The statistical model:

\(y_t = \beta_0 + \beta_1 * (Elevation_s)_t + \beta_2 * Slope_t + (b_s)_t + \epsilon_t\)

Where:

• \(\beta_0\) is the mean response when both Elevation and Slope are 0
• \(\beta_1\) is the change in mean response for a 1-unit change in elevation. Elevation is measured at the stand level, so all plots in a stand share a single value in elevation.
• \(\beta_2\) is the change in mean response for a 1-unit change in slope. Slope is measured at the plot level, so every plot potentially has a unique value of slope.

Let’s define the parameters:

• the intercept \((\beta_0)\) will be -1
• the coefficient for elevation \((\beta_1)\) will be set to 0.005
• the coefficient for slope \((\beta_2)\) will be set to 0.1

nstand = 5
nplot = 4
b0 = -1
b1 = .005
b2 = .1
sds = 2
sd = 1

Simulate other variables:

set.seed(16)
stand = rep(LETTERS[1:nstand], each = nplot)
standeff = rep( rnorm(nstand, 0, sds), each = nplot)
ploteff = rnorm(nstand*nplot, 0, sd)

Simulate elevation and slope:

elevation = rep( runif(nstand, 1000, 1500), each = nplot)
slope = runif(nstand*nplot, 2, 75)

Simulate response variable:

resp2 = b0 + b1*elevation + b2*slope + standeff + ploteff 

Your tasks (complete each task in its’ own code chunk, make sure to use echo=TRUE so I can see your code):

1. Fit a linear mixed model with the response variable as a function of elevation and slope with stand as a random effect. Are the estimated parameters similar to the intial parameters as we defined them?

##Answer The estimated parameter for elevation (0.02060) is higher than initial parameter (0.005); the estimated parameter for slope (0.09511) is slightly lower than initial parameter (0.1)

library(lme4)

## Loading required package: Matrix

library(Matrix)
mymodel = lmer(resp2 ~ 1 + elevation + slope + (1|stand))
mymodel

## Linear mixed model fit by REML ['lmerMod']
## Formula: resp2 ~ 1 + elevation + slope + (1 | stand)
## REML criterion at convergence: 81.9874
## Random effects:
##  Groups   Name        Std.Dev.
##  stand    (Intercept) 1.099   
##  Residual             1.165   
## Number of obs: 20, groups:  stand, 5
## Fixed Effects:
## (Intercept)    elevation        slope  
##   -21.31463      0.02060      0.09511

2. Create a function for your model and run 1000 simulations of that model.

model_sl <- function(n) {
  nstand = n
  nplot = 4
  b0 = -1
  b1 = .005
  b2 = .1
  sds = 2
  sd = 1

  stand = rep(LETTERS[1:nstand], each = nplot)
  standeff = rep(rnorm(nstand, 0, sds), each = nplot)
  ploteff = rnorm(nstand * nplot, 0, sd)
  elevation = rep(runif(nstand, 1000, 1500), each = nplot)
  slope = runif(nstand * nplot, 2, 75)
  resp2 = b0 + b1*elevation + b2*slope + standeff + ploteff 
  res <- lmer(resp2 ~ elevation + slope + (1|stand))
  return(res)
}

model_rp <- replicate(1000, model_sl (10))

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.350468
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00542153
## (tol = 0.002, component 1)

3. Extract the stand and residual variances from this simulation run. Print the first 6 rows of the data.

library(broom)
library(tidyverse)

## Registered S3 methods overwritten by 'ggplot2':
##   method         from 
##   [.quosures     rlang
##   c.quosures     rlang
##   print.quosures rlang

## Registered S3 method overwritten by 'rvest':
##   method            from
##   read_xml.response xml2

## ── Attaching packages ─────────────────────────────────────────────────────────────────────────────────────── tidyverse 1.2.1 ──

## ✔ ggplot2 3.1.1       ✔ purrr   0.3.2  
## ✔ tibble  2.1.1       ✔ dplyr   0.8.0.1
## ✔ tidyr   0.8.3       ✔ stringr 1.4.0  
## ✔ readr   1.3.1       ✔ forcats 0.4.0

## ── Conflicts ────────────────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ tidyr::expand() masks Matrix::expand()
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()

extract <- model_rp %>% map_dfr(tidy, effects = "ran_pars", scales = "vcov")
extract[1:6,]

## # A tibble: 6 x 3
##   term                     group    estimate
##   <chr>                    <chr>       <dbl>
## 1 var_(Intercept).stand    stand       3.55 
## 2 var_Observation.Residual Residual    1.12 
## 3 var_(Intercept).stand    stand       5.11 
## 4 var_Observation.Residual Residual    1.23 
## 5 var_(Intercept).stand    stand       4.36 
## 6 var_Observation.Residual Residual    0.990

4. Choose three different sample sizes (your choice) and run 1000 model simulations with each sample size. Create 3 visualizations that compare distributions of the variances for each of the 3 sample sizes. Make sure that the axes are labelled correctly. What do these graphs say about the relationship between sample size and variance?

##Answer: As sample size goes up, the variance went up.

simulation_1 <- replicate(1000, model_sl(n=5))

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00551362
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00406233
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.0032079
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00561834
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00438273
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00230691
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00280278
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00651067
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00555055
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

simulation_2 <- replicate(1000, model_sl(n=15))

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00846648
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.0129416
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00232099
## (tol = 0.002, component 1)

simulation_3 <- replicate(1000, model_sl(n=20))

extract_1 <- as.data.frame(t(sapply(simulation_1, function(x) as.data.frame(VarCorr(x))[,4])))
extract_2 <- as.data.frame(t(sapply(simulation_2, function(x) as.data.frame(VarCorr(x))[,4])))
extract_3 <- as.data.frame(t(sapply(simulation_3, function(x) as.data.frame(VarCorr(x))[,4])))

mydata <- rbind(extract_1, extract_2, extract_3)

colnames(mydata) <- c("Stand_Variances", "Residual Variances")
mydata$Sample_Size <- c(rep("5", 1000), rep("15", 1000), rep("20", 1000))

library(ggplot2)

ggplot(mydata, aes(x=Stand_Variances)) +
  geom_density() +
  facet_wrap( ~ Sample_Size) +
  geom_vline(xintercept = 4) +
  theme_bw()

5. Plot the coefficients of the estimates of elevation and slope. Hint: the x-axis should have 1000 values. Discuss the graphs.

##Answer: When the simulation times are small, the estimation could differ from actual. However, when the simulation is large enough, the mean of estimation is close to actual.

library(ggplot2)
library(dplyr)
library(purrr)

coefficients <- model_rp %>% 
  map(tidy, effects = "fixed") %>% 
  bind_rows()

coefficients %>% 
  dplyr::filter(term %in% c("elevation")) %>% 
  group_by(term) %>% 
  mutate(x = 1 : 1000) %>%
  ungroup() %>% 
  mutate(value = ifelse(term == "elevation", 0.005)) %>% 
  ggplot(aes(x = x, y = estimate)) +
  geom_line() +
  facet_wrap(~term) +
  geom_hline(aes(yintercept = value, color = term), linetype = 3, size = 0.3)

coefficients %>% 
  dplyr::filter(term %in% c("slope")) %>% 
  group_by(term) %>% 
  mutate(x = 1 : 1000) %>%
  ungroup() %>% 
  mutate(value = ifelse(term == "slope", 0.1)) %>% 
  ggplot(aes(x = x, y = estimate)) +
  geom_line() +
  facet_wrap(~term) +
  geom_hline(aes(yintercept = value, color = term), linetype = 3, size = 0.3)

6. Submit a link to this document in R Pubs to your Moodle. This assignment is worth 25 points.