Assignment #4

Questions

5E1. Which of the linear models below are multiple linear regressions? \[\begin{align} {μ_i = α + βx_i} \tag{1}\\ μ_i = β_xx_i + β_zz_i \tag{2} \\ μ_i = α + β(x_i − z_i) \tag{3} \\ μ_i = α + β_xx_i + β_zz_i \tag{4} \\ \end{align}\]

# 2, 3, 4  are multiple linear regressions

5E2. Write down a multiple regression to evaluate the claim: Animal diversity is linearly related to latitude, but only after controlling for plant diversity. You just need to write down the model definition.

#Model Definition: Here x is animal diversity, y is plant diversity and μ is the latitude
#μ_i = α+ β_x * x_i+ β_y * y_i

5E3. Write down a multiple regression to evaluate the claim: Neither amount of funding nor size of laboratory is by itself a good predictor of time to PhD degree; but together these variables are both positively associated with time to degree. Write down the model definition and indicate which side of zero each slope parameter should be on.

#Model Definition: Here fund is funding and lab_size is size of laboratory. In this definition β_f and β_s slope parameters should be on right side of zero, positive side
#μ_i = α +β_fund * fund_i + β_lab_size * lab_size_i

5E4. Suppose you have a single categorical predictor with 4 levels (unique values), labeled A, B, C and D. Let Ai be an indicator variable that is 1 where case i is in category A. Also suppose Bi, Ci, and Di for the other categories. Now which of the following linear models are inferentially equivalent ways to include the categorical variable in a regression? Models are inferentially equivalent when it’s possible to compute one posterior distribution from the posterior distribution of another model. \[\begin{align} μ_i = α + β_AA_i + β_BB_i + β_DD_i \tag{1} \\ μ_i = α + β_AA_i + β_BB_i + β_CC_i + β_DD_i \tag{2} \\ μ_i = α + β_BB_i + β_CC_i + β_DD_i \tag{3} \\ μ_i = α_AA_i + α_BB_i + α_CC_i + α_DD_i \tag{4} \\ μ_i = α_A(1 − B_i − C_i − D_i) + α_BB_i + α_CC_i + α_DD_i \tag{5} \\ \end{align}\]

#Linear models 1, 3, 4 and 5 are equivalent inferentially

5M1. Invent your own example of a spurious correlation. An outcome variable should be correlated with both predictor variables. But when both predictors are entered in the same model, the correlation between the outcome and one of the predictors should mostly vanish (or at least be greatly reduced).

#For a correlation between exercise  results in better sleep after some healthy diet. 
n = 10000
exercise <- rnorm(n, mean = 0, sd = 1)
diet <- rnorm(n, mean = exercise, sd = 2)
sleep <- rnorm(n, mean = exercise, sd = 1)
df <- data.frame(exercise, diet, sleep)
pairs(df)

#l1 <-map(
 # alist(
  #  sleep ~ dnorm(mu, sigma),
   # mu<-e1+p1*phy,
    #e1 ~ dnorm(0, 5),
    #p1 ~ dnorm(0,2),
    #sigma ~ dunif(0,5)
#  ),data=df)
#precis(l1)

5M2. Invent your own example of a masked relationship. An outcome variable should be correlated with both predictor variables, but in opposite directions. And the two predictor variables should be correlated with one another.

# I use the approach from page 141 to invent an example of a  relationship involving the prediction of  housing prices from the location at sea  and the amount of crime in the area.
N <- 100
rho <- 0.6
crime <- rnorm(n = N, mean = 0, sd = 1)
sea_location <- rnorm(n = N, mean = rho * crime, sd = sqrt(1 - rho^2))
prices <- rnorm(n = N, mean = sea_location - crime, sd = 1)
df <- data.frame(prices, crime, sea_location)
pairs(df)

#m <- map(
 # alist(
  #  prices ~ dnorm(mu, sigma),
   # mu <- a + ba * crime,
    #a ~ dnorm(0, 5),
    #ba ~ dnorm(0, 5),
    #sigma ~ dunif(0, 5)
  #),
  #data = df
#)
#precis(m)

5M3. It is sometimes observed that the best predictor of fire risk is the presence of firefighters— States and localities with many firefighters also have more fires. Presumably firefighters do not cause fires. Nevertheless, this is not a spurious correlation. Instead fires cause firefighters. Consider the same reversal of causal inference in the context of the divorce and marriage data. How might a high divorce rate cause a higher marriage rate? Can you think of a way to evaluate this relationship, using multiple regression?

#To hypothesize that a high divorce rate causes a higher marriage rate, we can introduce more unmarried individuals. 
# The possibility can be evaluated by using multiple regression by regressing marriage rate on both divorce rate and re-marriage rate. For instance, the rate of non-first marriages or marriages following divorces. If the divorce rate no longer predicts marriage rate even when the re-marriage rate is known, it can support the hypothesis.

5M4. In the divorce data, States with high numbers of members of the Church of Jesus Christ of Latter-day Saints (LDS) have much lower divorce rates than the regression models expected. Find a list of LDS population by State and use those numbers as a predictor variable, predicting divorce rate using marriage rate, median age at marriage, and percent LDS population (possibly standardized). You may want to consider transformations of the raw percent LDS variable.

# LDS population by State from Wikipedia and add it as a new column in the WaffleDivorce data frame (minus Nevada as it is apparently missing from WaffleDivorce). Given the positive skew of these percentages, I will log-transform them prior to standardization.
#data("WaffleDivorce")
#set.seed(5)

#df <- WaffleDivorce
#df$LDS <- c(0.0077, 0.0453, 0.0610, 0.0104, 0.0194, 0.0270, 0.0044, 0.0057, 0.0041, 0.0075, 0.0082, 0.0520, 0.2623, 0.0045, 0.0067, 0.0090, 0.0130, 0.0079, 0.0064, 0.0082, 0.0072, 0.0040, 0.0045, 0.0059, 0.0073, 0.0116, 0.0480, 0.0130, 0.0065, 0.0037, 0.0333, 0.0041, 0.0084, 0.0149, 0.0053, 0.0122, 0.0372, 0.0040, 0.0039, 0.0081, 0.0122, 0.0076, 0.0125, 0.6739, 0.0074, 0.0113, 0.0390, 0.0093, 0.0046, 0.1161)
#df$logLDS <- log(df$LDS)
#df$logLDS.s <- (df$logLDS - mean(df$logLDS)) / sd(df$logLDS)
#simplehist(df$LDS)
#simplehist(df$logLDS)
#simplehist(df$logLDS.s)
#m <- map(
 # alist(
  #  Divorce ~ dnorm(mu, sigma),
   # mu <- a + bm * Marriage + ba * MedianAgeMarriage + bl * logLDS.s,
    #a ~ dnorm(10, 20),
  #  bm ~ dnorm(0, 10),
   # ba ~ dnorm(0, 10),
    #bl ~ dnorm(0, 10),
    #sigma ~ dunif(0, 5)
  #),
  #data = df
#)
#precis(m)

# So the slopes of median age at marriage and percentage of LDS population were negative which means with older median age at marriage or higher percentages of Mormons had lower divorce rates.

5M5. One way to reason through multiple causation hypotheses is to imagine detailed mechanisms through which predictor variables may influence outcomes. For example, it is sometimes argued that the price of gasoline (predictor variable) is positively associated with lower obesity rates (outcome variable). However, there are at least two important mechanisms by which the price of gas could reduce obesity. First, it could lead to less driving and therefore more exercise. Second, it could lead to less driving, which leads to less eating out, which leads to less consumption of huge restaurant meals. Can you outline one or more multiple regressions that address these two mechanisms? Assume you can have any predictor data you need.

# To address the two mechanisms, variables that will capture them are needed. 
# For the first one, a variable corresponding to time spent exercising would be a reasonable start. For the second one, a variable corresponding to frequency of eating out at restaurant would be a reasonable start. So, I would propose the following multiple regression model:
# μ_i = α + β_G G_i + β_E E_i + β_R R_i
# where G represents the price of gasoline, E represents one exercise-related variable, and R represents one restaurant-related variable.

Assignment #4

Kshitij Deshpande

2021-02-23

Chapter 5 - Many Variables and Spurious Waffles

Questions