Chapter 26 Exercise Solutions

These are the solutions to the exercises used in the course.

26.1 Californian bird diversity

  1. First import the data. Check that the columns look as they should. (e.g. use summary or str functions). Tip: use the “Wizard” in RStudio to guide you.
birds <- read.csv("CourseData/suburbanBirds.csv")
  1. What is the mean, minimum, and maximum number of species seen? (there is more than one way to do this)*
mean(birds$nSpecies)
## [1] 9.647059
min(birds$nSpecies)
## [1] 3
max(birds$nSpecies)
## [1] 15
range(birds$nSpecies)
## [1]  3 15
summary(birds$nSpecies)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   3.000   6.000  11.000   9.647  13.000  15.000
  1. How old are the youngest and oldest suburbs? (hint: the survey was carried out in 1975, do the math!)
1975 - min(birds$Year)
## [1] 29
1975 - max(birds$Year)
## [1] 1
  1. Plot the relationship between Year and nSpecies as a scatterplot using base-R graphics (using the plot function).
plot(birds$Year, birds$nSpecies)

  1. The pattern might be easier to see if you could replace YearBuilt with suburb age. Create a new vector in your data frame for this variable (e.g. birds$Age <- 1975 - birds$Year)). Re-plot your results.
birds$Age <- 1975 - birds$Year
plot(birds$Age, birds$nSpecies)

  1. What do the data show? What might be the mechanisms for the patterns you see? Do they match your expectations?

If you recall that the average species richness pre-development was about 3.5 species, the data show that suburban development is actually good for bird species. This could be surprising, but a possible explanation is that the gardens, parks, trees etc. that come with development represent additional habitats that would not normally be there. Therefore, these areas attract new species.

  1. Export your plots and paste them into a Word Document.

You can do this with several methods. My favourite quick method is to click the Export button > Copy to Clipboard, resize the plot so it looks nice, then click Copy Plot. Finally, paste into Word with Ctrl (or Cmd) + V.

26.2 Wrangling the Amniote Life History Database

  1. When you have imported the data, use dim to check the dimensions of the whole data frame (you should see that there are 36 columns and 21322 rows). Use names to look at the names of all columns in the data in amniote.
amniote <- read.csv("CourseData/Amniote_Database_Aug_2015.csv",
  na.strings = "-999"
)
dim(amniote)
## [1] 21322    36
names(amniote)
##  [1] "class"                                
##  [2] "order"                                
##  [3] "family"                               
##  [4] "genus"                                
##  [5] "species"                              
##  [6] "subspecies"                           
##  [7] "common_name"                          
##  [8] "female_maturity_d"                    
##  [9] "litter_or_clutch_size_n"              
## [10] "litters_or_clutches_per_y"            
## [11] "adult_body_mass_g"                    
## [12] "maximum_longevity_y"                  
## [13] "gestation_d"                          
## [14] "weaning_d"                            
## [15] "birth_or_hatching_weight_g"           
## [16] "weaning_weight_g"                     
## [17] "egg_mass_g"                           
## [18] "incubation_d"                         
## [19] "fledging_age_d"                       
## [20] "longevity_y"                          
## [21] "male_maturity_d"                      
## [22] "inter_litter_or_interbirth_interval_y"
## [23] "female_body_mass_g"                   
## [24] "male_body_mass_g"                     
## [25] "no_sex_body_mass_g"                   
## [26] "egg_width_mm"                         
## [27] "egg_length_mm"                        
## [28] "fledging_mass_g"                      
## [29] "adult_svl_cm"                         
## [30] "male_svl_cm"                          
## [31] "female_svl_cm"                        
## [32] "birth_or_hatching_svl_cm"             
## [33] "female_svl_at_maturity_cm"            
## [34] "female_body_mass_at_maturity_g"       
## [35] "no_sex_svl_cm"                        
## [36] "no_sex_maturity_d"
  1. We are interested in longevity (lifespan) and body size and reproductive effort and how this might vary depending on the taxonomy (specifically, with Class). Use select to pick relevent columns of the dataset and discard the others. Call the new data frame x. The relevant columns are the taxonomic variables (class, species) and longevity_y, litter_or_clutch_size_n, litters_or_clutches_per_y, and adult_body_mass_g.
x <- amniote %>%
  select(
    class, genus, species,
    longevity_y, adult_body_mass_g,
    litter_or_clutch_size_n, litters_or_clutches_per_y
  )
  1. Take a look at the first few entries in the species column. You will see that it is only the epithet, the second part of the Genus_species name, that is given.
    Use mutate and paste to convert the species column to a Genus_species by pasting the data in genus and species together. To see how this works, try out the following command, paste(1:3, 4:6). After you have created the new column, remove the genus column (using select and -genus).
x <- x %>%
  mutate(species = paste(genus, species)) %>%
  select(-genus)
head(x)
##   class               species longevity_y adult_body_mass_g
## 1  Aves Accipiter albogularis          NA           251.500
## 2  Aves      Accipiter badius          NA           140.000
## 3  Aves     Accipiter bicolor          NA           345.000
## 4  Aves  Accipiter brachyurus          NA           142.000
## 5  Aves    Accipiter brevipes          NA           203.500
## 6  Aves Accipiter castanilius          NA           159.375
##   litter_or_clutch_size_n litters_or_clutches_per_y
## 1                      NA                        NA
## 2                    3.25                         1
## 3                    2.70                        NA
## 4                      NA                        NA
## 5                    4.00                         1
## 6                      NA                        NA
  1. What is the longest living species in the record? Use arrange to sort the data from longest to shortest longevity (longevity_y), and then look at the top of the file using head to find out. (hint: you will need to use reverse sort (-)). Cut and paste the species name into Google to find out more!
x <- x %>% arrange(-longevity_y)
head(x)
##      class                 species longevity_y adult_body_mass_g
## 1 Reptilia Chelonoidis duncanensis       177.0                NA
## 2 Reptilia  Aldabrachelys gigantea       152.0            117200
## 3 Reptilia          Testudo graeca       127.0              1430
## 4 Mammalia            Homo sapiens       122.5             62035
## 5 Mammalia   Balaenoptera physalus        95.0          38800000
## 6 Mammalia            Orcinus orca        90.0           4300000
##   litter_or_clutch_size_n litters_or_clutches_per_y
## 1                      NA                        NA
## 2                    13.5                  2.000000
## 3                     5.0                  3.200993
## 4                     1.0                  0.485000
## 5                     1.0                  0.400000
## 6                     1.0                  0.210000
  1. Do the same thing but this time find the shortest lived species.
x <- x %>% arrange(longevity_y)
head(x)
##      class              species longevity_y adult_body_mass_g
## 1 Mammalia    Lepus nigricollis  0.08333333          2196.875
## 2 Mammalia  Notoryctes caurinus  0.08333333            34.000
## 3 Mammalia Allactaga balikunica  0.08333333                NA
## 4 Mammalia    Allactaga bullata  0.08333333                NA
## 5 Mammalia       Geomys pinetis  0.08333333           195.750
## 6 Mammalia          Mus sorella  0.08333333            12.535
##   litter_or_clutch_size_n litters_or_clutches_per_y
## 1                    1.59                     7.150
## 2                    1.50                        NA
## 3                    2.52                        NA
## 4                    2.52                        NA
## 5                    1.77                     1.865
## 6                    5.20                     3.000
  1. Use summarise and group_by to make a table summarising min, median and max life spans (longevity_y) for the three taxonomic classes in the database. Remember that you need to tell R to remove the NA values using a na.rm = TRUE argument.
x %>%
  group_by(class) %>%
  summarise(
    min = min(longevity_y, na.rm = TRUE),
    median = median(longevity_y, na.rm = TRUE),
    max = max(longevity_y, na.rm = TRUE)
  )
## # A tibble: 3 × 4
##   class       min median   max
##   <chr>     <dbl>  <dbl> <dbl>
## 1 Aves     0.75    12.5    75 
## 2 Mammalia 0.0833   8.67  122.
## 3 Reptilia 0.2     11.0   177
  1. Body size is thought to be associated with life span. Let’s treat that as a hypothesis and test it graphically. Sketch what would the graph would look like if the hypothesis were true, and if it was false. Plot adult_body_mass_g vs. longevity_y (using base R graphics). You should notice that this looks a bit messy.
plot(x$adult_body_mass_g, x$longevity_y)

  1. Use mutate to create a new log-transformed variables, logMass and logLongevity. Use these to make a “log-log” plot. You should see that makes the relationship more linear, and easier to “read”.
x <- x %>%
  mutate(
    logMass = log(adult_body_mass_g),
    logLongevity = log(longevity_y)
  )

plot(x$logMass, x$logLongevity)

  1. Is there a trade-off between reproductive effort and life span? Think about this as a hypothesis - sketch what would the graph would look like if that were true, and if it was false. Now use the data to test that hypothesis: Use mutate to create a variable called logOffspring which is the logarithm of number of litters/clutches per year multiplied by the number of babies in each litter/clutch . Then plot logOffspring vs. logLongevity.
x <- x %>%
  mutate(logOffspring = log(
    litter_or_clutch_size_n * litters_or_clutches_per_y
  ))

plot(x$logOffspring, x$logLongevity)

  1. To answer the final question (differences between taxonomic classes) you could now use filter to subset to particular classes and repeat the plot to see whether the relationships holds universally.
aves <- x %>%
  filter(class == "Aves")

plot(aves$logOffspring, aves$logLongevity)
title("Aves")

mammalia <- x %>%
  filter(class == "Mammalia")

plot(mammalia$logOffspring, mammalia$logLongevity)
title("Mammalia")

reptilia <- x %>%
  filter(class == "Reptilia")

plot(reptilia$logOffspring, reptilia$logLongevity)
title("Reptilia")

26.3 Temperature effects on egg laying dates

  1. Import the data and take a look at it with head or str.
eggDates <- read.csv("CourseData/eggDates.csv")
head(eggDates)
##   boxNumber y2013 y2014 y2016 y2017 y2018 y2019
## 1         1   116   103    NA   107   111    NA
## 2         2    NA    NA    NA   114   118    NA
## 3         3    NA   102   108    NA    NA    NA
## 4         4   121   103   121   155   111   110
## 5         5   135   100   108   102   106   108
## 6         6   122   113   122    NA   124   149
  1. Use pivot_longer to reformat the data. This might take a bit of trial and error - don’t give up!

Maybe this will help: The first argument in the pivot_longer command (cols) tells R which columns contain the data you are interested in (in this case, these are y2013,y2014 etc). Then the names_to argument tells R what you want to name the new column from this data (in this case, Year). Then, the values_to argument tells R what the data column should be called (e.g. Day). In addition, there is a useful argument called names_prefix that will remove the part of the column name (e.g. the y of y2013)

You should also make sure that the Year column is recognised as being a numeric variable rather than a character string. You can do this by adding a command using mutate and as.numeric, like this mutate(Year = as.numeric(Year))

You should end up with a dataset with three columns as described above.

eggDates <- eggDates %>% pivot_longer(
  cols = starts_with("y"),
  names_to = "Year", values_to = "day"
)

head(eggDates)
## # A tibble: 6 × 3
##   boxNumber Year    day
##       <int> <chr> <int>
## 1         1 y2013   116
## 2         1 y2014   103
## 3         1 y2016    NA
## 4         1 y2017   107
## 5         1 y2018   111
## 6         1 y2019    NA
  1. Ensure that year is coded as numeric variable using mutate. [Hint, you can use the command as.numeric, but first remove the “y” in the name using gsub].
eggDates <- eggDates %>%
  mutate(Year = gsub(pattern = "y", replacement = "", Year)) %>%
  mutate(Year = as.numeric(Year))
head(eggDates)
## # A tibble: 6 × 3
##   boxNumber  Year   day
##       <int> <dbl> <int>
## 1         1  2013   116
## 2         1  2014   103
## 3         1  2016    NA
## 4         1  2017   107
## 5         1  2018   111
## 6         1  2019    NA
  1. Calculate the mean egg date per year using summarise (remember to group_by the year first). Take a look at the data.
meanEgg <- eggDates %>%
  group_by(Year) %>%
  summarise(meanEggDate = mean(day, na.rm = TRUE))

meanEgg
## # A tibble: 6 × 2
##    Year meanEggDate
##   <dbl>       <dbl>
## 1  2013        125.
## 2  2014        108.
## 3  2016        115.
## 4  2017        112.
## 5  2018        117.
## 6  2019        111.

Preparing the weather data

  1. Import the weather data and take a look at it with head or str.
weather <- read.csv("CourseData/AarslevTemperature.csv")
str(weather)
## 'data.frame':    3994 obs. of  4 variables:
##  $ YEAR : int  2012 2012 2012 2012 2012 2012 2012 2012 2012 2012 ...
##  $ MONTH: int  4 4 4 4 4 4 4 4 4 4 ...
##  $ DAY  : int  11 12 13 14 15 16 17 18 19 20 ...
##  $ TEMP : num  8.1 6.6 5.9 6.2 6.5 4.3 4.3 7 8.9 6.5 ...
  1. Use filter subset to the months of interest (February-April) and then summarise the data to calculate the mean temperature in this period (remember to group_by year). Look at the data. You should end up with a dataset with two columns - YEAR and meanSpringTemp.
weather <- weather %>%
  filter(MONTH %in% 2:4) %>%
  group_by(YEAR) %>%
  summarise(meanAprilTemp = mean(TEMP))

head(weather)
## # A tibble: 6 × 2
##    YEAR meanAprilTemp
##   <int>         <dbl>
## 1  2012          7.86
## 2  2013          1.60
## 3  2014          6.03
## 4  2015          4.63
## 5  2016          4.50
## 6  2017          4.77

Bringing it together

  1. Join the two datasets together using left_join. You should now have a dataset with columns nestNumber, Year, dayNumber and meanAprilTemp
joinedData <- left_join(meanEgg, weather, c("Year" = "YEAR"))
head(joinedData)
## # A tibble: 6 × 3
##    Year meanEggDate meanAprilTemp
##   <dbl>       <dbl>         <dbl>
## 1  2013        125.          1.60
## 2  2014        108.          6.03
## 3  2016        115.          4.50
## 4  2017        112.          4.77
## 5  2018        117.          3.07
## 6  2019        111.          6.12

Plot the data

  1. plot a graph of meanAprilTemp on the x-axis and dayNumber on the y-axis.
plot(joinedData$meanAprilTemp, joinedData$meanEggDate)

Now you should be able to answer the question we started with: is laying date associated with spring temperatures? Yes, there looks to be a negative relationship between temperature and egg laying date.

26.4 Virtual dice

Let’s try the same kind of thing with the roll of (virtual) dice.

Here’s how to do one roll of the dice:

diceRoll <- 1:6
sample(diceRoll, 1)

[1] 3

  1. Simulate 10 rolls of the dice, and make a table of results.
result <- sample(diceRoll, 10, replace = TRUE)
table(result)
## result
## 2 3 4 5 6 
## 2 3 1 1 3

Your table will probably look different to this, because it is a random process. You may notice that some numbers in the table are missing if some numbers were never rolled by our virtual dice.

  1. Now simulate 90 rolls of the dice, and plot the results as a bar plot using geom_bar in ggplot. Add a horizontal line using geom_abline to show the expected result based on what you know about probability.
n <- 90
result <- data.frame(result = sample(diceRoll, n, replace = TRUE))

ggplot(result, aes(x = result)) +
  geom_bar() +
  geom_abline(intercept = n / 6, slope = 0)
Barplot of 90 simulated dice throws

Figure 26.1: Barplot of 90 simulated dice throws

  1. Try adjusting the code to simulate dice rolls with small (say, 30) and large (say, 600, or 6000, or 9000) samples. Observe what happens to the proportions, and compare them to the expected value. What does this tell you about the importance of sample size when trying to estimate real phenomena?

You only need to edit the n <- 90 line in the code above

The main message here is that as sample size increases you are more likely to obtain a good estimate of the true value of the phenomenon. You may also notice that, what would be considered a good sample size for the coin flipping (i.e. it recovers the true probability of 0.5 reasonably well) is NOT adequate for getting a good estimate of the probabilities for the dice.

This is because of the different number of possibilities: as the range of possible outcomes increases, the sample size requirements increase. In other words, choosing a good sample size is context-dependent.

Barplots of 30 and 9000 simulated dice throws

Figure 26.2: Barplots of 30 and 9000 simulated dice throws

26.5 Sexual selection in Hercules beetles

  1. What is your null hypothesis?

Null Hypothesis - There is no difference in the widths of the species.

  1. What is your alternative hypothesis?

Alternative Hypothesis - Males have larger widths than females.

  1. Import the data.
hercules <- read.csv("CourseData/herculesBeetle.csv")
  1. Calculate the mean for each sex (either using tapply or using dplyr tools)
# With dplyr
hercules %>%
  group_by(sex) %>%
  summarise(mean = mean(width))
## # A tibble: 2 × 2
##   sex     mean
##   <chr>  <dbl>
## 1 Female  15.0
## 2 Male    17.4
# with tapply
tapply(hercules$width, hercules$sex, mean)
##   Female     Male 
## 15.02825 17.36568
  1. Plot the data as a histogram.
# Let's look at the male and female data
(plot1 <- ggplot(hercules, aes(x = width, fill = sex)) +
  geom_histogram(position = "identity", alpha = .7))

  1. Add vertical lines to the plot to indicate the mean values.
hercules %>%
  group_by(sex) %>%
  summarise(mean = mean(width)) %>%
  pull(mean) -> meanVals

plot1 + geom_vline(xintercept = meanVals, colour = c("red", "blue"))

  1. Now calculate the difference between the mean values using dplyr tools, or tapply.
# with dplyr
hercules %>%
  group_by(sex) %>%
  summarise(mean = mean(width)) %>%
  pull(mean) %>%
  diff() -> observedDiff
observedDiff
## [1] 2.337433
# with tapply
diff(as.vector(tapply(hercules$width, hercules$sex, mean)))
## [1] 2.337433
  1. Use sample to randomise the sex column of the data, and recalculate the difference between the mean.
# with dplyr
hercules %>%
  mutate(sex = sample(sex)) %>%
  group_by(sex) %>%
  summarise(mean = mean(width)) %>%
  pull(mean) %>%
  diff()
## [1] -1.900753
# with tapply
diff(as.vector(tapply(hercules$width, sample(hercules$sex), mean)))
## [1] 0.3205587
  1. Use replicate to repeat this 10 times (to ensure that you code works).
# with dplyr
replicate(
  10,
  hercules %>%
    mutate(sex = sample(sex)) %>%
    group_by(sex) %>%
    summarise(mean = mean(width)) %>%
    pull(mean) %>%
    diff()
)
##  [1]  0.3885126 -0.5711667 -0.2898130 -0.3766535 -0.4209082 -0.4678595
##  [7] -2.1980346 -1.5961324  0.8836404  1.2863489
# with tapply
replicate(
  10,
  diff(as.vector(tapply(
    hercules$width,
    sample(hercules$sex), mean
  )))
)
##  [1]  1.2267611 -0.2954281  0.3783011 -0.6666253 -1.4024738  0.6233728
##  [7] -0.5334274  0.2076405 -0.4760695 -0.6967478
  1. When your code is working, use replicate again, but this time with 1000 replicates and pass the results into a data frame.
# with dplyr
diffs <- data.frame(
  diffs =
    replicate(
      1000,
      hercules %>%
        mutate(sex = sample(sex)) %>%
        group_by(sex) %>%
        summarise(mean = mean(width)) %>%
        pull(mean) %>%
        diff()
    )
)

# with tapply
diffs <- data.frame(
  diffs =
    replicate(
      1000,
      diff(as.vector(
        tapply(hercules$width, sample(hercules$sex), mean)
      ))
    )
)
  1. Use ggplot to plot the null distribution you have just created, and add the observed difference.
ggplot(diffs, aes(x = diffs)) +
  geom_histogram() +
  geom_vline(xintercept = observedDiff)

  1. Obtain the p-value for the hypothesis test described above. (1) how many of the shuffled differences are more extreme than the observed distribution (2) what is this expressed as a proportion of the number of replicates.
sum(diffs$diffs >= observedDiff)
## [1] 14
sum(diffs$diffs >= observedDiff) / 1000
## [1] 0.014
  1. Summarise your result as in a report. Describe the method, followed by the result and conclusion.

“I used a randomisation test to estimate the significance of the observed difference of 2.337 (mean values: female=17.366; male = 15.028) in mean widths of the sexes. To do this I generated a null distribution of differences between sexes using 1000 replicates. I found that only 14 of the differences in the null distribution were as extreme as the observed difference. Thus the p-value is 0.014: I therefore reject the null hypothesis that there is no difference between the sexes and accept the alternative hypothesis that males are significantly larger than females.”

26.6 Sex differences in fine motor skills

Some people have suggested that there might be sex differences in fine motor skills in humans. Use the data collected on the class to address this topic using t-tests. The relevant data set is called classData.csv, and the columns of interest are Gender and Precision.

Carry out a two-sample t-test.

  1. Plot the data (e.g. with a box plot, or histogram)
classData <- read.csv("CourseData/classData.csv") %>%
  filter(Gender %in% c("Male", "Female")) %>% 
  filter(Year == 2019) #you may use a different filter

ggplot(classData, aes(x = Gender, y = Precision)) +
  geom_boxplot() +
  geom_jitter(width = 0.2)

  1. Formulate null and alternative hypotheses.

Null hypotheses - the differences in precision between male and female are due to chance alone. Alternative hypothesis - there is a significant difference between male and female precision scores.

  1. Use the t.test function to do the test.
t.test(Precision ~ Gender, data = classData)
## 
##  Welch Two Sample t-test
## 
## data:  Precision by Gender
## t = -1.1836, df = 22.601, p-value = 0.2489
## alternative hypothesis: true difference in means between group Female and group Male is not equal to 0
## 95 percent confidence interval:
##  -6.873689  1.873689
## sample estimates:
## mean in group Female   mean in group Male 
##             26.16667             28.66667
  1. Write a sentence or two describing the results.

There was no significant difference in mean precision between the two genders (t-test: t = -1.18, df = 22.60, p = 0.249). The 95% confidence interval for the difference between genders overlapped 0 (95% CI = -6.874-1.874). I therefore fail to reject the null hypothesis that the observed differences are due to chance alone.

26.7 Therapy for anorexia

A study was carried out looking at the effect of cognitive behavioural therapy on weight of people with anorexia. Weight was measured in week 1 and again in week 8. Use a paired t-test to assess whether the treatment is effective.

The data is called anorexiaCBT.csv

The data are in “wide format”. You may wish to convert it to “long format” depending on how you use the data. You can do that with the pivot_longer function, which rearranges the data:

  1. Plot the data (e.g. with an interaction plot).
anorexiaCBT <- read.csv("CourseData/anorexiaCBT.csv",
  header = TRUE
)

anorexiaCBT_long <- anorexiaCBT %>%
  pivot_longer(
    cols = starts_with("Week"), names_to = "time",
    values_to = "weight"
  )
ggplot(anorexiaCBT_long, aes(
  x = time, y = weight,
  group = Subject
)) +
  geom_line() +
  geom_point() +
  xlab("Time") +
  ylab("Weight (kg)")

  1. Formulate a null and alternative hypothesis.

Null = The difference in weight between the two times is no different than random chance. Alternative Hypothesis = There is a significant change in weight between the two time points.

  1. Use t.test to conduct a paired t-test.

The method here depends on whether you use the “long” data or not:

t.test(anorexiaCBT$Week01, anorexiaCBT$Week08, paired = TRUE)
## 
##  Paired t-test
## 
## data:  anorexiaCBT$Week01 and anorexiaCBT$Week08
## t = -2.2156, df = 28, p-value = 0.03502
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -5.7869029 -0.2268902
## sample estimates:
## mean difference 
##       -3.006897

or

anorexiaCBT_long <- anorexiaCBT_long %>%
  arrange(Subject, time)

t.test(weight ~ time, data = anorexiaCBT_long, paired = TRUE)
## 
##  Paired t-test
## 
## data:  weight by time
## t = -2.2156, df = 28, p-value = 0.03502
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -5.7869029 -0.2268902
## sample estimates:
## mean difference 
##       -3.006897
  1. Write a couple of sentences to report your result.
x <- t.test(anorexiaCBT$Week01, anorexiaCBT$Week08, paired = TRUE)

There was a significant difference in weight of -3.007kg between week 1 and week 8 (t.test: t = -2.22, df = 28.00, p = 0.035). The 95% confidence interval for the difference between weeks was between -5.787 and -0.227. Therefore, I reject the null hypotheses, that the difference is due to chance alone, and accept the alternative hypothesis.

26.8 Compare t-tests with randomisation tests

Try re-fitting some of these tests as randomisation tests (or analyse the randomisation test data using t.test). Do they give approximately the same results?

Then try answering the question - “are people who prefer dogs taller than those who prefer cats?” using the classData.csv. Can you think of any problems with this analysis?

The problem with the analysis is that it is “confounded”. That is to say, gender is correlated with height, so you would not be sure whether any difference you found would be due to height, or gender.

26.9 Apple tree crop yield

Import the data (apples.csv) and analyse it using the techniques you have learned in the ANOVA lecture, and the previous worksheet, to answer the question “What is the effect of tree spacing on apple yields?”

  1. Import and look at the data (e.g. summary or str or head)
apples <- read.csv("CourseData/apples.csv")
summary(apples)
##     spacing       yield      
##  Min.   : 6   Min.   : 64.1  
##  1st Qu.: 6   1st Qu.:108.2  
##  Median :10   Median :147.1  
##  Mean   :10   Mean   :145.4  
##  3rd Qu.:14   3rd Qu.:176.5  
##  Max.   :14   Max.   :282.3  
##               NA's   :28
  1. Ensure that the explanatory variable (spacing) is defined as a categorical variable (i.e. a “factor”, in R-speak). You can use mutate and as.factor functions for this.
apples <- apples %>%
  mutate(spacing = as.factor(spacing))
  1. Plot the data using ggplot (a box plot with (optionally) added jittered points would be good).
ggplot(apples, aes(x = spacing, y = yield)) +
  geom_boxplot() +
  geom_jitter(width = 0.2) +
  ylab("Yield (kg per tree)")

  1. Fit an ANOVA model using lm.
apple_mod <- lm(yield ~ spacing, data = apples)
  1. Check the model using a diagnostic plot (i.e. using autoplot from the ggfortify package).
library(ggfortify)
autoplot(apple_mod)

  1. Use anova to get the ANOVA summary.
anova(apple_mod)
## Analysis of Variance Table
## 
## Response: yield
##           Df Sum Sq Mean Sq F value    Pr(>F)    
## spacing    2  35801 17900.3  9.3851 0.0002003 ***
## Residuals 89 169750  1907.3                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  1. You should see that there are differences among treatments. But where are those differences? Use summary on your model to find out.
summary(apple_mod)
## 
## Call:
## lm(formula = yield ~ spacing, data = apples)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -92.389 -30.577  -3.516  33.192 117.628 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  120.566      7.382  16.332  < 2e-16 ***
## spacing10     35.924     11.073   3.244 0.001659 ** 
## spacing14     44.107     10.966   4.022 0.000121 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 43.67 on 89 degrees of freedom
##   (28 observations deleted due to missingness)
## Multiple R-squared:  0.1742, Adjusted R-squared:  0.1556 
## F-statistic: 9.385 on 2 and 89 DF,  p-value: 0.0002003
  1. Use a Tukey test to make all the pairwise comparisons among the treatments.
library(agricolae)
HSD.test(apple_mod, "spacing", console = TRUE)
## 
## Study: apple_mod ~ "spacing"
## 
## HSD Test for yield 
## 
## Mean Square Error:  1907.304 
## 
## spacing,  means
## 
##       yield      std  r  Min   Max
## 10 156.4893 45.60411 28 64.1 248.6
## 14 164.6724 50.41401 29 72.3 282.3
## 6  120.5657 35.32755 35 66.4 185.5
## 
## Alpha: 0.05 ; DF Error: 89 
## Critical Value of Studentized Range: 3.370849 
## 
## Groups according to probability of means differences and alpha level( 0.05 )
## 
## Treatments with the same letter are not significantly different.
## 
##       yield groups
## 14 164.6724      a
## 10 156.4893      a
## 6  120.5657      b
  1. Write a few sentences that summarise your findings. What biological processes do you think drive the effects that you have detected?

There was a significant effect of spacing on apple yields (Figure XX, ANOVA: F = 9.385, d.f. = 2 and 89, p = 0.0002).

Then: The pairwise comparisons in the ANOVA model showed that means of the 6 and 10 foot spacing treatment were significantly different (t= 3.244, p = 0.0017), as were those of 6 and 14 (t=4.022, p = 0.0001), but the 10 foot - 14 foot comparison showed no significant difference (t= 0.707, p= 0.4813)19.

Or, more simply: The 6-10ft and 6-14ft comparisons showed significant differences (Tukey HSD: p<0.05), but the 10-14ft comparison showed no significant difference (Tukey HSD: p>0.05)

  1. Optional. Instead of using a Tukey test, use the alternative “relevel” approach to make the missing comparison.
apples2 <- apples %>%
  mutate(spacing = relevel(spacing, ref = "10"))
apple_mod2 <- lm(yield ~ spacing, data = apples2)
summary(apple_mod2)
## 
## Call:
## lm(formula = yield ~ spacing, data = apples2)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -92.389 -30.577  -3.516  33.192 117.628 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  156.489      8.253  18.961  < 2e-16 ***
## spacing6     -35.924     11.073  -3.244  0.00166 ** 
## spacing14      8.183     11.571   0.707  0.48128    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 43.67 on 89 degrees of freedom
##   (28 observations deleted due to missingness)
## Multiple R-squared:  0.1742, Adjusted R-squared:  0.1556 
## F-statistic: 9.385 on 2 and 89 DF,  p-value: 0.0002003

If you get this far, try using the ANOVA approach on one of the previous t-test examples (remember that ANOVA can be used when your single explanatory variable has TWO or more levels). You should notice that the results are the same whether you use the t.test function or the ANOVA approach with lm.

26.10 Chirping crickets

  1. Import the data
chirps <- read.csv("CourseData/chirps.csv")
  1. Use mutate to convert Fahrenheit to Celsius (Google it)
chirps <- chirps %>%
  mutate(temperatureC = (temperature - 32) * (5 / 9))
head(chirps)
##   chirps temperature temperatureC
## 1   20.0        88.6     31.44444
## 2   16.0        71.6     22.00000
## 3   19.8        93.3     34.05556
## 4   18.4        84.3     29.05556
## 5   17.1        80.6     27.00000
## 6   15.5        75.2     24.00000
  1. Plot the data
(A <- ggplot(chirps, aes(x = temperatureC, y = chirps)) +
  geom_point())

  1. Fit a linear regression model with lm
chirp_mod <- lm(chirps ~ temperatureC, data = chirps)
  1. Look at diagnostic plots to evaluate the model
library(ggfortify)
autoplot(chirp_mod)

  1. Use anova to figure out if the effect of temperature is statistically significant.
anova(chirp_mod)
## Analysis of Variance Table
## 
## Response: chirps
##              Df Sum Sq Mean Sq F value    Pr(>F)    
## temperatureC  1 28.435 28.4348  29.248 0.0001195 ***
## Residuals    13 12.639  0.9722                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  1. Use summary to obtain information about the coefficients and \(R^2\)-value.
summary(chirp_mod)
## 
## Call:
## lm(formula = chirps ~ temperatureC, data = chirps)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.6181 -0.6154  0.0916  0.7669  1.5549 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   6.95531    1.79534   3.874 0.001918 ** 
## temperatureC  0.36540    0.06756   5.408 0.000119 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.986 on 13 degrees of freedom
## Multiple R-squared:  0.6923, Adjusted R-squared:  0.6686 
## F-statistic: 29.25 on 1 and 13 DF,  p-value: 0.0001195
  1. Summarise the model in words.

There is a statistically significant association between temperature and chirp frequency (F = 29.2482, d.f. = 1,13, p < 0.001) The equation of the fitted model is: Chirp Freq = 6.96(\(\pm\) 1.80) \(\times\) Temp + 0.37(\(\pm\) 0.07). The model explains 69% of the variation in chirp frequency (\(R^2\) = 0.692).

  1. Add model fit line to the plot.
A + geom_smooth(method = "lm")

  1. Can I use cricket chirp frequency as a kind of thermometer?

Yes, using the equation from the model. Ask an instructor if you can’t figure this out.

26.11 Fish behaviour

  1. Import the data set, fishPersonality.csv
fishPersonality <- read.csv("CourseData/fishPersonality.csv")
  1. Plot the data (e.g. as a box plot)
ggplot(fishPersonality, aes(
  x = colouration, y = spawning,
  fill = personality
)) +
  geom_boxplot()

  1. Fit an ANOVA model using lm.
mod_A <- lm(spawning ~ personality + colouration +
  personality:colouration, data = fishPersonality)

  1. Look at diagnostic plots from this model (autoplot)

  2. Use anova to get an Analysis of Variance summary table, and interpret the results.

anova(mod_A)
## Analysis of Variance Table
## 
## Response: spawning
##                         Df Sum Sq Mean Sq F value    Pr(>F)    
## personality              1   7781    7781  5.5032   0.02629 *  
## colouration              1  55862   55862 39.5074 8.519e-07 ***
## personality:colouration  1   4118    4118  2.9123   0.09898 .  
## Residuals               28  39591    1414                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  1. Get the coefficient summary (summary) and interpret the output.
summary(mod_A)
## 
## Call:
## lm(formula = spawning ~ personality + colouration + personality:colouration, 
##     data = fishPersonality)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -61.625 -23.344  -5.375  26.313  60.125 
## 
## Coefficients:
##                                            Estimate Std. Error t value Pr(>|t|)
## (Intercept)                                  218.50      13.29  16.435  6.5e-16
## personalityReactive                          -53.88      18.80  -2.865  0.00781
## colourationHomogeneous                        60.88      18.80   3.238  0.00309
## personalityReactive:colourationHomogeneous    45.38      26.59   1.707  0.09898
##                                               
## (Intercept)                                ***
## personalityReactive                        ** 
## colourationHomogeneous                     ** 
## personalityReactive:colourationHomogeneous .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 37.6 on 28 degrees of freedom
## Multiple R-squared:  0.6312, Adjusted R-squared:  0.5917 
## F-statistic: 15.97 on 3 and 28 DF,  p-value: 3.021e-06
  1. Do post-hoc Tukey tests (e.g. using HSD.test from the agricolae package). Interpret the results.
library(agricolae)
HSD.test(mod_A, c("personality", "colouration"), console = TRUE)
## 
## Study: mod_A ~ c("personality", "colouration")
## 
## HSD Test for spawning 
## 
## Mean Square Error:  1413.951 
## 
## personality:colouration,  means
## 
##                         spawning      std r Min Max
## Proactive:Heterogeneous  218.500 34.92850 8 161 268
## Proactive:Homogeneous    279.375 40.40133 8 219 339
## Reactive:Heterogeneous   164.625 39.86383 8 103 220
## Reactive:Homogeneous     270.875 34.84840 8 233 331
## 
## Alpha: 0.05 ; DF Error: 28 
## Critical Value of Studentized Range: 3.861244 
## 
## Minimun Significant Difference: 51.33332 
## 
## Treatments with the same letter are not significantly different.
## 
##                         spawning groups
## Proactive:Homogeneous    279.375      a
## Reactive:Homogeneous     270.875      a
## Proactive:Heterogeneous  218.500      b
## Reactive:Heterogeneous   164.625      c
  1. Sum up your findings with reference to the initial research questions.
  • Homogeneous coloured fish seem to do better than heterogeneous ones overall (from the plot)
  • The anova table shows that personality and colouration are important variables (p <0.05); Colouration seems to be more important than personality overall (based on the sum of squares in the anova summary). The interaction between personality and colouration is not significant (p>0.05), which indicates that the effect of colour does not depend on the personality (and that the effect of personality does not depend on colour).
  • The Tukey test table shows that - (1) Personality is associated with spawning, but only for heterogeneous coloured fish. (2) In the heterogeneous coloured fish the proactive fish spawn significantly more than the reactive ones. (3) In the homogeneous coloured fish there is no significant difference between the personalities.

26.12 Maze runner

  1. Import the data and graph it (geom_boxplot). Try adding the points to the ggplot using the new (to you) function geom_dotplot(binaxis = "y", stackdir = "center").
maze <- read.csv("CourseData/maze.csv", stringsAsFactors = TRUE)
head(maze)
##     Age nErrors
## 1 Child       2
## 2 Child       4
## 3 Child       2
## 4 Child       5
## 5 Child       6
## 6 Child       0
(A <- ggplot(maze, aes(x = Age, y = nErrors)) +
  geom_boxplot() +
  geom_dotplot(binaxis = "y", stackdir = "center"))

  1. Fit an appropriate GLM.
mod_A <- glm(nErrors ~ Age, data = maze, family = poisson)
  1. Examine the diagnostic plots of the model (autoplot).
library(ggfortify)
autoplot(mod_A)

  1. Get the analysis of variance (deviance) table (anova). What does this tell you?
anova(mod_A, test = "Chi")
## Analysis of Deviance Table
## 
## Model: poisson, link: log
## 
## Response: nErrors
## 
## Terms added sequentially (first to last)
## 
## 
##      Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    
## NULL                    19     36.672              
## Age   1   11.511        18     25.161 0.0006917 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  1. Obtain the summary table. What does this tell you?
summary(mod_A)
## 
## Call:
## glm(formula = nErrors ~ Age, family = poisson, data = maze)
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  0.09531    0.30151   0.316   0.7519   
## AgeChild     1.09861    0.34815   3.156   0.0016 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for poisson family taken to be 1)
## 
##     Null deviance: 36.672  on 19  degrees of freedom
## Residual deviance: 25.161  on 18  degrees of freedom
## AIC: 71.845
## 
## Number of Fisher Scoring iterations: 5
  1. Use the coefficient information in the summary table to get the model predictions for average number of mistakes (plus/minus 95% Confidence interval). Remember that (i) the model summary is on the scale of the linear predictor, and (ii) the 95% CI can be calculated as 1.96 times the standard error. You can do these calculations “by hand”, or using the predict function. Ask for help if you get stuck.
newData <- data.frame(Age = c("Child", "Adult"))
pv <- predict(mod_A, newData, se.fit = TRUE)

newData <- newData %>%
  mutate(nErrors_LP = pv$fit) %>%
  mutate(lowerCI_LP = pv$fit - 1.96 * pv$se.fit) %>%
  mutate(upperCI_LP = pv$fit + 1.96 * pv$se.fit)

inverseFunction <- family(mod_A)$linkinv

newData <- newData %>%
  mutate(nErrors = inverseFunction(nErrors_LP)) %>%
  mutate(lowerCI = inverseFunction(lowerCI_LP)) %>%
  mutate(upperCI = inverseFunction(upperCI_LP))


A + geom_point(
  data = newData, aes(x = Age, y = nErrors),
  colour = "red"
) +
  geom_segment(data = newData, aes(
    x = Age, xend = Age,
    y = lowerCI, yend = upperCI
  ), colour = "red")

26.13 Snails on the move

Simulate a t-test-based analysis in R to figure out what sample size would result in 80% power.

sampleSize <- 40 # vary this until power > 0.8
result <- replicate(
  1000,
  t.test(
    rnorm(sampleSize, mean = 7.4, sd = 2.76),
    rnorm(sampleSize, mean = 7.4 * 1.25, sd = 2.76)
  )$p.value
)
sum(result < 0.05) / 1000
## [1] 0.83

You should find that you need a sample size of about 40 per group.

26.14 Mouse lemur strength

What difference in strength could you reliably detect (with power >80%) with these sample sizes?

diff <- 185 # Vary this to give a difference to 2nd group t-test
result <- replicate(
  1000,
  t.test(
    rnorm(25, mean = 600, sd = 145),
    rnorm(8, mean = 600 - diff, sd = 145)
  )$p.value
)
sum(result < 0.05) / 1000
## [1] 0.812

You should find that you could detect a difference of about 185g or a mean for old animals of 415g which is 70% of the young animals i.e. a 30% reduction.

  1. These values come from the summary tables for the ANOVA model, and the releveled ANOVA model↩︎