Intro to STAN Homework Part #1

After our Intro to Stan lecture I think it would be valuable to have you go through a similar exercise. Let’s test a second research question.

Research question: Is sea ice extent declining in the Southern Hemisphere over time? Is the same pattern happening in the Antarctic as in the Arctic? Fit a Stan model to find out!

Make sure you follow the steps we used in class.

What do your Stan model results indicate so far?

1. Load and Inspect Data

#place the code here
seaice <- read.csv("C:/Users/sharo/Desktop/seaice.csv", stringsAsFactors=F)
colnames(seaice) <- c("year", "extent_north", "extent_south")
head(seaice)
##   year extent_north extent_south
## 1 1979       12.328       11.700
## 2 1980       12.337       11.230
## 3 1981       12.127       11.435
## 4 1982       12.447       11.640
## 5 1983       12.332       11.389
## 6 1984       11.910       11.454

2. Plot the data

#plot data
plot(extent_south ~ year, pch = 20, data = seaice)

3. Run a general linear model using lm()

#write the code
lm1 <- lm(extent_south ~ year, data = seaice)
summary(lm1)
## 
## Call:
## lm(formula = extent_south ~ year, data = seaice)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.23372 -0.18142  0.01587  0.18465  0.88814 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)  
## (Intercept) -14.199551  10.925576  -1.300   0.2018  
## year          0.012953   0.005468   2.369   0.0232 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.3843 on 37 degrees of freedom
## Multiple R-squared:  0.1317, Adjusted R-squared:  0.1082 
## F-statistic: 5.611 on 1 and 37 DF,  p-value: 0.02318

4. Index the data, re-run the lm(), extract summary statistics and turn the indexed data into a dataframe to pass into Stan

#write the code here
x <- I(seaice$year - 1978)
y <- seaice$extent_south
N <- length(seaice$year)

lm1 <- lm(y ~ x)
summary(lm1)
## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.23372 -0.18142  0.01587  0.18465  0.88814 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.421555   0.125490  91.015   <2e-16 ***
## x            0.012953   0.005468   2.369   0.0232 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.3843 on 37 degrees of freedom
## Multiple R-squared:  0.1317, Adjusted R-squared:  0.1082 
## F-statistic: 5.611 on 1 and 37 DF,  p-value: 0.02318
lm_alpha <- summary(lm1)$coeff[1]; lm_alpha
## [1] 11.42155
lm_beta <- summary(lm1)$coeff[2]; lm_beta 
## [1] 0.01295304
lm_sigma <- sigma(lm1); lm_sigma
## [1] 0.384331

5. Write the Stan model

#write the code
stan_data <- list(N = N, x = x, y = y)
 write("// Stan model for simple linear regression

data {
 int < lower = 1 > N; // Sample size
 vector[N] x; // Predictor
 vector[N] y; // Outcome
}

parameters {
 real alpha; // Intercept
 real beta; // Slope (regression coefficients)
 real < lower = 0 > sigma; // Error SD
}

model {
 y ~ normal(alpha + x * beta , sigma);
}

generated quantities {
} // The posterior predictive distribution",

"stan_model1.stan")

#Let's save that file path
stan_model1 <- "stan_model1.stan"

6. Run the Stan model and inspect the results

#code here
fit <- stan(file = stan_model1, data = stan_data, warmup = 500, iter = 1000, chains = 4, cores = 2, thin = 1)
fit
## Inference for Stan model: stan_model1.
## 4 chains, each with iter=1000; warmup=500; thin=1; 
## post-warmup draws per chain=500, total post-warmup draws=2000.
## 
##        mean se_mean   sd  2.5%   25%   50%   75% 97.5% n_eff Rhat
## alpha 11.42    0.00 0.13 11.14 11.33 11.42 11.50 11.68   760    1
## beta   0.01    0.00 0.01  0.00  0.01  0.01  0.02  0.02   786    1
## sigma  0.40    0.00 0.05  0.32  0.37  0.39  0.43  0.50   808    1
## lp__  16.27    0.05 1.24 13.04 15.73 16.56 17.20 17.71   757    1
## 
## Samples were drawn using NUTS(diag_e) at Tue Mar 24 19:34:14 2020.
## For each parameter, n_eff is a crude measure of effective sample size,
## and Rhat is the potential scale reduction factor on split chains (at 
## convergence, Rhat=1).

7. Extract the posterior estimates into a list so we can plot them

#code here
posterior <- extract(fit)
str(posterior)
## List of 4
##  $ alpha: num [1:2000(1d)] 11.4 11.7 11.4 11.2 11.4 ...
##   ..- attr(*, "dimnames")=List of 1
##   .. ..$ iterations: NULL
##  $ beta : num [1:2000(1d)] 0.01072 0.00668 0.01405 0.01875 0.01576 ...
##   ..- attr(*, "dimnames")=List of 1
##   .. ..$ iterations: NULL
##  $ sigma: num [1:2000(1d)] 0.458 0.468 0.403 0.384 0.424 ...
##   ..- attr(*, "dimnames")=List of 1
##   .. ..$ iterations: NULL
##  $ lp__ : num [1:2000(1d)] 16.4 14.7 17.7 16.1 17.3 ...
##   ..- attr(*, "dimnames")=List of 1
##   .. ..$ iterations: NULL

8. Compare your results to our results to “lm”

#code here
plot(y ~ x, pch = 20, main="Comaring the resut of Linear and Stan Model")
abline(lm1, col = "green", pch=22, lty = 2, lw = 3)
abline( mean(posterior$alpha), mean(posterior$beta), col = "yellow", lw = 1)
legend("topleft",c("Linear Model","Stan Model"),fill=c("green","yellow"))

#The results are identical to our “lm” model.

9. Plot multiple estimates from the posterior

#code here
plot(y ~ x, pch = 20, main="Comaring the result of Linear and 'All' Stan Models")

for (i in 1:500) {
 abline(posterior$alpha[i], posterior$beta[i], col = "blue", lty = 1)
}

abline(lm1, col = "green", pch=22, lty = 2, lw = 3)
abline( mean(posterior$alpha), mean(posterior$beta), col = "yellow", lw = 1)
legend("topleft",c("Linear Model","Mean Stan Model", "All stan Models"),fill=c("green","yellow","blue"))