Chapter 8 - Conditional Manatees

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. Make sure to include plots if the question requests them.

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.

Questions

8-1. 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?

#In the tulips example temperature is the fundamental factor and shade and water factors depend on temperature. We could to have either a three way association(water, temp and shade) or two way (water-shade, water- Temp, shade- temp).Hence When temperature is too hot it will not help flowers to bloom.

8-2. Can you invent a regression equation that would make the bloom size zero, whenever the temperature is hot?

##Bloom_Size = \(\alpha\) + (\(\beta\)_Water * Water) + (\(\beta\)_Shade * Shade) + (\(\beta\)_Temp * Temp) + (\(\beta\)_WS * Water * Shade) + (\(\beta\)_WT * Water * Temp) + (\(\beta\)_ST * Shade * Temp) + (??_WST * Water * Shade * Temp)

##Temperature should be a binary variable, with 0 being hot and 1 being cold. Water and shade are both binary variables, with the presence of water equaling 1 and the presence of shade equaling 1.

##when temperature is hot

##Bloom_size = \(\alpha\) + (\(\beta\)_Water * Water) + (\(\beta\)_Shade * Shade) + (\(\beta\)_ws * Water * Shade)

##When water and shade is present

##Bloom_Size = \(\alpha\) + \(\beta\)_Water + \(\beta\)_Shade + \(\beta\)_WS

##Now, in order for Bloom Size to be 0, the water*shade interaction coefficient must be 0:

##\(\beta\)_ws = -(\(\alpha\) + \(\beta\)_water + \(\beta\)_shade)

##As a result, in hot temperatures, the bloom size will be zero if the preceding equation holds true.

8-3. 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 and visualize. What do these prior assumptions mean for the interaction prior, if anything?

library(rethinking)

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] 0.6091
a <- rnorm( 1e4 , 0.5 , 0.25 ); 
sum( a < 0 | a > 1 ) / length( a )
## [1] 0.0462
model1 <- 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(model1, depth=2)
##            mean         sd       5.5%     94.5%
## a     0.3587649 0.03021852 0.31046986 0.4070599
## bw    0.2050348 0.03688909 0.14607888 0.2639907
## bs    0.1125309 0.03687523 0.05359714 0.1714646
## sigma 0.1581523 0.02144307 0.12388217 0.1924225
par(mfrow=c(1,3)) # 3 plots in 1 row
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( model1 , 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) )
}

8-4. 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.

library(rethinking)
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)

# Dummy variables
d$bedb <- d$bed == "b"
d$bedc <- d$bed == "c"

# Index variable
d$bedi <- coerce_index(d$bed)

#Using dummy variables
model2 <- map(
  alist(
    blooms ~ dnorm(mu, sigma),
    mu <- a + (bW * water_cent) + (bS * shade_cent) + (bWS * water_cent * shade_cent) + (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(model2)
##            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
model3 <- map(
  alist(
    blooms ~ dnorm(mu, sigma),
    mu <- a + (bW * water_cent) + (bS * shade_cent) + (bWS * water_cent * shade_cent),
    a ~ dnorm(130, 100),
    bW ~ dnorm(0, 100),
    bS ~ dnorm(0, 100),
    bWS ~ dnorm(0, 100),
    sigma ~ dunif(0, 100)
  ),
  data = d,
  start = list(a = mean(d$blooms), bW = 0, bS = 0, bWS = 0, sigma = sd(d$blooms))
)

precis(model3, depth=2)
##            mean        sd      5.5%     94.5%
## a     129.00797  8.670771 115.15041 142.86554
## bW     74.95946 10.601997  58.01542  91.90350
## bS    -41.14054 10.600309 -58.08188 -24.19920
## bWS   -51.87265 12.948117 -72.56625 -31.17906
## sigma  45.22497  6.152982  35.39132  55.05863

8-5. Use WAIC to compare the model from 8-4 to a model that omits bed. What do you infer from this comparison? Can you reconcile the WAIC results with the posterior distribution of the bed coefficients?

set.seed(11)
compare( model2, model3, func=WAIC )
##            WAIC       SE     dWAIC      dSE    pWAIC    weight
## model2 295.1020 10.18953 0.0000000       NA 9.958808 0.5957375
## model3 295.8774 10.47512 0.7754712 8.034999 6.541469 0.4042625
post <- extract.samples(model2)
post.a <- post$a
post.b <- post$a + post$bBb
post.c <- post$a + post$bBc
dens(post.a, col = "Red", xlim = c(50, 200), ylim = c(0, 0.035))
dens(post.b, col = "dark blue", add = TRUE)
dens(post.c, col = "Yellow", add = TRUE)

# Model 2 seem to show better performance when compared to model3 since the WAIC value is better. But the difference between the WAIC values of both models is very less hence we can say that there is no much effect of bed when compared to temp and shade.