linear_model_3_exercise.Rmd

---
title: "Exercises"
output:
  html_document:
    toc: no
    code_folding: hide
---


```{r setup, echo=FALSE, purl = FALSE}
knitr::opts_chunk$set(echo=TRUE, message = FALSE, warning = FALSE, eval = FALSE, cache = FALSE)
SOLUTIONS <- FALSE
```

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## Linear model with an interaction between a continuous and a categorical variable

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This exercise builds on the previous exercises where you fitted a linear model with either a single continuous explanatory variable or a single categorical explanatory variable. In this exercise you will fit a model with both a continuous and categorical explanatory variable and allow their effects to interact (i.e. the effect of one explanatory variable on the response variable changes with the value of the another explanatory variable). This is the third of four complementary exercises, based on the `loyn` data set.

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1\. As in previous exercises, either create a new R script (perhaps call it 'linear_model_3') or continue with your previous R script in your RStudio Project. Again, make sure you include any metadata you feel is appropriate (title, description of task, date of creation etc) and  don't forget to comment out your metadata with a `#` at the beginning of the line. 

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2\. Import the data file 'loyn.txt' into R and take a look at the structure of this dataframe using the `str()` function. We know that the abundance of birds `ABUND` increases with the log~10~ transformed area of the forest patch (`LOGAREA` variable). We also know that bird abundance changes with the grazing intensity (`FGRAZE` variable) with forest patches with higher grazing intensity having fewer birds on average. But how do these effects combine together? Would a small patch with low grazing intensity have more birds than a larger patch with high grazing intensity? Could the fit of the `ABUND ~ LOGAREA` model for the large patches be improved if we accounted for grazing intensity in the patches?

```{r Q2, eval=SOLUTIONS, echo=SOLUTIONS, results=SOLUTIONS, collapse=TRUE}
loyn <- read.table("data/loyn.txt", header = TRUE, 
                   stringsAsFactors = TRUE)
str(loyn)
```

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3\. As previously we want to treat `AREA` as a log~10~-transformed area to limit the influence of the couple of disproportionately large forest patches, and `GRAZE` as a categorical variable with five levels. So the first thing we need to do is create the corresponding variables in the `loyn` dataframe, called `LOGAREA` and `FGRAZE`.

```{r Q3, eval=SOLUTIONS, echo=SOLUTIONS, results=SOLUTIONS, collapse=TRUE}
loyn$LOGAREA <- log10(loyn$AREA)
# create factor GRAZE as it was originally coded as an integer
loyn$FGRAZE <- factor(loyn$GRAZE)
```

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4\. Explore the relationship between bird abundance and log transformed forest patch area for each level of grazing. You could manually create a separate plot for each graze level but it's more efficient to use a conditional scatterplot (aka coplot, see [section 4.2.6](https://intro2r.com/simple-base-r-plots.html#coplots) of the R book or the help page for the function `coplot()`). A coplot lets you visualise the relationship between `ABUND` and `LOGAREA` for each level of `FGRAZE`, with `FGRAZE` levels increasing from the bottom-left panel (Graze level 1) to the top-right panel (Graze level 5). What patterns do you see? Is it okay to assume that the relationship between `ABUND` and `LOGAREA` is the same for all grazing levels or does the relationship change? This is effectively asking if the slopes of the relationship between `ABUND`  and `LOGAREA` are different for each `FGRAZE` level - this is called an interaction.

```{r Q4, eval=SOLUTIONS, echo=SOLUTIONS, results=SOLUTIONS, collapse=TRUE}
coplot(ABUND ~ LOGAREA | FGRAZE, data = loyn)

# or
# library(lattice)
# xyplot(ABUND ~ LOGAREA | FGRAZE, data = loyn)

# - Within a grazing level, abundance seems to increase with the log patch area
#   in a more or less linear fashion
# - Overall, the mean abundance seems to decrease as grazing levels increase.
#   This is most noticeable in the highest grazing level.
# - Some of the slopes of the relationships (imagine a straight line) appear to be 
#   somewhat different for the different graze levels. The slopes for graze levels
#   1 and 2 are similar, but different for graze levels 3, 4, and 5. We will 
#   need to test this with a model.
```

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5\. Fit an appropriate linear model in R to explain the variation in the response variable `ABUND` with the explanatory variables `LOGAREA` and `FGRAZE`. Also include the interaction between `LOGAREA` and `FGRAZE`. Hint: `:` is the interaction symbol! Remember to use the `data =` argument. Assign this linear model to an appropriately named object, like `birds_inter1`. Optional: Can you remember how to specify the model using the 'shortcut' version (with `*`) instead? 

```{r Q5, eval=SOLUTIONS, echo=SOLUTIONS, results=SOLUTIONS, collapse=TRUE}
birds_inter1 <- lm(ABUND ~ FGRAZE + LOGAREA + FGRAZE:LOGAREA, data = loyn)

# Or use the 'shortcut' - it's equivalent to the model above
# birds.inter.1 <- lm(ABUND ~ FGRAZE * LOGAREA, data = loyn)
```

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6\. As conscientious researchers, let's first check the assumptions of our linear model by creating plots of the residuals. Remember, that you can split your plotting device into 2 rows and 2 columns using the `par()` function before you create the plots. Check each of the assumptions using these plots and report whether your model meets these assumptions.

```{r Q6, eval=SOLUTIONS, echo=SOLUTIONS, results=SOLUTIONS, collapse=TRUE}
# first split the plotting device into 2 rows and 2 columns
par(mfrow = c(2,2))

# now create the residuals plots
plot(birds_inter1)

# To test the normality of residuals assumption we use the Normal Q-Q plot. 
# The central residuals are not too far from the Q-Q line but the extremes
# are too extreme (the tails of the distribution are too long). Some
# observations, both high and low, are poorly explained by the model.

# The plot of the residuals against the fitted values suggests these
# extreme residuals happen for intermediate fitted values.

# Looking at the homogeneity of variance assumption (Residuals vs
# Fitted and Scale-Location plot),
# the graphs are mostly messy, with no clear pattern emerging.

# The observations with the highest leverage don't appear to be overly
# influential, according to the Cook's distances in the Residuals vs
# Leverage plot (all < 0.5).  
```

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7\. Use the `anova()` function to produce the ANOVA table of your linear model. Remember, this ANOVA table is based on sequential sums of squares (the order of the explanatory variables matters). The only P value that we should interpret is the one that occurs in the second to last row of the table (the one above 'Residuals'). This is the interaction term between `FGRAZE` and `LOGAREA` (`FGRAZE:LOGAREA`). What is the null hypothesis associated with this interaction term? 

Do we reject or fail to reject this hypothesis? 

What is the biological interpretation of this interaction term? Can we say anything about the hypotheses for the main effects of `FGRAZE` or `LOGAREA`?    

```{r Q7, eval=SOLUTIONS, echo=SOLUTIONS, results=SOLUTIONS, collapse=TRUE}
anova(birds_inter1)

# The null hypothesis is that there is no significant interaction between 
# FGRAZE and LOGAREA. 
# As the P value is smaller than our cutoff of 0.05 (p = 0.005) we reject the 
# null hypothesis and conclude that there is a significant interaction. 

# This means that there is a significant relationship between bird abundance 
# and log area, and that this relationship is different for different levels of 
# graze (at least one of them is different). Put another way, the slopes of the 
# relationship between abundance and log area for each level of graze are different.

# As there is a significant interaction, it's difficult to interpret the main
# effects of FGRAZE and LOGAREA as by definition the effect of one variable is 
# dependent on the value of the other variable.
```

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8\. OK, now for the part you all know and love! Use the `summary()` function on your model object to produce the table of parameter estimates. Using this output, take each line in turn and answer the following questions: 

* what does this parameter estimate? 
* What is the biological interpretation of the corresponding estimate? 
* What is the null hypothesis associated with it? 
* Do you reject or fail to reject this hypothesis? 

Also compare the multiple R^2^ from this model the models you created in the previous two exercises. 

Please ask one of the course instructors to take you through this if you're confused :). 

```{r Q8, eval=SOLUTIONS, echo=SOLUTIONS, results=SOLUTIONS, collapse=TRUE}
summary(birds_inter1)

# (Intercept)
# Here the Intercept (baseline) is the predicted ABUND when LOGAREA = 0,
# for FGRAZE level 1.
# the null hypothesis for the intercept is that the intercept = 0
# As the P value < 0.05 (7.09e-08) we reject this null hypothesis.

# LOGAREA 
# Represents the slope of the relationship between ABUND and LOGAREA, 
# specific to FGRAZE = 1.
# The null hypothesis is that the slope of the relationship
# between LOGAREA and ABUND = 0, for level FGRAZE = 1 only. 
# So, for graze 1 the slope is 4.14. This means, that for a 1 unit increase in LOGAREA
# we get a corresponding increase of 4.4 birds on average.
# As the P value (0.022) is < 0.05 we conclude that the slope is
# significantly different than 0.

# FGRAZE2
# Is the estimated difference (contrasts) between the *intercept* of FGRAZE level
# 2 and the reference level intercept, FGRAZE = 1.
# The null hypothesis associated with this estimate is that the difference
# in the intercepts between graze level 1 and graze level 2 = 0.
# As the P value (0.118) is > 0.05 we fail to reject this null hypothesis and
# conclude that the intercepts for graze level 1 and graze level 2 are the same.

# FGRAZE3, FGRAZE4, FGRAZE5
# The parameter estimates have the same interpretation as for FGRAZE2 (above). 
# They are all estimates of the difference between FGRAZE at the appropriate level 
# and FGRAZE 1 (Intercept).

# FGRAZE2:LOGAREA
# This represents the difference in the slope of the relationship between ABUND 
# and LOGAREA between graze level 2 and graze level 1.
# The null hypothesis is that the difference in slopes between graze level 1 
# and 2 = 0 (i.e. no difference).
# As the P value (0.082) > 0.05 we conclude that the slopes are not different 
# (i.e. they are the same).

# FGRAZE3:LOGAREA
# This represents the difference in the slope of the relationship between ABUND 
# and LOGAREA between graze level 3 and graze level 1.
# The null hypothesis is that the difference in slopes between graze level 1 
# and 3 = 0 (i.e. no difference).
# As the P value (0.004) < 0.05 we conclude that the slopes are different (and
# therefore the relationship is different).

# FGRAZE4:LOGAREA, # FGRAZE5:LOGAREA
# These parameter estimates and null hypotheses are interpreted in the same way 
# as for FGRAZE4:LOGAREA and FGRAZE5:LOGAREA


# The Multiple R-square value is 0.79, so 79% of the variation in the data is 
# explained by the model. This is quite a bit more than the models with only
# LOGAREA and FGRAZE as single explanatory variables.
```

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9\. Right, now hang onto your hat! Let's plot the predictions from your model to figure out how it really fits the data (and help us understand the output from the `summary()` function :). Here's a general recipe, using the `predict()` function.

* plot the raw data, using a different colour for points from each `FGRAZE` level

* for each `FGRAZE` level in turn:
    + create a sequence of `LOGAREA` from the minimum value to the maximum within the grazing level (unless you wish to predict outside the range of observed values, probably best not too!).
    + store it in a data frame (i.e. `dat4pred`) containing the variables `FGRAZE` and `LOGAREA.` Remember that `FGRAZE` is a factor, so its value need to be placed in quotes (i.e. `FGRAZE = "1"`).
    + Create a vector of predicted bird abundances using our new dataframe (`dat4pred`) using the `predict()` function.
    + Add the predicted values to the plot using the `lines()` function with the appropriate colour (same points colour for each `FGRAZE` level above).
  
See the solutions for one of many ways of doing this. Now this might seem like a huge amount of code just to plot your predicted values (and admittedly it is!) but most of the code is just repeated for each level of graze. Just work through it slowly and logically and hopefully it will make sense (please ask if you are confused). I will also show you some alternative (easier?) ways to do this in the solutions.

```{r Q9a, eval=SOLUTIONS, echo=SOLUTIONS, collapse=FALSE}
par(mfrow= c(1, 1))
plot(ABUND ~ LOGAREA, data = loyn, col = GRAZE, pch = 16)
# Note: # colour 1 means black in R
# colour 2 means red in R
# colour 3 means green in R
# colour 4 means blue in R
# colour 5 means cyan in R

# FGRAZE1
# create a sequence of increasing LOGAREA within the observed range
LOGAREA.seq <- seq(from = min(loyn$LOGAREA[loyn$FGRAZE == 1]),
					to = max(loyn$LOGAREA[loyn$FGRAZE == 1]),
					length = 20)

# create data frame for prediction
dat4pred <- data.frame(FGRAZE = "1", LOGAREA = LOGAREA.seq)

# predict for new data
P1 <- predict(birds_inter1, newdata = dat4pred)

# add the predictions to the plot of the data
lines(dat4pred$LOGAREA, P1, col = 1, lwd = 2)

# FGRAZE2
LOGAREA.seq <- seq(from = min(loyn$LOGAREA[loyn$FGRAZE == 2]),
					to = max(loyn$LOGAREA[loyn$FGRAZE == 2]),
					length = 20)

dat4pred <- data.frame(FGRAZE = "2", LOGAREA = LOGAREA.seq)

P2 <- predict(birds_inter1, newdata = dat4pred)

lines(dat4pred$LOGAREA, P2, col = 2, lwd = 2)

# FGRAZE3
LOGAREA.seq <- seq(from = min(loyn$LOGAREA[loyn$FGRAZE == 3]),
					to = max(loyn$LOGAREA[loyn$FGRAZE == 3]),
					length = 20)

dat4pred <- data.frame(FGRAZE = "3", LOGAREA = LOGAREA.seq)

P3 <- predict(birds_inter1, newdata = dat4pred)

lines(dat4pred$LOGAREA, P3, col = 3, lwd = 2)

# FGRAZE4
LOGAREA.seq <- seq(from = min(loyn$LOGAREA[loyn$FGRAZE == 4]),
					to = max(loyn$LOGAREA[loyn$FGRAZE == 4]),
					length = 20)

dat4pred <- data.frame(FGRAZE = "4", LOGAREA = LOGAREA.seq)

P4 <- predict(birds_inter1, newdata = dat4pred)

lines(dat4pred$LOGAREA, P4, col = 4, lwd = 2)

# FGRAZE5
LOGAREA.seq <- seq(from = min(loyn$LOGAREA[loyn$FGRAZE == 5]),
					to = max(loyn$LOGAREA[loyn$FGRAZE == 5]),
					length = 20)

dat4pred <- data.frame(FGRAZE = "5", LOGAREA = LOGAREA.seq)

P5 <- predict(birds_inter1, newdata = dat4pred)

lines(dat4pred$LOGAREA, P5, col = 5, lwd = 2)

legend("topleft", 
 legend = paste("Graze = ", 5:1), 
 col = c(5:1), bty = "n",
 lty = c(1, 1, 1), 
 lwd = c(1, 1, 1))
```

**(Optional, for the geeks)** Alternative method given in the solutions using a`for` loop. Just keep this code in case you ever want to do something like this in the future.

```{r Q9b, eval=SOLUTIONS, echo=SOLUTIONS, collapse=FALSE}
# Okay, that was a long-winded way of doing this.
# If, like me, you prefer more compact code and less risks of errors,
# you can use a loop, to save repeating the sequence 5 times:
par(mfrow = c(1, 1))
plot(ABUND ~ LOGAREA, data = loyn, col = GRAZE, pch = 16)

for(g in levels(loyn$FGRAZE)){ # g will take the values "1", "2",..., "5" in turn
	LOGAREA.seq <- seq(from = min(loyn$LOGAREA[loyn$FGRAZE == g]),
										to = max(loyn$LOGAREA[loyn$FGRAZE == g]),
														length = 20)
	dat4pred <- data.frame(FGRAZE = g, LOGAREA = LOGAREA.seq)
	predicted <- predict(birds_inter1, newdata = dat4pred)
	lines(dat4pred$LOGAREA, predicted, col = as.numeric(g), lwd = 2)
}
legend("topleft", 
 legend = paste("Graze = ", 5:1), 
 col = c(5:1), bty= "n",
 lty = c(1, 1, 1), 
 lwd = c(1, 1, 1))
```

**(Optional, for the lazy!)** And, if you want an even easier way, then we can use the `ggolot2` package (code in the solutions). Note: you will need to install the `ggplot2` package first if you don't already have it. 

```{r Q9c, eval=SOLUTIONS, echo=SOLUTIONS, collapse=FALSE}
# install.packages('ggplot2', dep = TRUE)
library(ggplot2)

ggplot(loyn, aes(x = LOGAREA, y = ABUND, colour = FGRAZE) ) +
     geom_point() +
     geom_smooth(method = "lm", se = FALSE)
```

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End of the Linear model with continuous and categorical explanatory variables exercise