Assignment #4
Jiang Lan
2020-06-14
5E1. Which of the linear models below are multiple linear regressions? \[\begin{align} {μ_i = α + βx_i} \tag{1}\\ μ_i = β_xx_i + β_zz_i \tag{2} \\ μ_i = β_xx_i + β_zz_i \tag{3} \\ μ_i = α + β(x_i − z_i) \tag{4} \\ μ_i = α + β_xx_i + β_zz_i \tag{5} \\ \end{align}\]
In sum, models 2, 3, and 4 are multiple regressions.Becuase the first one even dones’t have multiple viarables.
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.
In this question, I assuming the latitude is the outcoue, so to keep the two diversity variables together.
animal diversity i ∼ Normal(μi, σ)
μi=α+βAAi+βPPi
In above, A is animal diversity and P is plant diversity.
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.
Following textbook (page 139), the linear model in math form:
μi = α + βFFi + βSSi
In above definition, F is amount of funding and S is size of laboratory.
Given the question, both slope parameters, βf and βs, should be positive.
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. \[\begin{align} μ_i = α + β_AA_i + β_BB_i + β_DD_i \tag{1} \\ μ_i = α + β_AA_i + β_BB_i + β_CC_i + β_DD_i \tag{2} \\ μ_i = α + β_BB_i + β_CC_i + β_DD_i \tag{3} \\ μ_i = α_AA_i + α_BB_i + α_CC_i + α_DD_i \tag{4} \\ μ_i = α_A(1 − B_i − C_i − D_i) + α_BB_i + α_CC_i + α_DD_i \tag{5} \\ \end{align}\]
The option 1,the model includes a single intercept (for category C) and slopes for A, B, and D.
The option 2, the model is a non-identifiable one.
The option 3, the model includes a single intercept (for category A) and slopes for B, C, and D.
The option 4, the model uses the unique index approach.
The option 5, the model uses the reparameterized approach.
Therefor, models 1, 3, 4, and 5 are inferentially equivalent.
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).
N <- 100
income <- rnorm(n = 100, mean = 0, sd = 1)
operas <- rnorm(n = N, mean = income, sd = 2)
scores <- rnorm(n = N, mean = income, sd = 1)
d <- data.frame(scores, operas, income)
pairs(d)
m <- map(
alist(
scores ~ dnorm(mu, sigma),
mu <- a + bo * operas,
a ~ dnorm(0, 5),
bo ~ dnorm(0, 5),
sigma ~ dunif(0, 5)
),
data = d
)
precis(m)## mean sd 5.5% 94.5%
## a 0.1400297 0.15240299 -0.1035397 0.3835991
## bo 0.2582661 0.06070769 0.1612435 0.3552887
## sigma 1.4994918 0.10602990 1.3300355 1.6689481m <- map(
alist(
scores ~ dnorm(mu, sigma),
mu <- a + bo * operas + bi * income,
a ~ dnorm(0, 5),
bo ~ dnorm(0, 5),
bi ~ dnorm(0, 5),
sigma ~ dunif(0, 5)
),
data = d
)
precis(m)## mean sd 5.5% 94.5%
## a 0.227373846 0.10180477 0.06467015 0.39007754
## bo -0.007618675 0.04687858 -0.08253970 0.06730235
## bi 1.136617454 0.10146310 0.97445983 1.29877508
## sigma 0.998459083 0.07060136 0.88562447 1.111293695M2. 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.
N <- 100
rho <- 0.6
alcohol <- rnorm(n = N, mean = 0, sd = 1)
illness <- rnorm(n = N, mean = rho * alcohol, sd = sqrt(1 - rho^2))
happiness <- rnorm(n = N, mean = alcohol - illness, sd = 1)
d <- data.frame(happiness, alcohol, illness)
pairs(d)
The bivariate associations should be weak/widely variable
m <- map(
alist(
happiness ~ dnorm(mu, sigma),
mu <- a + ba * alcohol,
a ~ dnorm(0, 5),
ba ~ dnorm(0, 5),
sigma ~ dunif(0, 5)
),
data = d
)
precis(m)## mean sd 5.5% 94.5%
## a -0.02043711 0.12293045 -0.21690371 0.1760295
## ba 0.27782668 0.12563380 0.07703961 0.4786138
## sigma 1.21678232 0.08603923 1.07927500 1.3542896m <- map(
alist(
happiness ~ dnorm(mu, sigma),
mu <- a + bi * illness,
a ~ dnorm(0, 5),
bi ~ dnorm(0, 5),
sigma ~ dunif(0, 5)
),
data = d
)
precis(m)## mean sd 5.5% 94.5%
## a -0.06926054 0.11939288 -0.2600734 0.1215524
## bi -0.38841483 0.12975360 -0.5957861 -0.1810435
## sigma 1.19383425 0.08441706 1.0589195 1.3287490m <- map(
alist(
happiness ~ dnorm(mu, sigma),
mu <- a + ba * alcohol + bi * illness,
a ~ dnorm(0, 5),
ba ~ dnorm(0, 5),
bi ~ dnorm(0, 5),
sigma ~ dunif(0, 5)
),
data = d
)
precis(m)## mean sd 5.5% 94.5%
## a 0.04524784 0.10007387 -0.1146895 0.2051852
## ba 0.92370056 0.13556198 0.7070463 1.1403548
## bi -1.03012554 0.14269810 -1.2581847 -0.8020664
## sigma 0.98633786 0.06974485 0.8748721 1.09780365M3. 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?
In the question of divorce and marriage data: How might a high divorce rate cause a higher marriage rate?
A high divorce rate could suggest that there may be also a higher number of people that can marry again consequently, so raising the marriage rate. One way probably could use to test this relationship would be to include the remarriage rate in a multivariate regression.
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.
data("WaffleDivorce")
d <- WaffleDivorce
d$LDS <- c(0.0077, 0.0453, 0.0610, 0.0104, 0.0194, 0.0270, 0.0044, 0.0057, 0.0041, 0.0075, 0.0082, 0.0520, 0.2623, 0.0045, 0.0067, 0.0090, 0.0130, 0.0079, 0.0064, 0.0082, 0.0072, 0.0040, 0.0045, 0.0059, 0.0073, 0.0116, 0.0480, 0.0130, 0.0065, 0.0037, 0.0333, 0.0041, 0.0084, 0.0149, 0.0053, 0.0122, 0.0372, 0.0040, 0.0039, 0.0081, 0.0122, 0.0076, 0.0125, 0.6739, 0.0074, 0.0113, 0.0390, 0.0093, 0.0046, 0.1161)
d$logLDS <- log(d$LDS)
d$logLDS.s <- (d$logLDS - mean(d$logLDS)) / sd(d$logLDS)
simplehist(d$LDS)
hist(d$LDS)
d$log.LDS <- log(d$LDS)
d$log.LDS.s <- ( d$log.LDS - mean(d$log.LDS) ) / sd(d$log.LDS)
d$LDS.s <- (d$LDS - mean(d$LDS)) / sd(d$LDS)
hist(d$log.LDS.s)
Also we could standardize the other variables:
d$MedianAgeMarriage.s <- (d$MedianAgeMarriage - mean(d$MedianAgeMarriage)) / sd(d$MedianAgeMarriage)
d$Marriage.s <- (d$Marriage - mean(d$Marriage)) / sd(d$Marriage)Then the model:
mod4 <- map(
alist(
Divorce ~ dnorm( mu, sigma),
mu <- a + bMA*MedianAgeMarriage.s + bMR*Marriage.s + bL*log.LDS.s,
a ~ dnorm( 0, 10 ),
c(bMA, bMR, bL) ~ dnorm( 0, 1),
sigma ~ dunif(0, 10)
), data=d
)
precis( mod4 )## mean sd 5.5% 94.5%
## a 9.6843192 0.1949168 9.3728045 9.9958338
## bMA -1.2938299 0.2799905 -1.7413088 -0.8463509
## bMR 0.1291169 0.3008025 -0.3516235 0.6098574
## bL -0.5433776 0.2761815 -0.9847689 -0.1019862
## sigma 1.3785216 0.1383412 1.1574257 1.5996176plot(precis( mod4))
mu <- link(mod4)
mu.mean <- apply(mu, 2, mean)
mu.PI <- apply(mu, 2, PI)
divorce.sim <- sim( mod4, n=1e4 )
divorce.PI <- apply(divorce.sim, 2, PI)
# residual plot showing the mean prediction error
# compute residuals
divorce.resid <- d$Divorce - mu.mean
# get ordering by divorce rate
o <- order(divorce.resid)
# make the plot
dotchart( divorce.resid[o], labels=d$Loc[o], xlim=c(-6,5), cex=0.6 )
abline(v=0, col=col.alpha("black", 0.2))
for (i in 1:nrow(d) ) {
j <- o[i] # which State in order
lines( d$Divorce[j] - c(mu.PI[1, j], mu.PI[2, j]), rep(i,2) )
points( d$Divorce[j] - c(divorce.PI[1,j], divorce.PI[2, j]), rep(i,2) ,
pch=3, cex=0.6, col="gray")
}
divorce.resid[d$Loc == "ID"]## [1] -3.5927055M5. 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 solve these two mechanisms, we needs some variable to capture them.
μi = α + βGGi + βEEi + βRRi
in above, G represents the gasoline price, E represents one exercise-related variable, and R represents one restaurant-related variable.