This chapter introduced interactions, which allow for the association between a predictor and an outcome to depend upon the value of another predictor. While you can’t see them in a DAG, interactions can be important for making accurate inferences. Interactions can be difficult to interpret, and so the chapter also introduced triptych plots that help in visualizing the effect of an interaction. No new coding skills were introduced, but the statistical models considered were among the most complicated so far in the book.
Place each answer inside the code chunk (grey box). The code chunks should contain a text response or a code that completes/answers the question or activity requested. Problems are labeled Easy (E), Medium (M), and Hard(H).
Finally, upon completion, name your final output .html file as: YourName_ANLY505-Year-Semester.html and publish the assignment to your R Pubs account and submit the link to Canvas. Each question is worth 5 points.
8E1. For each of the causal relationships below, name a hypothetical third variable that would lead to an interaction effect:
#1. Temperature
#2. Family's income
#3. Type of engine8E2. Which of the following explanations invokes an interaction?
#1 heat and dryness
#3 parents and friends
#4 sociality abd manipulative appendage8E3. For each of the explanations in 8E2, write a linear model that expresses the stated relationship.
#μi=βHHi+βDDi+βHDHiDi
#μi=βCCi+βQQi
#μi=βTPTiPi+βTFTiFi
#μi=βSSi+βAAi+βSASiAi8M1. 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?
#Between water, shage and temperature, there would be a single three-way interaction - WST and three two-way interactions - WS, WT, ST.8M2. Can you invent a regression equation that would make the bloom size zero, whenever the temperature is hot?
#μi=α+βWWi+βSSi−αTi–βWSWiSiTi+βWSWiSi–βWWiTi–βSSiTi
#when Ti = 0, α+βWWi+βSSi equals to -βWSWiSi8M3. 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?
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':
##
## rstudentN <- 100
rPW <- 0.6 #correlation 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
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': 100 obs. of 3 variables:
## $ raven: num -2.4291 -1.6702 0.4644 -0.0201 1.355 ...
## $ prey : num -0.9395 0.5966 0.4804 -1.3395 -0.0869 ...
## $ wolf : num -1.143 -0.412 -0.916 -1.871 -0.316 ...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 -0.07258048 0.11095773 -0.24991236 0.1047514
## bP 0.08493447 0.10973275 -0.09043967 0.2603086
## bW 0.25385246 0.12535378 0.05351290 0.4541920
## bPW 0.37724968 0.08983843 0.23367052 0.5208288
## sigma 1.01228688 0.07158151 0.89788581 1.1266880#biological interaction can be linear. 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$shade)
bw_d<- abs(rnorm(nrow(d),0,0.25))
bs_d<- (-abs(rnorm(nrow(d),0,0.25)))
m_tulip <- quap(
alist(
blooms_std ~ dnorm( mu , sigma ) ,
mu <- a + bw*water_cent + bs*shade_cent + bws*water_cent*shade_cent,
a ~ dnorm( 0.5, 0.25) ,
bw ~ dnorm(bw_d) ,
bs ~ dnorm(bs_d) ,
bws ~ dnorm( 0 , 0.25 ),
sigma ~ dexp( 1 )
) ,
data=d )
precis(m_tulip)## mean sd 5.5% 94.5%
## a 0.3579803 0.02391749 0.31975550 0.39620504
## bw 0.2090538 0.02908693 0.16256724 0.25554030
## bs -0.1181241 0.02909813 -0.16462854 -0.07161969
## bws -0.1431615 0.03567770 -0.20018138 -0.08614168
## sigma 0.1248380 0.01693786 0.09776798 0.151907938H1. Return to the data(tulips) example in the chapter. Now include the bed variable as a predictor in the interaction model. Don’t interact bed with the other predictors; just include it as a main effect. Note that bed is categorical. So to use it properly, you will need to either construct dummy variables or rather an index variable, as explained in Chapter 5.
d <- tulips
d$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$bed)
m_dummy <- map(
alist(
blooms ~ dnorm(mu, sigma),
mu <- a + bW*water.c + bS*shade.c + bWS*water.c*shade.c + bBb*bedb + bBc*bedc,
a ~ dnorm(130, 100),
bW ~ dnorm(0, 100),
bS ~ dnorm(0, 100),
bWS ~ dnorm(0, 100),
bBb ~ dnorm(0, 100),
bBc ~ dnorm(0, 100),
sigma ~ dunif(0, 100)
),
data = d,
start = list(a = mean(d$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 47.72067m_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] 97.73527 12.953863 77.03250 118.43805
## a[2] 142.43417 12.952872 121.73298 163.13536
## a[3] 146.98007 12.953061 126.27857 167.68156
## bW 75.19309 9.199152 60.49107 89.89512
## bS -41.19366 9.198345 -55.89439 -26.49293
## bWS -52.18859 11.242481 -70.15625 -34.22094
## sigma 39.18857 5.336212 30.66027 47.71687