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.

#1. Time. Yeast needs time to reproduce and exert effect on the rise of bread dough.
#2. Industry. Different industries have different income levels.
#3. The type of gasoline. The higher the quality of gasoline, the better for the car and therefore better performance.

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.).

#1. It invokes and interaction between heat and dryness.
#2. No specific interaction is invoked.
#3. The interaction between people and their parents/friends and the effect that parents/friends have on one's political beliefs is an interaction.
#4. There is some sort of relationship between the intelligence and the social/manipulative appendages of an animal specie. The intelligent animal species, however, are not both highly social and have manipulative appendages.

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

#yi = βH * Hi + βD * Di + βHD * Hi * Di

#yi = βC * Ci + βF * Fi

#yi = βP * Pi + βF * Pi * Fi

#yi = 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?

#The result with hot temperature did not develop any blooms because influence of hot temperature on blooms was dependent upon on the interaction between temperature, water and shade or the interaction between temperature and water or water and shade or temperature and shade. Either of these interactions could have resulted in plants not blooming at all under the hot temperature.

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

#μi = α + βt*xt + βw*xw + βs*xs + βtws*xt*xw*xs + βtw*xt*xw + βws*xw*xs + βts*xt*xs
#μi is the bloom size, for μi = 0 when xt is hot, depends on other variables

#After invention, the regression equation that would make the bloom size zero, whenever the temperature is hot
# α + βt + βw*xw + βs*xs + βtws*xw*xs + βtw*xw + βws*xw*xs + βts*xs = 0

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?

data(tulips)
data(rugged)
data(nettle)

set.seed(123)
# We consider a linear model
N <- 500 # simulation size
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
# 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':    500 obs. of  3 variables:
##  $ raven: num  -1.017 -1.095 1.994 -0.024 -2.697 ...
##  $ prey : num  -0.5605 -0.2302 1.5587 0.0705 0.1293 ...
##  $ wolf : num  -0.818 -0.933 1.757 0.643 -1.13 ...

m <- quap(
  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     0.03740266 0.04855042 -0.04019029 0.1149956
## bP    0.32213161 0.05466076  0.23477316 0.4094901
## bW    0.16898993 0.05480987  0.08139316 0.2565867
## bPW   0.47507099 0.03840570  0.41369126 0.5364507
## sigma 0.98839478 0.03126046  0.93843452 1.0383550

plot(m)

#Conclusion: It seems the biological interaction could 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

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

## [1] 0.0457

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 )
precis(m8.4)

##            mean         sd       5.5%     94.5%
## a     0.3587758 0.03021898 0.31048005 0.4070716
## bw    0.2050361 0.03688957 0.14607948 0.2639928
## bs    0.1125399 0.03687568 0.05360543 0.1714744
## sigma 0.1581545 0.02144380 0.12388316 0.1924258

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))}

#Conclusion: Both water and shades would affect tulip. Water has more effect when there's much light, as the tulips need more water to do photosynthesis. Water has a smaller effect when there's not enough light, as the tulips are not as active and would not need that much water.

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.

data(tulips)
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  47.72067

Assignment #7

Shidong Li

2021-01-23

Chapter 8 - Conditional Manatees

Questions