## 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?

# I'm not able to conclude if the temperature has direct impact or positive impact on blooms. The hot temperature does not develop any blooms because 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.

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

# The regression equation would be: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
# α + βt + βw*xw + βs*xs + βtws*xw*xs + βtw*xw + βws*xw*xs + βts*xs = 0is the equation with bloom size = 0 and temperature is hot. 

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?

data("tulips")
d <- data()

d$blooms_std <- d$blooms / max(d$blooms) d$water_cent <- d$water - mean(d$water)
d$shade_cent <- d$shade - mean(d$shade) #bw_d<- abs(rnorm(nrow(d),0,0.25)) #bs_d<- (-abs(rnorm(nrow(d),0,0.25))) #a <- rnorm(1e4, 0.5, 0.25);sum(a<0|a>1)/length(a) ## [1] 0.0457 #m8.4 <- quap(alist(blooms_std ~ dnorm(mu , sigma), mu <- a + bw*water_cent - bs*shade_cent, # a ~ dnorm(0.5, 0.25), bw ~ dnorm(0, 0.25), bs ~ dnorm(0, 0.25), #sigma ~ dexp(1)) , data=d ) #precis(m8.4) ## mean sd 5.5% 94.5% ## a 0.3587758 0.03021898 0.31048005 0.4070716 ## bw 0.2050361 0.03688957 0.14607948 0.2639928 ## bs 0.1125399 0.03687568 0.05360543 0.1714744 ## sigma 0.1581545 0.02144380 0.12388316 0.1924258 # From the result, I can conclude that both water and shades affect tulip. However, water has more impact with more light because tulips require more water to convert light energy into chemical energy.  8-4. Return to the data(tulips) example in the chapter. Now include the bed variable as a predictor in the interaction model. Don’t interact bed with the other predictors; just include it as a main effect. Note that bed is categorical. So to use it properly, you will need to either construct dummy variables or rather an index variable, as explained in Chapter 5. #data("tulips") #d <- data #d$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$blooms))) #precis(m_dummy) ## mean sd 5.5% 94.5% ## a 99.36131 12.757521 78.97233 119.75029 ## bW 75.12433 9.199747 60.42136 89.82730 ## bS -41.23103 9.198481 -55.93198 -26.53008 ## bWS -52.15060 11.242951 -70.11901 -34.18219 ## bBb 42.41139 18.039255 13.58118 71.24160 ## bBc 47.03141 18.040136 18.19979 75.86303 ## sigma 39.18964 5.337920 30.65862 47.72067 8-5. Use WAIC to compare the model from 8-4 to a model that omits bed. What do you infer from this comparison? Can you reconcile the WAIC results with the posterior distribution of the bedcoefficients? #m_omit <- map( # alist( # blooms ~ dnorm(mu, sigma), # mu <- a + bW*water.c + bS*shade.c + bWS*water.c*shade.c, # a ~ dnorm(130, 100), # bW ~ dnorm(0, 100), # bS ~ dnorm(0, 100), # bWS ~ dnorm(0, 100), # sigma ~ dunif(0, 100) # ), # data = d, # start = list(a = mean(d$blooms), bW = 0, bS = 0, bWS = 0, sigma = sd(d\$blooms))
#)
#precis(m_omit)

#            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

## comparing now using WAIC: compare(m_dummy, m_omit)
##             WAIC       SE    dWAIC      dSE     pWAIC    weight
## m_dummy 296.0376 10.69663 0.000000       NA 10.447507 0.6978782
## m_omit  297.7120 10.83022 1.674429 7.886955  7.361526 0.3021218

#The model has dummy variables that omitted the bed variable. Based on this, there were variability in bloom size between the beds.