Complete all Exercises, and submit answers to Questions on the Coursera platform.

R Packages

library(dplyr)
## Warning: package 'dplyr' was built under R version 3.3.3
library(ggplot2)
## Warning: package 'ggplot2' was built under R version 3.3.3
library(statsr)

Dataset 1: Dr. Arbuthnot’s Baptism Records

To get you started, run the following command to load the data.

data(arbuthnot)

The Arbuthnot baptism counts for boys and girls. The Arbuthnot data set refers to Dr. John Arbuthnot, an 18th century physician, writer, and mathematician. He was interested in the ratio of newborn boys to newborn girls, so he gathered the baptism records for children born in London for every year from 1629 to 1710. We can take a look at the data by typing its name into the console.

arbuthnot
## # A tibble: 82 x 3
##     year  boys girls
##    <int> <int> <int>
##  1  1629  5218  4683
##  2  1630  4858  4457
##  3  1631  4422  4102
##  4  1632  4994  4590
##  5  1633  5158  4839
##  6  1634  5035  4820
##  7  1635  5106  4928
##  8  1636  4917  4605
##  9  1637  4703  4457
## 10  1638  5359  4952
## # ... with 72 more rows

What you should see are four columns of numbers, each row representing a different year: the first entry in each row is simply the row number (an index we can use to access the data from individual years if we want), the second is the year, and the third and fourth are the numbers of boys and girls baptized that year, respectively. Use the scrollbar on the right side of the console window to examine the complete data set.

You can see the dimensions of this data frame by typing:

dim(arbuthnot)
## [1] 82  3

This command should output [1] 82 3, indicating that there are 82 rows and 3 columns. You can see the names of these columns (or variables) by typing:

names(arbuthnot)
## [1] "year"  "boys"  "girls"
  1. How many variables are included in this data set?
    1. 2
    2. 3
    3. 4
    4. 82
    5. 1710

Exercise: What years are included in this dataset? Hint: Take a look at the year variable in the Data Viewer to answer this question.

arbuthnot$year
##  [1] 1629 1630 1631 1632 1633 1634 1635 1636 1637 1638 1639 1640 1641 1642
## [15] 1643 1644 1645 1646 1647 1648 1649 1650 1651 1652 1653 1654 1655 1656
## [29] 1657 1658 1659 1660 1661 1662 1663 1664 1665 1666 1667 1668 1669 1670
## [43] 1671 1672 1673 1674 1675 1676 1677 1678 1679 1680 1681 1682 1683 1684
## [57] 1685 1686 1687 1688 1689 1690 1691 1692 1693 1694 1695 1696 1697 1698
## [71] 1699 1700 1701 1702 1703 1704 1705 1706 1707 1708 1709 1710

Some Exploration

Let’s start to examine the data a little more closely. We can access the data in a single column of a data frame separately using a command like

arbuthnot$boys
##  [1] 5218 4858 4422 4994 5158 5035 5106 4917 4703 5359 5366 5518 5470 5460
## [15] 4793 4107 4047 3768 3796 3363 3079 2890 3231 3220 3196 3441 3655 3668
## [29] 3396 3157 3209 3724 4748 5216 5411 6041 5114 4678 5616 6073 6506 6278
## [43] 6449 6443 6073 6113 6058 6552 6423 6568 6247 6548 6822 6909 7577 7575
## [57] 7484 7575 7737 7487 7604 7909 7662 7602 7676 6985 7263 7632 8062 8426
## [71] 7911 7578 8102 8031 7765 6113 8366 7952 8379 8239 7840 7640

This command will only show the number of boys baptized each year. The dollar sign basically says “go to the data frame that comes before me, and find the variable that comes after me”.

  1. What command would you use to extract just the counts of girls born?
    1. arbuthnot$boys
    2. arbuthnot$girls
    3. girls
    4. arbuthnot[girls]
    5. $girls 
arbuthnot$girls
##  [1] 4683 4457 4102 4590 4839 4820 4928 4605 4457 4952 4784 5332 5200 4910
## [15] 4617 3997 3919 3395 3536 3181 2746 2722 2840 2908 2959 3179 3349 3382
## [29] 3289 3013 2781 3247 4107 4803 4881 5681 4858 4319 5322 5560 5829 5719
## [43] 6061 6120 5822 5738 5717 5847 6203 6033 6041 6299 6533 6744 7158 7127
## [57] 7246 7119 7214 7101 7167 7302 7392 7316 7483 6647 6713 7229 7767 7626
## [71] 7452 7061 7514 7656 7683 5738 7779 7417 7687 7623 7380 7288

R has some powerful functions for making graphics. We can create a simple plot of the number of girls baptized per year with the command

ggplot(data = arbuthnot, aes(x = year, y = girls)) +
  geom_point()

Before we review the code for this plot, let’s summarize the trends we see in the data.

There is initially an increase in the number of girls baptised, which peaks around 1640. After 1640 there is a decrease in the number of girls baptised, but the number begins to increase again in 1660. Overall the trend is an increase in the number of girls baptised.

Back to the code… We use the ggplot() function to build plots. If you run the plotting code in your console, you should see the plot appear under the Plots tab of the lower right panel of RStudio. Notice that the command above again looks like a function, this time with arguments separated by commas.

  • The first argument is always the dataset.
  • Next, we provide the variables from the dataset to be assigned to aesthetic elements of the plot, e.g. the x and the y axes.
  • Finally, we use another layer, separated by a + to specify the geometric object for the plot. Since we want to scatterplot, we use geom_point.

You might wonder how you are supposed to know the syntax for the ggplot function. Thankfully, R documents all of its functions extensively. To read what a function does and learn the arguments that are available to you, just type in a question mark followed by the name of the function that you’re interested in. Try the following in your console:

?ggplot
## starting httpd help server ... done

Notice that the help file replaces the plot in the lower right panel. You can toggle between plots and help files using the tabs at the top of that panel.

More extensive help for plotting with the ggplot2 package can be found at http://docs.ggplot2.org/current/. The best (and easiest) way to learn the syntax is to take a look at the sample plots provided on that page, and modify the code bit by bit until you get achieve the plot you want.

R as a big calculator

Now, suppose we want to plot the total number of baptisms. To compute this, we could use the fact that R is really just a big calculator. We can type in mathematical expressions like

5218 + 4683
## [1] 9901

to see the total number of baptisms in 1629. We could repeat this once for each year, but there is a faster way. If we add the vector for baptisms for boys to that of girls, R will compute all sums simultaneously.

arbuthnot$boys + arbuthnot$girls
##  [1]  9901  9315  8524  9584  9997  9855 10034  9522  9160 10311 10150
## [12] 10850 10670 10370  9410  8104  7966  7163  7332  6544  5825  5612
## [23]  6071  6128  6155  6620  7004  7050  6685  6170  5990  6971  8855
## [34] 10019 10292 11722  9972  8997 10938 11633 12335 11997 12510 12563
## [45] 11895 11851 11775 12399 12626 12601 12288 12847 13355 13653 14735
## [56] 14702 14730 14694 14951 14588 14771 15211 15054 14918 15159 13632
## [67] 13976 14861 15829 16052 15363 14639 15616 15687 15448 11851 16145
## [78] 15369 16066 15862 15220 14928

What you will see are 82 numbers (in that packed display, because we aren’t looking at a data frame here), each one representing the sum we’re after. Take a look at a few of them and verify that they are right.

Adding a new variable to the data frame

We’ll be using this new vector to generate some plots, so we’ll want to save it as a permanent column in our data frame.

arbuthnot <- arbuthnot %>%
  mutate(total = boys + girls)
## Warning: package 'bindrcpp' was built under R version 3.3.3

What in the world is going on here? The %>% operator is called the piping operator. Basically, it takes the output of the current line and pipes it into the following line of code.

A note on piping: Note that we can read these three lines of code as the following:

“Take the arbuthnot dataset and pipe it into the mutate function. Using this mutate a new variable called total that is the sum of the variables called boys and girls. Then assign this new resulting dataset to the object called arbuthnot, i.e. overwrite the old arbuthnot dataset with the new one containing the new variable.”

This is essentially equivalent to going through each row and adding up the boys and girls counts for that year and recording that value in a new column called total.

Where is the new variable? When you make changes to variables in your dataset, click on the name of the dataset again to update it in the data viewer.

You’ll see that there is now a new column called total that has been tacked on to the data frame. The special symbol <- performs an assignment, taking the output of one line of code and saving it into an object in your workspace. In this case, you already have an object called arbuthnot, so this command updates that data set with the new mutated column.

We can make a plot of the total number of baptisms per year with the following command.

ggplot(data = arbuthnot, aes(x = year, y = total)) +
  geom_line()

Note that using geom_line() instead of geom_point() results in a line plot instead of a scatter plot. You want both? Just layer them on:

ggplot(data = arbuthnot, aes(x = year, y = total)) +
  geom_line() +
  geom_point()

Exercise: Now, generate a plot of the proportion of boys born over time. What do you see?

# type your code for the Exercise here, and Knit
ggplot(data = arbuthnot, aes(x=year,y=boys)) + geom_line() + geom_point()

Finally, in addition to simple mathematical operators like subtraction and division, you can ask R to make comparisons like greater than, >, less than, <, and equality, ==. For example, we can ask if boys outnumber girls in each year with the expression

arbuthnot <- arbuthnot %>%
  mutate(more_boys = boys > girls)

This command add a new variable to the arbuthnot data frame containing the values of either TRUE if that year had more boys than girls, or FALSE if that year did not (the answer may surprise you). This variable contains different kind of data than we have considered so far. All other columns in the arbuthnot data frame have values are numerical (the year, the number of boys and girls). Here, we’ve asked R to create logical data, data where the values are either TRUE or FALSE. In general, data analysis will involve many different kinds of data types, and one reason for using R is that it is able to represent and compute with many of them.

Dataset 2: Present birth records

In the previous few pages, you recreated some of the displays and preliminary analysis of Arbuthnot’s baptism data. Next you will do a similar analysis, but for present day birth records in the United States. Load up the present day data with the following command.

data(present)

The data are stored in a data frame called present which should now be loaded in your workspace.

  1. How many variables are included in this data set?
    1. 2
    2. 3
    3. 4
    4. 74
    5. 2013 
# type your code for Question 4 here, and Knit
View(present)

Exercise: What years are included in this dataset? Hint: Use the range function and present$year as its argument.

# type your code for Exercise here, and Knit
range(present$year)
## [1] 1940 2013
  1. Calculate the total number of births for each year and store these values in a new variable called total in the present dataset. Then, calculate the proportion of boys born each year and store these values in a new variable called prop_boys in the same dataset. Plot these values over time and based on the plot determine if the following statement is true or false: The proportion of boys born in the US has decreased over time.
    1. True
    2. False 
# type your code for Question 5 here, and Knit
present <- present %>% mutate(total = boys + girls)
present <- present %>% mutate(prop_boys = boys/total)
ggplot(present, aes(x=year,y=prop_boys)) + geom_line() + geom_point()

  1. Create a new variable called more_boys which contains the value of either TRUE if that year had more boys than girls, or FALSE if that year did not. Based on this variable which of the following statements is true?
    1. Every year there are more girls born than boys.
    2. Every year there are more boys born than girls.
    3. Half of the years there are more boys born, and the other half more girls born. 
# type your code for Question 6 here, and Knit
present <- present %>% mutate(more_boys = boys > girls)
present$more_boys
##  [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [15] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [29] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [43] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [57] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [71] TRUE TRUE TRUE TRUE
  1. Calculate the boy-to-girl ratio each year, and store these values in a new variable called prop_boy_girl in the present dataset. Plot these values over time. Which of the following best describes the trend?
    1. There appears to be no trend in the boy-to-girl ratio from 1940 to 2013.
    2. There is initially an increase in boy-to-girl ratio, which peaks around 1960. After 1960 there is a decrease in the boy-to-girl ratio, but the number begins to increase in the mid 1970s.
    3. There is initially a decrease in the boy-to-girl ratio, and then an increase between 1960 and 1970, followed by a decrease.
    4. The boy-to-girl ratio has increased over time.
    5. There is an initial decrease in the boy-to-girl ratio born but this number appears to level around 1960 and remain constant since then. 
# type your code for Question 7 here, and Knit
present <- present %>% mutate(prop_boy_girl = boys/girls)
ggplot(data = present, aes(x=year,y=prop_boy_girl)) + geom_line() + geom_point()

  1. In what year did we see the most total number of births in the U.S.? Hint: Sort your dataset in descending order based on the total column. You can do this interactively in the data viewer by clicking on the arrows next to the variable names. Or to arrange the data in a descenting order with new function: descr (for descending order).
    1. 1940
    2. 1957
    3. 1961
    4. 1991
    5. 2007 
# type your code for Question 8 here
# sample code is provided below, edit as necessary, uncomment, and then Knit
#present %>%
#  mutate(total = ?) %>%
#  arrange(desc(total))
present$year[which.max(present$total)]
## [1] 2007