118  71.24160
## bBc    47.03141 18.040136  18.19979  75.86303
## sigma  39.18964  5.337920  30.65862  47.72067

#repeat with index variable
m_index <- map(
  alist(
    blooms ~ dnorm(mu, sigma),
    mu <- a[bedx] + bW*water.c + bS*shade.c + bWS*water.c*shade.c,
    a[bedx] ~ dnorm(130, 100),
    bW ~ dnorm(0, 100),
    bS ~ dnorm(0, 100),
    bWS ~ dnorm(0, 100),
    sigma ~ dunif(0, 200)
  ),
  data = d
)
precis(m_index, depth = 2)

##            mean        sd      5.5%     94.5%
## a[1]   97.68467 12.952811  76.98358 118.38577
## a[2]  142.41151 12.952004 121.71170 163.11131
## a[3]  146.91841 12.952292 126.21815 167.61868
## bW     75.14892  9.198740  60.44756  89.85028
## bS    -41.23507  9.197587 -55.93459 -26.53555
## bWS   -52.12330 11.242001 -70.09019 -34.15641
## sigma  39.18587  5.335335  30.65897  47.71276

#The results are very similar; the main differences are the form of the bed-specific intercepts. 
coeftab(m_dummy, m_index)

##       m_dummy m_index
## a       99.36      NA
## bW      75.12   75.15
## bS     -41.23  -41.24
## bWS    -52.15  -52.12
## bBb     42.41      NA
## bBc     47.03      NA
## sigma   39.19   39.19
## a[1]       NA   97.68
## a[2]       NA  142.41
## a[3]       NA  146.92
## nobs       27      27

Assignment #7

Jiayuan Shi

2020-09-10

Chapter 8 - Conditional Manatees

Questions