Pre course assignment of Applied Statics
This is a pre course homework of Applied Statics. In this file, we need to complete software installation, demonstrate competency in Markdown, and in the basics of ggplot.
Task 1: A brief biography
Hi, my name is Xinyue Zhang and I am from China.
Here is some basic information about me.
- Undergraduate : Shanghai University of Finance and Economics, School of Information Management and Engineering
- Graduate : London Business School
- Hometown: Changchun, China
- Birthday: March 7,1999
As for work experience, I have great passion in product management and data analysis. Please refer to my LinkedIn Profilefor more details.
This is the photo of LBS campus.
Task 2: gapminder country comparison
You have seen the gapminder dataset that has data on life expectancy, population, and GDP per capita for 142 countries from 1952 to 2007. To get a glimpse of the dataframe, namely to see the variable names, variable types, etc., we use the glimpse function. We also want to have a look at the first 20 rows of data.
glimpse(gapminder)## Rows: 1,704
## Columns: 6
## $ country <fct> "Afghanistan", "Afghanistan", "Afghanistan", "Afghanistan", ~
## $ continent <fct> Asia, Asia, Asia, Asia, Asia, Asia, Asia, Asia, Asia, Asia, ~
## $ year <int> 1952, 1957, 1962, 1967, 1972, 1977, 1982, 1987, 1992, 1997, ~
## $ lifeExp <dbl> 28.801, 30.332, 31.997, 34.020, 36.088, 38.438, 39.854, 40.8~
## $ pop <int> 8425333, 9240934, 10267083, 11537966, 13079460, 14880372, 12~
## $ gdpPercap <dbl> 779.4453, 820.8530, 853.1007, 836.1971, 739.9811, 786.1134, ~head(gapminder, 20) # look at the first 20 rows of the dataframe## # A tibble: 20 x 6
## country continent year lifeExp pop gdpPercap
## <fct> <fct> <int> <dbl> <int> <dbl>
## 1 Afghanistan Asia 1952 28.8 8425333 779.
## 2 Afghanistan Asia 1957 30.3 9240934 821.
## 3 Afghanistan Asia 1962 32.0 10267083 853.
## 4 Afghanistan Asia 1967 34.0 11537966 836.
## 5 Afghanistan Asia 1972 36.1 13079460 740.
## 6 Afghanistan Asia 1977 38.4 14880372 786.
## 7 Afghanistan Asia 1982 39.9 12881816 978.
## 8 Afghanistan Asia 1987 40.8 13867957 852.
## 9 Afghanistan Asia 1992 41.7 16317921 649.
## 10 Afghanistan Asia 1997 41.8 22227415 635.
## 11 Afghanistan Asia 2002 42.1 25268405 727.
## 12 Afghanistan Asia 2007 43.8 31889923 975.
## 13 Albania Europe 1952 55.2 1282697 1601.
## 14 Albania Europe 1957 59.3 1476505 1942.
## 15 Albania Europe 1962 64.8 1728137 2313.
## 16 Albania Europe 1967 66.2 1984060 2760.
## 17 Albania Europe 1972 67.7 2263554 3313.
## 18 Albania Europe 1977 68.9 2509048 3533.
## 19 Albania Europe 1982 70.4 2780097 3631.
## 20 Albania Europe 1987 72 3075321 3739.Your task is to produce two graphs of how life expectancy has changed over the years for the country and the continent you come from.
I have created the country_data and continent_data with the code below.
country_data <- gapminder %>%
filter(country == "Greece") # just choosing Greece, as this is where I come from
continent_data <- gapminder %>%
filter(continent == "Asia")First, create a plot of life expectancy over time for the single country you chose. Map year on the x-axis, and lifeExp on the y-axis. You should also use geom_point() to see the actual data points and geom_smooth(se = FALSE) to plot the underlying trendlines. You need to remove the comments # from the lines below for your code to run.
plot1 <- ggplot(data = country_data, mapping = aes(x = year , y = lifeExp))+
geom_point() +
geom_smooth(se = FALSE)+
NULL
plot1## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
Next we need to add a title. Create a new plot, or extend plot1, using the labs() function to add an informative title to the plot.
plot1<- plot1 +
labs(title = " Life expectancy of Greece ",
x = " Year ",
y = " Life expectancy in years") +
NULL
plot1## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
Secondly, produce a plot for all countries in the continent you come from. (Hint: map the country variable to the colour aesthetic. You also want to map country to the group aesthetic, so all points for each country are grouped together).
ggplot(continent_data, mapping = aes(x = year , y = lifeExp , colour= country , group = country))+
geom_point() +
geom_smooth(se = FALSE) +
NULL## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
Finally, using the original gapminder data, produce a life expectancy over time graph, grouped (or faceted) by continent. We will remove all legends, adding the theme(legend.position="none") in the end of our ggplot.
ggplot(data = gapminder , mapping = aes(x = year , y = lifeExp , colour= country))+
geom_point() +
geom_smooth(se = FALSE) +
facet_wrap(~continent) +
theme(legend.position="none") + #remove all legends
NULL## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
Given these trends, what can you say about life expectancy since 1952? Again, don’t just say what’s happening in the graph. Tell some sort of story and speculate about the differences in the patterns.
Type your answer after this blockquote.
Since 1952, life expectancy has increased all over the world. However,after 1985, life expectancy in several African countries sharply decreased. This could be caused by wrong record of data or sever war.Developed countries in Europe and Oceania had a higher life expectancy in 1952 and steadily grows to around 80 during this period. And there is still one country in Asia that has a life expectancy for only 40.
Task 3: Brexit vote analysis
We will have a look at the results of the 2016 Brexit vote in the UK. First we read the data using read_csv() and have a quick glimpse at the data
brexit_results <- read_csv(here::here("data","brexit_results.csv"))
glimpse(brexit_results)## Rows: 632
## Columns: 11
## $ Seat <chr> "Aldershot", "Aldridge-Brownhills", "Altrincham and Sale W~
## $ con_2015 <dbl> 50.592, 52.050, 52.994, 43.979, 60.788, 22.418, 52.454, 22~
## $ lab_2015 <dbl> 18.333, 22.369, 26.686, 34.781, 11.197, 41.022, 18.441, 49~
## $ ld_2015 <dbl> 8.824, 3.367, 8.383, 2.975, 7.192, 14.828, 5.984, 2.423, 1~
## $ ukip_2015 <dbl> 17.867, 19.624, 8.011, 15.887, 14.438, 21.409, 18.821, 21.~
## $ leave_share <dbl> 57.89777, 67.79635, 38.58780, 65.29912, 49.70111, 70.47289~
## $ born_in_uk <dbl> 83.10464, 96.12207, 90.48566, 97.30437, 93.33793, 96.96214~
## $ male <dbl> 49.89896, 48.92951, 48.90621, 49.21657, 48.00189, 49.17185~
## $ unemployed <dbl> 3.637000, 4.553607, 3.039963, 4.261173, 2.468100, 4.742731~
## $ degree <dbl> 13.870661, 9.974114, 28.600135, 9.336294, 18.775591, 6.085~
## $ age_18to24 <dbl> 9.406093, 7.325850, 6.437453, 7.747801, 5.734730, 8.209863~The data comes from Elliott Morris, who cleaned it and made it available through his DataCamp class on analysing election and polling data in R.
Our main outcome variable (or y) is leave_share, which is the percent of votes cast in favour of Brexit, or leaving the EU. Each row is a UK parliament constituency.
To get a sense of the spread, or distribution, of the data, we can plot a histogram, a density plot, and the empirical cumulative distribution function of the leave % in all constituencies.
# histogram
ggplot(brexit_results, aes(x = leave_share)) +
geom_histogram(binwidth = 2.5)+
labs(x='Percent of votes cast in favour of Brexit (%)',
title='Histogram'
)+
NULL
# density plot-- think smoothed histogram
ggplot(brexit_results, aes(x = leave_share)) +
geom_density()+
labs(x='Percent of votes cast in favour of Brexit (%)',
title='Density Plot'
)+
NULL
# The empirical cumulative distribution function (ECDF)
ggplot(brexit_results, aes(x = leave_share)) +
stat_ecdf(geom = "step", pad = FALSE) +
scale_y_continuous(labels = scales::percent)+
labs(x='Percent of votes cast in favour of Brexit (%)',
y='Cumulative percent',
title='Cumulative distribution '
)+
NULL
One common explanation for the Brexit outcome was fear of immigration and opposition to the EU’s more open border policy. We can check the relationship (or correlation) between the proportion of native born residents (born_in_uk) in a constituency and its leave_share. To do this, let us get the correlation between the two variables
brexit_results %>%
select(leave_share, born_in_uk) %>%
cor()## leave_share born_in_uk
## leave_share 1.0000000 0.4934295
## born_in_uk 0.4934295 1.0000000The correlation is almost 0.5, which shows that the two variables are positively correlated.
We can also create a scatterplot between these two variables using geom_point. We also add the best fit line, using geom_smooth(method = "lm").
ggplot(brexit_results, aes(x = born_in_uk, y = leave_share)) +
geom_point(alpha=0.3) +
# add a smoothing line, and use method="lm" to get the best straight-line
geom_smooth(method = "lm") +
# use a white background and frame the plot with a black box
theme_bw() +
labs(x='Proportion of native born residents (%)',
y=' Rercent of votes cast in favour of Brexit (%)',
title='Relationship between the UK born residents and Berxit',
subtitle='UK born residents tend to vote for Brexit')+
NULL## `geom_smooth()` using formula 'y ~ x'
You have the code for the plots, I would like you to revisit all of them and use the labs() function to add an informative title, subtitle, and axes titles to all plots.
What can you say about the relationship shown above? Again, don’t just say what’s happening in the graph. Tell some sort of story and speculate about the differences in the patterns.
Type your answer after, and outside, this blockquote.
Residents born in UK tend to vote for Brexit since they were fear of immigration and opposition to the EU’s more open border policy.They tend to believe that continous immigrants will bring continous labour and it will be harder for UK residents to find a job.
Task 4: Animal rescue incidents attended by the London Fire Brigade
The London Fire Brigade attends a range of non-fire incidents (which we call ‘special services’). These ‘special services’ include assistance to animals that may be trapped or in distress. The data is provided from January 2009 and is updated monthly. A range of information is supplied for each incident including some location information (postcode, borough, ward), as well as the data/time of the incidents. We do not routinely record data about animal deaths or injuries.
Please note that any cost included is a notional cost calculated based on the length of time rounded up to the nearest hour spent by Pump, Aerial and FRU appliances at the incident and charged at the current Brigade hourly rate.
url <- "https://data.london.gov.uk/download/animal-rescue-incidents-attended-by-lfb/8a7d91c2-9aec-4bde-937a-3998f4717cd8/Animal%20Rescue%20incidents%20attended%20by%20LFB%20from%20Jan%202009.csv"
animal_rescue <- read_csv(url,
locale = locale(encoding = "CP1252")) %>%
janitor::clean_names()
glimpse(animal_rescue)## Rows: 7,873
## Columns: 31
## $ incident_number <chr> "139091", "275091", "2075091", "2872091"~
## $ date_time_of_call <chr> "01/01/2009 03:01", "01/01/2009 08:51", ~
## $ cal_year <dbl> 2009, 2009, 2009, 2009, 2009, 2009, 2009~
## $ fin_year <chr> "2008/09", "2008/09", "2008/09", "2008/0~
## $ type_of_incident <chr> "Special Service", "Special Service", "S~
## $ pump_count <chr> "1", "1", "1", "1", "1", "1", "1", "1", ~
## $ pump_hours_total <chr> "2", "1", "1", "1", "1", "1", "1", "1", ~
## $ hourly_notional_cost <dbl> 255, 255, 255, 255, 255, 255, 255, 255, ~
## $ incident_notional_cost <chr> "510", "255", "255", "255", "255", "255"~
## $ final_description <chr> "Redacted", "Redacted", "Redacted", "Red~
## $ animal_group_parent <chr> "Dog", "Fox", "Dog", "Horse", "Rabbit", ~
## $ originof_call <chr> "Person (land line)", "Person (land line~
## $ property_type <chr> "House - single occupancy", "Railings", ~
## $ property_category <chr> "Dwelling", "Outdoor Structure", "Outdoo~
## $ special_service_type_category <chr> "Other animal assistance", "Other animal~
## $ special_service_type <chr> "Animal assistance involving livestock -~
## $ ward_code <chr> "E05011467", "E05000169", "E05000558", "~
## $ ward <chr> "Crystal Palace & Upper Norwood", "Woods~
## $ borough_code <chr> "E09000008", "E09000008", "E09000029", "~
## $ borough <chr> "Croydon", "Croydon", "Sutton", "Hilling~
## $ stn_ground_name <chr> "Norbury", "Woodside", "Wallington", "Ru~
## $ uprn <chr> "NULL", "NULL", "NULL", "100021491149.00~
## $ street <chr> "Waddington Way", "Grasmere Road", "Mill~
## $ usrn <chr> "20500146.00", "NULL", "NULL", "21401484~
## $ postcode_district <chr> "SE19", "SE25", "SM5", "UB9", "RM3", "RM~
## $ easting_m <chr> "NULL", "534785", "528041", "504689", "N~
## $ northing_m <chr> "NULL", "167546", "164923", "190685", "N~
## $ easting_rounded <dbl> 532350, 534750, 528050, 504650, 554650, ~
## $ northing_rounded <dbl> 170050, 167550, 164950, 190650, 192350, ~
## $ latitude <chr> "NULL", "51.39095371", "51.36894086", "5~
## $ longitude <chr> "NULL", "-0.064166887", "-0.161985191", ~One of the more useful things one can do with any data set is quick counts, namely to see how many observations fall within one category. For instance, if we wanted to count the number of incidents by year, we would either use group_by()... summarise() or, simply count()
animal_rescue %>%
dplyr::group_by(cal_year) %>%
summarise(count=n())## # A tibble: 13 x 2
## cal_year count
## <dbl> <int>
## 1 2009 568
## 2 2010 611
## 3 2011 620
## 4 2012 603
## 5 2013 585
## 6 2014 583
## 7 2015 540
## 8 2016 604
## 9 2017 539
## 10 2018 610
## 11 2019 604
## 12 2020 758
## 13 2021 648animal_rescue %>%
count(cal_year, name="count")## # A tibble: 13 x 2
## cal_year count
## <dbl> <int>
## 1 2009 568
## 2 2010 611
## 3 2011 620
## 4 2012 603
## 5 2013 585
## 6 2014 583
## 7 2015 540
## 8 2016 604
## 9 2017 539
## 10 2018 610
## 11 2019 604
## 12 2020 758
## 13 2021 648Let us try to see how many incidents we have by animal group. Again, we can do this either using group_by() and summarise(), or by using count()
animal_rescue %>%
group_by(animal_group_parent) %>%
#group_by and summarise will produce a new column with the count in each animal group
summarise(count = n()) %>%
# mutate adds a new column; here we calculate the percentage
mutate(percent = round(100*count/sum(count),2)) %>%
# arrange() sorts the data by percent. Since the default sorting is min to max and we would like to see it sorted
# in descending order (max to min), we use arrange(desc())
arrange(desc(percent))## # A tibble: 28 x 3
## animal_group_parent count percent
## <chr> <int> <dbl>
## 1 Cat 3783 48.0
## 2 Bird 1631 20.7
## 3 Dog 1230 15.6
## 4 Fox 373 4.74
## 5 Unknown - Domestic Animal Or Pet 201 2.55
## 6 Horse 195 2.48
## 7 Deer 136 1.73
## 8 Unknown - Wild Animal 94 1.19
## 9 Squirrel 68 0.86
## 10 Unknown - Heavy Livestock Animal 50 0.64
## # ... with 18 more rowsanimal_rescue %>%
#count does the same thing as group_by and summarise
# name = "count" will call the column with the counts "count" ( exciting, I know)
# and 'sort=TRUE' will sort them from max to min
count(animal_group_parent, name="count", sort=TRUE) %>%
mutate(percent = round(100*count/sum(count),2))## # A tibble: 28 x 3
## animal_group_parent count percent
## <chr> <int> <dbl>
## 1 Cat 3783 48.0
## 2 Bird 1631 20.7
## 3 Dog 1230 15.6
## 4 Fox 373 4.74
## 5 Unknown - Domestic Animal Or Pet 201 2.55
## 6 Horse 195 2.48
## 7 Deer 136 1.73
## 8 Unknown - Wild Animal 94 1.19
## 9 Squirrel 68 0.86
## 10 Unknown - Heavy Livestock Animal 50 0.64
## # ... with 18 more rowsDo you see anything strange in these tables?
- There are two ‘cat’ in animal group.
Finally, let us have a loot at the notional cost for rescuing each of these animals. As the LFB says,
Please note that any cost included is a notional cost calculated based on the length of time rounded up to the nearest hour spent by Pump, Aerial and FRU appliances at the incident and charged at the current Brigade hourly rate.
There is two things we will do:
- Calculate the mean and median
incident_notional_costfor eachanimal_group_parent - Plot a boxplot to get a feel for the distribution of
incident_notional_costbyanimal_group_parent.
Before we go on, however, we need to fix incident_notional_cost as it is stored as a chr, or character, rather than a number.
# what type is variable incident_notional_cost from dataframe `animal_rescue`
typeof(animal_rescue$incident_notional_cost)## [1] "character"# readr::parse_number() will convert any numerical values stored as characters into numbers
animal_rescue <- animal_rescue %>%
# we use mutate() to use the parse_number() function and overwrite the same variable
mutate(incident_notional_cost = parse_number(incident_notional_cost))
# incident_notional_cost from dataframe `animal_rescue` is now 'double' or numeric
typeof(animal_rescue$incident_notional_cost)## [1] "double"Now that incident_notional_cost is numeric, let us quickly calculate summary statistics for each animal group.
animal_rescue %>%
# group by animal_group_parent
group_by(animal_group_parent) %>%
# filter resulting data, so each group has at least 6 observations
filter(n()>6) %>%
# summarise() will collapse all values into 3 values: the mean, median, and count
# we use na.rm=TRUE to make sure we remove any NAs, or cases where we do not have the incident cos
summarise(mean_incident_cost = mean (incident_notional_cost, na.rm=TRUE),
median_incident_cost = median (incident_notional_cost, na.rm=TRUE),
sd_incident_cost = sd (incident_notional_cost, na.rm=TRUE),
min_incident_cost = min (incident_notional_cost, na.rm=TRUE),
max_incident_cost = max (incident_notional_cost, na.rm=TRUE),
count = n()) %>%
# sort the resulting data in descending order. You choose whether to sort by count or mean cost.
arrange(desc(mean_incident_cost))## # A tibble: 16 x 7
## animal_group_parent mean_incident_co~ median_incident_~ sd_incident_cost
## <chr> <dbl> <dbl> <dbl>
## 1 Horse 740. 596 541.
## 2 Cow 599. 436 451.
## 3 Unknown - Wild Animal 416. 333 322.
## 4 Deer 415. 333 282.
## 5 Unknown - Heavy Livesto~ 374. 260 263.
## 6 Fox 374. 328 205.
## 7 Snake 356. 339 105.
## 8 Dog 347. 298 168.
## 9 Bird 344. 328 134.
## 10 Cat 344. 326 160.
## 11 Unknown - Domestic Anim~ 326. 295 116.
## 12 cat 324. 290 94.1
## 13 Hamster 315. 290 95.0
## 14 Squirrel 314. 326 56.7
## 15 Ferret 309. 333 39.4
## 16 Rabbit 309. 326 32.2
## # ... with 3 more variables: min_incident_cost <dbl>, max_incident_cost <dbl>,
## # count <int>Compare the mean and the median for each animal group. what do you think this is telling us? Anything else that stands out? Any outliers? - Horse and Cow are especially expensive to rescue in incidents and people should have an expectation about this.One outliers in horse incident cost 3480.
Finally, let us plot a few plots that show the distribution of incident_cost for each animal group.
# base_plot
base_plot <- animal_rescue %>%
group_by(animal_group_parent) %>%
filter(n()>6) %>%
ggplot(aes(x=incident_notional_cost))+
facet_wrap(~animal_group_parent, scales = "free")+
theme_bw()
base_plot + geom_histogram()
base_plot + geom_density()
base_plot + geom_boxplot()
base_plot + stat_ecdf(geom = "step", pad = FALSE) +
scale_y_continuous(labels = scales::percent)
Which of these four graphs do you think best communicates the variability of the incident_notional_cost values? Also, can you please tell some sort of story (which animals are more expensive to rescue than others, the spread of values) and speculate about the differences in the patterns.
- The boxplot best communicates the variability of the
incident_notional_costvalues.
Horses are more expensive to rescue than others and have many outliers. When watching Rick and Morty I realize there is a job called horse surgeon. I assume that other animals are rescued by normal vets and horse is different. Therefore the cost is higher.
Submit the assignment
Knit the completed R Markdown file as an HTML document (use the “Knit” button at the top of the script editor window) and upload it to Canvas.
Details
If you want to, please answer the following
- Who did you collaborate with: Myself
- Approximately how much time did you spend on this problem set: 6 hours
- What, if anything, gave you the most trouble: Memorizing the ggplot functions are difficult.