Read Chapter 6 to help you complete the questions in this exercise.
Although this short course is primarily focussed on introducing you to R, it wouldn’t be complete if we didn’t have a peek at some of R’s statistical roots. Having said that, this will be a very brief overview with very little in the way of theory so don’t worry if you get a little lost - this is just a taster, the main course is still to come!
1. Download the datafile ‘prawnGR.CSV’ from the Data link
and save it to the data directory. Import these data into R
and assign to a variable with an appropriate name. These data were
collected from an experiment to investigate the difference in growth
rate of the giant tiger
prawn (Penaeus monodon) fed either an artificial or natural
diet. Have a quick look at the structure of this dataset and plot the
growth rate versus the diet using an appropriate plot. How many
observations are there in each diet treatment?
prawns <- read.table('data/prawnGR.CSV', sep = ",", header = TRUE,
stringsAsFactors = TRUE)
# or
prawns <- read.csv("data/prawnGR.CSV", stringsAsFactors = TRUE)
# take a look at the data
str(prawns)
# 'data.frame': 60 obs. of 2 variables:
# $ GRate: num 9.77 10.29 10.05 10.08 9.31 ...
# $ diet : Factor w/ 2 levels "Artificial","Natural":...
summary(prawns)
# GRate diet
# Min. : 7.395 Artificial:30
# 1st Qu.: 9.550 Natural :30
# Median : 9.943
# Mean : 9.920
# 3rd Qu.:10.344
# Max. :11.632
# how many replicates for each level of diet
table(prawns$diet)
# Artificial Natural
# 30 30
# or use xtabs
xtabs(~ diet, data = prawns)
# produce a boxplot
boxplot(GRate ~ diet, data = prawns, xlab = "Diet", ylab = "Growth Rate")
2. You want to compare the difference in growth rate between the two
diets using a two sample t-test. Before you conduct the test, you need
to assess the normality and equal variance assumptions. Use the function
shapiro.test() to assess normality of growth rate for each
diet separately (Hint: use your indexing skills to extract the growth
rate for each diet GRate[diet=='Natural'] first). Use the
function var.test() to test for equal variance (see
?var.test for more information or Section
6.1 of the book for more details). Are your data normally
distributed and have equal variances? Note: We don’t really advocate
using these ‘approaches’ for assessing the normality and equal variance
assumptions assumptions but include them here as many people will still
want to use them. A much better way to assess assumptions is to use
diagnostic plots of the residuals (see Q6 for an example).
# test normality assumption
# Do not perform test on all data together, i.e.
shapiro.test(prawns$GRate) # this is wrong!!
# Need to test for departures from normality for each group
# separately. Remember your indexing [ ]
shapiro.test(prawns$GRate[prawns$diet == "Artificial"])
# Shapiro-Wilk normality test
#
# data: prawns$GRate[prawns$diet == "Artificial"]
# W = 0.9486, p-value = 0.1553
shapiro.test(prawns$GRate[prawns$diet == "Natural"])
# Shapiro-Wilk normality test
#
# data: prawns$GRate[prawns$diet == "Natural"]
# W = 0.9598, p-value = 0.307
# Therefore no evidence to reject the Null hypothesis and data are normally distributed
# However much better assessment of normality is to use a quantile - quantile plot
# looking for points to lie along the line for normality
qqnorm(prawns$GRate[prawns$diet == "Artificial"])
qqline(prawns$GRate[prawns$diet == "Artificial"])
qqnorm(prawns$GRate[prawns$diet == "Natural"])
qqline(prawns$GRate[prawns$diet == "Natural"])
# test for equal variance
# if normal
# Null hypothesis Ho: variances are equal
var.test(prawns$GRate ~ prawns$diet)
# F test to compare two variances
# data: prawns$GRate by prawns$diet
# F = 1.9629, num df = 29, denom df = 29, p-value = 0.07445
# alternative hypothesis: true ratio of variances is not equal to 1
# 95 percent confidence interval:
# 0.9342621 4.1240043
# sample estimates:
# ratio of variances
# 1.962881
# No evidence to reject null hypothesis (P=0.07) therefore no
# difference in variance
3. Conduct a two sample t-test using the t.test()
function (Section
6.1of the book). Use the argument var.equal = TRUE to
perform the t-test assuming equal variances. What is the null hypothesis
you want to test? Do you reject or fail to reject the null hypothesis?
What is the value of the t statistic, degrees of freedom and p value?
How would you summarise these summary statistics in a report?
# conduct t-test assuming equal variances
# Null hypothesis Ho: no difference in growth rate
# between prawns fed on artificial diet or Natural diet
t.test(GRate ~ diet, var.equal = TRUE, data = prawn)
# Two Sample t-test
#
# data: prawns$GRate by prawns$diet
# t = -1.3259, df = 58, p-value = 0.1901
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
# -0.6319362 0.1283495
# sample estimates:
# mean in group Artificial mean in group Natural
# 9.794133 10.045927
#
# No evidence to reject the Null hypothesis, therefore no
# difference in growth rate of prawns fed on either artificial
# or natural diet (t = -1.33, df = 58, p = 0.19).
4. An alternative (but equivalent) way to compare the mean growth
rate between diets is to use a linear model. Use the lm()
function to fit a linear model with GRate as the response
variable and diet as an explanatory variable (see Section 6.3 for
a very brief introduction to linear modelling). Assign
(<-) the results of the linear model to a variable with
an appropriate name (i.e. growth.lm).
5. Produce an ANOVA table for the fitted model using the
anova() function i.e. anova(growth.lm).
Compare the ANOVA p value to the p value obtained using a t-test. What
do you notice? What is the value of the F statistics and degrees of
freedom? How would you summarise these results in a report?
# produce the ANOVA table
anova(growth.lm)
# Analysis of Variance Table
#
# Response: GRate
# Df Sum Sq Mean Sq F value Pr(>F)
# diet 1 0.951 0.95100 1.7579 0.1901
# Residuals 58 31.377 0.54098
# notice the p value is the same as for the t-test
# also if you square the t statistic from the t-test
# you get the F statistic from the linear model.
# They're the same test
6. Assess the normality and equal variance assumptions by plotting
the residuals of the fitted model (see Section 6.3 for
more details). Split the plotting device into 2 rows and 2 columns using
par(mfrow=c(2,2)) so you can fit four plots on a single
device. Use the plot() function on your fitted model
(plot(growth.lm)) to plot the graphs. Discuss with an
instructor how to interpret these plots. What are your conclusions?
# plot the residuals to assess normality and equal variance
# divide the plotting device into 2 rows and 2 columns to get all
# the graphs on one device
par(mfrow = c(2,2))
plot(growth.lm)
7. Download the datafile ‘Gigartina.CSV’ from the Data link
and save it to the data directory. Import the dataset into
R and assign the dataframe an appropriate name. These data were
collected from a study to examine the change in diameter of
red algae Mastocarpus
stellatus spores grown in three different diatom cultures and a
control group grown in artificial seawater (diatom.treat
variable). Use the function str() to examine the dataframe.
How many replicates are there per diatom treatment? Use an appropriate
plot to examine whether there are any obvious differences in diameter
between the treatments.
gigartina <- read.table("data/Gigartina.CSV", header = TRUE, sep = ",", stringsAsFactors = TRUE)
# or
gigartina <- read.csv("data/Gigartina.CSV", stringsAsFactors = TRUE)
str(gigartina)
# 'data.frame':\t40 obs. of 2 variables: $ diameter : int 110 115 110 108 109 101 101 98 120 ... $
# diatom.treat: Factor w/ 4 levels 'ASGM','Sdecl',..: 1 1...
table(gigartina$diatom.treat)
# ASGM Sdecl Sexpo Sstat 10 10 10 10
# or use xtabs
xtabs(~diatom.treat, data = gigartina)
# diatom.treat ASGM Sdecl Sexpo Sstat 10 10 10 10
# plot these data
boxplot(diameter ~ diatom.treat, data = gigartina, xlab = "diatom treatment", ylab = "diameter (um)")
# or if you want to do the fancy um symbol correctly
boxplot(diameter ~ diatom.treat, data = gigartina, xlab = "diatom treatment", ylab = expression(paste("diameter",
" (", mu, "m)")))
8. You wish to compare the mean diameter of Metacarpus grown
in the four treatment groups (ASGM, Sdecl,
Sexpo, Sstat) using a one-way analysis of
variance (ANOVA). What is your null hypothesis?
# The null hypothesis Ho: there is no difference in mean diameter
# of the spores between the different treatment groups
9. We will conduct the ANOVA using the linear model function
lm() once again. Make sure you know which of the variables
is your response variable and which is your explanatory variable (ask an
instructor if in doubt). Fit the linear model and assign the model
output to a variable with an appropriate name
(i.e. gigartina.lm).
10. Produce an ANOVA table using the anova() function.
What is the value of the p value? Do you reject or fail to reject the
null hypothesis? What is the value of the F statistic and
degrees of freedom? How would you report these summary statistics in a
report?
anova(gigartina.lm)
# Analysis of Variance Table
#
# Response: diameter
# Df Sum Sq Mean Sq F value Pr(>F)
# diatom.treat 3 1880.3 626.76 22.775 1.929e-08 ***
# Residuals 36 990.7 27.52
# ---
# reject the null hypothesis, therefore there is a significant
# difference in the diameter between the treatment groups
# (F_3,36 = 22.78, p < 0.001)
11. Assess the assumptions of normality and equal variance of the residuals by producing the residual plots as before. Don’t forget to split the plotting device into 2 rows and 2 columns before plotting. Discuss with an instructor whether the residuals meet these assumptions. Do you accept this model as acceptable?
12. Let’s compare the treatment group means to determine which treatment group is different from other treatment groups. In general, you should be careful with these types of post-hoc comparisons, especially if you have a large number of groups (There are much better ways to do this, but that’s for another course!). In this case we only have 4 groups, and therefore we will use Tukey’s Honest significant difference to perform the comparisons and control for type 1 error rate (rejecting a true null hypothesis).
13. We will use the function TukeyHSD() from the
mosaic package to perform these comparisons (you will need
to install this package first and then use library(mosaic)
to make the function available). Which groups are different from each
other if we use the p-value cutoff (alpha) of p < 0.05?
# what group mean is different from what? Post-hoc comparisons.
# we will use Tukey's Honest significant difference method
# to compare group means.
# install.packages('mosaic')
library(mosaic)
# compare the group means using TukeysHSD method
TukeyHSD(gigartina.lm)
# Tukey multiple comparisons of means
# 95% family-wise confidence level
#
# Fit: aov(formula = diameter ~ diatom.treat, data = gigartina)
#
# $diatom.treat
# diff lwr upr p adj
# Sdecl-ASGM -14.3 -20.6184102 -7.98159 0.0000030
# Sexpo-ASGM -8.9 -15.2184102 -2.58159 0.0029489
# Sstat-ASGM -18.3 -24.6184102 -11.98159 0.0000000
# Sexpo-Sdecl 5.4 -0.9184102 11.71841 0.1165421
# Sstat-Sdecl -4.0 -10.3184102 2.31841 0.3360087
# Sstat-Sexpo -9.4 -15.7184102 -3.08159 0.0016145
# the null hypothesis for each comparison is
# grp1 - grp2 = 0 (i.e. no difference)
# Sdecl-ASGM, Sexpo-ASGM, Sstat-ASGM and Sstat-Sexpo
# are significantly different
14. We can also produce a plot of the comparisons to help us
interpret the table of comparisons. Use the plot() function
with the TukeyHSD(gigartina.lm). Ask if you get stuck (or
Google it!).
15. Download the ‘TemoraBR.csv’ file from the Data link
and save it to the data directory. Import the dataset into
R and as usual assign it to a variable. These data are from an
experiment that was conducted to investigate the relationship between
temperature (temp) and the beat rate (Hz)
beat_rate of the copepod Temora
longicornis which had been acclimatised at three different
temperature regimes (acclimitisation_temp). Examine the
structure of the dataset. How many variables are there? What type of
variables are they? Which is the response (dependent) variable, and
which are the explanatory (independent) variables?
temora <- read.table("data/TemoraBR.CSV", header = TRUE, sep = ",", stringsAsFactors = TRUE)
# or
temora <- read.csv("data/TemoraBR.CSV", stringsAsFactors = TRUE)
str(temora)
# 'data.frame':\t45 obs. of 3 variables: $ temp : num 5 6 7 10 11 12 13 15 16 17 ... $ beat_rate : num
# 3.76 5.4 8 9.4 16.6 18.5 19... $ acclimitisation_temp: int 5 5 5 5 5 5 5 5 5 5 ...
16. What type of variable is acclimitisation_temp? Is it
a factor? Convert acclimitisation_temp to a factor and
store the result in a new column in your dataframe called
Facclimitisation_temp. Hint: use the function
factor(). Use an appropriate plot to visualise these data
(perhaps a coplot or similar?).
temora$Facclimitisation_temp <- factor(temora$acclimitisation_temp)
# boxplot of beat rate and acclimitisation temp
boxplot(beat_rate ~ Facclimitisation_temp, data = temora, xlab = "acclimitisation temp", ylab = "beat rate")
# scatter plot using the with function
with(temora, plot(beat_rate ~ temp, xlab = "temperature", ylab = "beat rate"))
# using a coplot
coplot(beat_rate ~ temp | Facclimitisation_temp, data = temora)
# scatter plot with different symbols and colours
with(temora, plot(beat_rate ~ temp, xlab = "temperature", ylab = "beat rate", col = as.numeric(Facclimitisation_temp),
pch = as.numeric(Facclimitisation_temp)))
legend("topleft", legend = c("5", "10", "20"), pch = unique(as.numeric(temora$Facclimitisation_temp)), col = unique(as.numeric(temora$Facclimitisation_temp)))
# or more flexibly
plot(beat_rate ~ temp, xlab = "temperature", ylab = "beat rate", type = "n", data = temora)
with(temora, points(beat_rate[Facclimitisation_temp == "5"] ~ temp[Facclimitisation_temp == "5"], pch = 1,
col = "black"))
with(temora, points(beat_rate[Facclimitisation_temp == "10"] ~ temp[Facclimitisation_temp == "10"], pch = 2,
col = "red"))
with(temora, points(beat_rate[Facclimitisation_temp == "20"] ~ temp[Facclimitisation_temp == "20"], pch = 3,
col = "blue"))
legend("topleft", legend = c("5", "10", "20"), col = c("black", "red", "blue"), pch = c(1, 2, 3))
17. We will analyse these data using an Analysis of Covariance
(ANCOVA) to compare the slopes and the intercepts of the relationship
between beat_rate and temp for each level of
Facclimatisation_temp. Take a look at the plot you produced
in Q16, do you think the relationships are different?
# the slope of the relationship between beat rate and temp
# look different for each acclimitisation temp
18. As usual we will fit the model using the lm()
function. You will need to fit the main effects of temp and
Facclimatisation_temp and the interaction between
temp and Facclimatisation_temp. You can do
this using either of the equivalent specifications:
temp + Facclimatisation_temp + temp:Facclimatisation_temp
or
temp * Facclimatisation_temp
temora.lm <- lm(beat_rate ~ temp + Facclimitisation_temp + temp:Facclimitisation_temp, data = temora)
# or equivalently
temora.lm <- lm(beat_rate ~ temp * Facclimitisation_temp, data = temora)
19. Produce the summary ANOVA table as usual. Is the interaction
between temp and Facclimatisation_temp
significant? What is the interpretation of the interaction term? Should
we interpret the main effects of temp and
Facclimatisation_temp as well?
anova(temora.lm)
# Analysis of Variance Table
# Response: beat_rate
# Df Sum Sq Mean Sq F value Pr(>F)
# temp 1 4293.7 4293.7 835.866 < 2.2e-16 ***
# Facclimitisation_temp 2 1197.7 598.8 116.576 < 2.2e-16 ***
# temp:Facclimitisation_temp 2 284.1 142.0 27.651 3.331e-08 ***
# Residuals 39 200.3 5.1
# there is a significant interaction between temp and
# Facclimitisation_temp therefore there is a significant
# relationship between beat_rate and temp, and this relationship
# is different depending on the level of Facclimitisation_temp.
# Therefore we should not interpret the main effect of temp
# or Facclimitisation_temp
20. Assess the assumptions of normality and equal variance of the residuals in the usual way. Do the residuals meet these assumptions? Discuss with a instructor.
par(mfrow = c(2,2))
plot(temora.lm)
# there is a hint of heterogeneity of variance (non equal variance)
# as the variance increases with the fitted values. This is typical
# of count data.
21. Write a short summary in you R script (don’t forget to comment
this out with #) describing the interpretation of this
model. Report the appropriate summary statistics such as F
values, degrees of freedom and p values.
22. (Optional) refit the model using the square root transformed
beat_rate as the response variable. Does the interpretation
of the model change? Do the validation plots of the residuals look
better?
# we could try square root transforming the variable
# beat_rate to stabilise the variance
# square root transform beat_rate and store in the dataframe
temora$SQRT_beatrate <- sqrt(temora$beat_rate)
# refit the model using the square root transformed data
temora.lm2 <- lm(SQRT_beatrate ~ temp * Facclimitisation_temp, data = temora)
par(mfrow = c(2,2))
plot(temora.lm2)
# Residuals look a bit better
# now lets look at the ANOVA table for our new model
anova(temora.lm2)
# Analysis of Variance Table
#
# Response: SQRT_beatrate
# Df Sum Sq Mean Sq F value Pr(>F)
# temp 1 67.916 67.916 712.1746 < 2.2e-16 ***
# Facclimitisation_temp 2 17.600 8.800 92.2782 1.632e-15 ***
# temp:Facclimitisation_temp 2 1.151 0.576 6.0353 0.005205 **
# Residuals 39 3.719 0.095
# model has the same interpretation but the p value for the
# interaction term is a bit larger.
End of Exercise 5