3  Homework 0: Basic Descriptive Statistics

Download this assignment — unzip it, open week0-homework.qmd in Positron, and answer inside the answer blocks.

This assignment is about summarizing a dataset numerically and visually: the mean, the standard deviation, and the frequency that something happens. You’ll calculate them by hand and in R, draw them, and ask each time whether the summary is a fair picture of what’s there.

Mean and Standard Deviation

Definitions

Below is a list of a thousand numbers, drawn as dots, with the two things we are about to define marked on it: their mean, and one standard deviation either side of it.

One thousand purple observations shown with their histogram, a solid blue mean line, and dashed lines one standard deviation below and above the mean.

If you have a list of numbers \(X_1, X_2, \ldots, X_n\), the mean (which we call \(\bar X\)), is the sum of the numbers divided by the number of numbers. \[ \bar X = \frac{1}{n}\sum_{i=1}^n X_i \]

The standard deviation (which we call \(\hat\sigma\)) is a measure of how spread out the numbers are. It’s meant to be what it sounds like: the standard (usual) deviation (distance) of number in the list from the list’s mean.

\[ \hat \sigma^2 = \frac{1}{n}\sum_{i=1}^n (X_i - \bar X)^2 \]

Here to come up with one number describing what’s ‘standard’, instead of just taking the average, we do something different. We square our deviations, take the average of squares, and use the square root of the result. This still gives us a number that measures the size of ‘a deviation’ instead of ‘a squared deviation’ because we’ve taken the square root after averaging. But by averaging the squares, we’re effectively making bigger deviations ‘count more’ than smaller ones.1

In the figure above, we visualize a list of n=1000 numbers as purple dots. The x-coordinates are their values \(X_i\) and their y-coordinates are their index \(i\) in the list. The solid blue line indicates their mean \(\bar X\) and the dashed lines one standard deviation away from the mean in either direction, i.e., \(\bar X \pm \hat\sigma\). We also include a histogram of the numbers to show the density of dots near different values of \(x\). As you think about the following exercises, it might make sense to think about what a visualization like this might look like for the lists you’re working with.

Calculations

Exercise 4.1  

Exercise

Each of the following lists has a mean of 50. For which is the standard deviation biggest? Smallest?

  1. 0, 20, 40, 50, 60, 80, 100.
  2. 0, 48, 49, 50, 51, 52, 100
  3. 0, 1, 2, 50, 98, 99, 100

Exercise 4.2  

Exercise

For the two lists below, calculate the mean and standard deviation of the numbers. Then compare the your answers for the two lists. Think about this comparison. How is the first list related to the second? How does this relationship carry over to the mean? The standard deviation?

  1. 1, 3, 4, 5, 7
  2. 6, 8, 9, 10, 12

Exercise 4.3  

Exercise

Repeat this exercise for a new pair of lists.

  1. 1, 3, 4, 5, 7
  2. 3, 9, 12, 15, 21

Exercise 4.4  

Exercise

Repeat it again.

  1. 5, -4, 3, -1, 7
  2. -5, 4, -3, 1, -7

Properties

Exercise 4.5  

Exercise

Can a standard deviation ever be negative? Explain.

Exercise 4.6  

Exercise

For a set of non-negative numbers, can the standard deviation ever be larger than the average? Explain.

Visualization

Consider the three histograms below.

Left-skewed beta-sample histogram concentrated near .7.

Histogram 1

Right-skewed beta-sample histogram concentrated near .3.

Histogram 2

Symmetric beta-sample histogram centered near .5.

Histogram 3

Exercise 4.7  

Exercise

The means of the samples we’ve histogrammed are approximately 0.3, 0.7, 0.5. Which histogram corresponds to which mean?

Exercise 4.8  

Exercise

True or false: the standard deviation of the sample summarized by Histogram 1 is a lot smaller than the one summarized by Histogram 2. Explain.

Exercise 4.9  

Exercise

By hand, on paper, sketch two histograms that have the same mean but clearly different standard deviations. Label which one has the larger standard deviation. Then photograph your sketch and include it in your answer below.

Supreme Court Justices

The Data

Start R and run this block to get the dataset we’ll be working with.

EMdata = read.csv("https://qtm285-1.github.io/assets/data/EMdata.csv")
  • This data is on 27 justices from the Warren (’53 - ’69), Burger (’69 - ’86), and Rehnquist (’86 - ’05) courts
  • The data can be interpreted as a census of justices for the 1953 - 2005 era. Each row is a justice and each column is a variable. The column ‘justice’ is the name of the justice. We’ll be looking at a few other variables.
    • CLlib: The percentage of votes in liberal direction for each justice in civil liberties cases
    • party: the political party that nominated the justice (Republican =0, Democrat=1)
    • ur: the justice is a member of an under-represented group, such as a racial or gender minority (under-represented group=1, not in under-represented group=0)
  • To get you started, I’m going to plot a histogram of the percentage of liberal votes, identifying the mean with a vertical line. You may want to edit this code to answer future questions.
CL.histogram = ggplot(EMdata) + 
           geom_histogram(aes(x=CLlib, y=after_stat(density)), 
                          bins=10, alpha=.3, color='black') +  
           geom_vline(aes(xintercept=mean(CLlib)), 
                          color="blue") +
           xlab("% Support for Liberal Position on Civil Liberties Cases (CLlib)")

CL.histogram

Histogram of Supreme Court justices' percentages of liberal civil-liberties votes, with the mean marked by a blue vertical line.

NoteHow that plot is put together

Here’s that plot one piece at a time. It starts with the dataset, and pieces are added onto it with +. ggplot calls these pieces layers.

ggplot(EMdata) says which dataset we’re drawing from. Every piece added after it looks up its columns there, which is why they can all say CLlib without saying EMdata again.

geom_histogram(aes(x=CLlib, y=after_stat(density)), bins=10, alpha=.3, color='black') draws the bars. A histogram chops the range of a variable into equal intervals—its bins—and draws a bar over each one, whose height says how much of the data falls in each.

  • x=CLlib says the variable from the dataset that we will be histogramming is CLlib.
  • y=after_stat(density) says what the height of the bars is. after_stat is there because that number comes from the histogram’s own calculation and not from a column of the dataset. You could omit y to use ggplot’s default choice, which is the count of justices in each bin. Instead, we ask it to plot density, which chooses the bar’s height so its area is the fraction of the justices that falls into the range described by the histogram bin. The implication is this—if you want to know what fraction of the justices voted the liberal way in between 40% and 60% of their civil liberties cases, you integrate the bar heights over that interval; the area under the histogram between 40 and 60 is that fraction.
  • bins=10 is how many bars to chop the range of CLlib into.
  • alpha=.3 is how opaque the bars are.
  • color='black' is their outline.

geom_vline(aes(xintercept=mean(CLlib)), color="blue") draws one vertical line.

  • xintercept=mean(CLlib) is where along the bottom to put it, here at the mean.
  • color="blue" is what makes it blue.

xlab("% Support for ...") is the text written under the x-axis.

By wrapping our x and y specifications up in an aes, we are specifying that the variables we refer to should be looked up in the dataset we passed in when we said ggplot(EMdata), not where R usually looks for variables. The other inputs—bins=10, alpha=.3, color='black'—sit outside the aes because we want them used directly instead of looked up.

The whole thing is saved as CL.histogram, so to add something to it later I can write CL.histogram + and one more layer, without repeating any of this.

Exercise 5.1  

Exercise

Write your own function to calculate the standard deviation of CLlib (i.e. not using “sd”) and report it. Use ‘sd’ to check your answer.2 Draw a new figure that adds, to the plot above, vertical lines indicating the mean plus and minus 1 and 2 standard deviations. What is the substantive interpretation of this mean?

Tip. To make the plot easier to read and talk about, style these lines differently. I tend to use dashed lines for one standard deviation and dotted lines for two. To do that, pass linetype="dashed" or linetype="dotted" after the color argument for geom_vline.

Exercise 5.2  

Exercise

Replicate the plot above, i.e. a histogram with lines for the mean plus or minus two standard deviations, for the variable ‘ur’. What is this mean? And what is its substantive interpretation? What do you think of your plot? Is it a good visualization of the data?

Tip. To point that same picture at a different variable, you change CLlib everywhere it appears—in both of the aes(...) and in the xlab.

Breaking Down the Data

The next question asks for one plot per party, so you need a copy of the data with only those justices in it. Two bits of notation do that.

To get at one column, put a $ after the dataset’s name. EMdata$party is the list of 27 zeros and ones, one per justice, so EMdata$party == 0 is TRUE for each justice a Republican nominated and FALSE for the rest.

To get at some of the rows, put a condition in front of the comma. EMdata[EMdata$party == 0, ] keeps the rows where that condition is TRUE. What comes before the comma picks rows and what comes after it picks columns, and leaving the part after it empty means all of them. What comes back is a dataset just like EMdata with fewer rows in it, so anything you can do to EMdata you can do to that.

Exercise 5.3  

Exercise

Draw two histograms of CLlib, one for Republican-nominated justices and one for Democrat-nominated justices. Calculate the mean of CLlib in each group. Which is larger, the mean among Republican-nominated justices or Democrat-nominated justices? Give a substantive interpretation of this difference.

You have just drawn the two parties as separate histograms. Here they are in one picture instead: every justice as a dot, the parties side by side, and each party’s mean marked with a line reaching two standard deviations either side of it.

mean_sd = function(x,mult=1) {
  data.frame(y=mean(x),
             ymin=mean(x)-mult*sd(x),
             ymax=mean(x)+mult*sd(x))
}

civil.liberties.plot = ggplot(EMdata) +
  geom_point(aes(x=party, y=CLlib),
             position=position_jitter(w=.1, h=0), alpha=.4) +
  stat_summary(aes(x=party, y=CLlib), geom="pointrange",
               fun.data=mean_sd, fun.args = list(mult=2)) +
  xlab("Nominating Party") +
  ylab("%Support for Liberal Position on CL Cases")

civil.liberties.plot

Liberal civil-liberties voting by nominating party, with individual justices jittered and each party mean plus or minus two standard deviations.

That scatter plot is put together the same way as the histogram was, out of two drawing pieces instead of one.

geom_point(aes(x=party, y=CLlib), position=position_jitter(w=.1, h=0), alpha=.4) draws one dot per justice.

  • x=party and y=CLlib put the nominating party along the bottom and the percentage up the side.
  • position=position_jitter(w=.1, h=0) shifts each dot sideways by a random amount of up to .1, so that justices nominated by the same party don’t land on top of one another. h=0 leaves the height alone, because that one is the justice’s actual percentage and moving it would be a lie.
  • alpha=.4 is how opaque the dots are.

stat_summary(aes(x=party, y=CLlib), geom="pointrange", fun.data=mean_sd, fun.args=list(mult=2)) draws a summary on top of them.

  • It collects the justices that share an x, hands that group’s CLlib values to the function named in fun.data, and draws whatever comes back. Which summary you get is decided entirely by that function.
  • mean_sd is the function written just above. Given a list of numbers it returns a single row with three columns: y is their mean, and ymin and ymax are mult standard deviations below and above it.
  • geom="pointrange" says how to draw that row—a dot at y, and a line running from ymin up to ymax. The columns have to carry those names, because that is what pointrange goes looking for.
  • fun.args=list(mult=2) is how the 2 reaches mean_sd, so the lines run two standard deviations either side of the mean rather than one.

ylab(...) is the label up the side, the way xlab(...) is the one along the bottom.

Exercise 5.4  

Exercise

Above, I’ve drawn a scatter plot of CLlib with the nominating party as the x-axis. I’ve used stat_summary to draw in the mean plus and minus two standard deviations for each group. Calculate these means and standard deviations and report the mean and endpoints of the intervals that have been drawn in. Check your answer visually for agreement with the plot. Can you give a substantive interpretation of these intervals?

Frequencies, Indicators and Means

Introduction

Look at this list of numbers. \[ 1, 2, 3, 4, 5 \]

How many of those numbers are greater than or equal to 3? 3/5 of them are. We pronounce this ‘3 out of 5’, but if we take the division sign in there seriously, we get \(3/5 = .6\). \(.6\)—often we say, equivalently, 60%—is the frequency that one of those five numbers is greater than or equal to 3.

We’ve been talking a lot about means so far and there is a connection. Frequencies are means. Let’s think of the list above as a sample of 5 numbers: \(Y_1=1, Y_2=2, \ldots\). And let’s define, in terms of these, a corresponding sample of zeros and ones, \(O_1 \ldots O_5\).

\[ O_i = \begin{cases} 1 & \text{ if } Y_i \ge 3 \\ 0 & \text{ otherwise } \end{cases} \]

We call these indicators: \(O_i\) indicates whether \(Y_i\) is greater than or equal to \(3\) by being one if it is and zero if it isn’t. And for our list specifically, the indicator list can be written as

\[ O_1 = 0, \ O_2 = 0, \ O_3 = 1, \ O_4 = 1, \ O_5 = 1 \]

What’s the mean of our indicators \(O_1 \ldots O_5\)? \(.6\), right? That’s not a coincidence. The frequency that something happens is the mean of indicators that it does happen. Thinking this way will come in handy because we’ll talk about means a lot in this class and this lets us use all the same ideas to think about frequencies. We do this so often that we have a special notation for indicators. Instead of \(O_i\), we’d usually write \(1_{\ge 3}(Y_i)\) so we don’t have to remember the meaning of a new letter—it’s all there. Indicators aren’t just for something being greater than equal to something else. We could, for example, talk about the indicators \(1_{=3}(Y_i)\) or \(1_{<0}(Y_i)\). I’ll leave it to you to work out what those mean.

Writing indicators this way makes it clear that what we’re doing is evaluating a function at \(Y_i\). A function that is defined like this. \[ 1_{\ge 3}(y) = \begin{cases} 1 & \text{ if } y \ge 3 \\ 0 & \text{ otherwise } \end{cases} \qqtext{ for any value of $y$} \]

\(1_{=3}\) and \(1_{<0}\) are, of course, also functions. We call them indicator functions.

Calculating Frequencies in R

The R code we tend to use to calculate frequencies uses these connections. Here’s one phrased exactly the way we’ve been talking about it, where we first evaluate the indicator function \(1_{\ge 3}\) at the sample \(Y_1 \ldots Y_5\), to get the indicator variables \(1_{\ge 3}(Y_1) \ldots 1_{\ge 3}(Y_5)\), then take their mean to get the frequency we want.

Y = c(1,2,3,4,5)
ge.3 = function(x) { ifelse(x >= 3, 1, 0) }
freq.X.ge.3 = mean(ge.3(Y))
freq.X.ge.3
1
This is the list of numbers we’re talking about. \(Y_1 \ldots Y_n\).
2
This defines the indicator function \(1_{\ge 3}\)
3
This evaluates it to get the the indicator variables \(1_{\ge 3}(Y_i)\) and takes their mean.
[1] 0.6

And here’s what we’d usually write in practice. It’s more compact.

mean(Y >= 3)
[1] 0.6

Visualization

Drawing indicator functions into our data visualizations can help us get a sense of what they mean in the context of the data. In particular, it helps us identify cases where a lot of observations are just outside the region where the indicator is 1. Or just inside. This matters because coarsening loses that distinction: after we’ve merged the rows, the result doesn’t tell us who was just inside and who was far in. It also matters because it’s often effective to talk about frequencies—they’re simple and a lot of people feel comfortable with them—but saying things like ‘only 15% of people in Georgia live below the poverty line’ can be a way of concealing the truth if another 35% are just above it. We’ll be working with income data a few weeks from now. We’ll get a chance to see whether it’s possible to use frequencies to tell two different stories about the same reality.

In the plot below, the dots show the sample \(Y_1 \ldots Y_5 = 1 \ldots 5\) we were talking about earlier. And the blue-shaded rectangle shows the indicator function \(1_{\ge 3}\). The indicator values \(1_{\ge 3}(Y_1) \ldots 1_{\ge 3}(Y_5)\) are 1 for the points inside the rectangle and 0 for the points outside. The frequency we’ve been talking about is represented visually by the proportion of points inside this rectangle.

freq.data = data.frame(Y = c(1,2,3,4,5), 
                       X = c(1,1,1,1,1))

ggplot(freq.data) +
  geom_point(aes(x=X, y = Y)) +
  annotate("rect", xmin = -Inf, xmax = Inf, ymax = Inf, ymin = 3,
           alpha = .1,fill = "blue")                                                         
1
To draw a ‘scatter plot’ when we have Ys but no Xs, we need to make up some Xs. Here we’ve just used 1s so they all appear in one column.
2
This is ‘ggplot’ for the indicator \(1_{\ge 3}\). More explicitly, it’s ggplot for the indicator \(1_{\in [-\infty, +\infty] \times [3, +\infty]}\), where \([-\infty, +\infty] \times [3, +\infty]\) is an infinitely wide ‘rectangle’ that starts at 3 on the y-axis and goes up to infinity.

Five values from one through five in a vertical column, with the region at or above three shaded blue to show the indicator function.

Exercises

All this—the R stuff and the visualizations — starts to get more useful when we have a larger list. We usually do. Let’s check that we’ve got all of this down by doing a few simple exercises using another list of five numbers, then move on to our supreme court data.

Exercise 6.1  

Exercise

Here’s a new list of five numbers. \[ 3, 0, 1, 2, -1 \]

If we call these \(Y_1 \ldots Y_5\), what are the values of the indicator variables \(1_{\le 0}(Y_1) \ldots 1_{\le 0}(Y_5)\)?

Exercise 6.2  

Exercise

Calculate the frequency of numbers less than or equal to 0 three ways: by counting, by writing out indicators and taking the mean, and by writing R code. Are they all the same? I’m looking for a yes/no answer to this question. I’m hoping for a yes. If it’s a no and you’re not sure why, ask me about it.

Exercise 6.3  

Exercise

Adapt the R code above to visualize this new list of numbers and our new indicator function \(1_{\le 0}\). It may help to sketch out what you want then translate your sketch into code. And when you’ve done that, check that the plot is, in fact, what you wanted to draw. Sometimes we mistranslate.

What I’m asking for here is the plot.

Hint. Assuming you’ve re-defined freq.data so that Y is the new list of numbers, all you’ve got to do is adjust the call to annotate to highlight the correct region.

Now that we’ve got all this down, let’s think about the frequency a few things happen in the supreme court. Suppose I want indicators that support for liberal position on civil liberties cases among both parties is greater than or equal to 25%. Denoting percent of support as \(Y_i\), these can be written as \[ 1_{\ge 25}(Y_i) \qfor 1_{\ge 25}(y) = \begin{cases} 1 & \text{ if } y \ge 25 \\ 0 & \text{ otherwise.} \end{cases} \]

Exercise 6.4  

Exercise

Calculate the frequency that support for the liberal position is greater than or equal to 25%. What is it?

If you want to avoid writing a very small amount of code, go ahead and do it using this plot.

CLindicator25.plot = ggplot(EMdata) +  
  geom_point(aes(x=party, y=CLlib), 
             position=position_jitter(w=.1, h=0), alpha=.4) + 
  stat_summary(aes(x=party, y=CLlib), geom="pointrange",
              fun.data=mean_sd, fun.args = list(mult=2)) + 
  annotate("rect", xmin = -Inf, xmax = Inf, ymax = Inf, ymin = 25, alpha = .1,fill = "blue") +
  xlab("Nominating Party") + 
  ylab("%Support for Liberal Position on CL Cases")

CLindicator25.plot

Liberal civil-liberties voting by nominating party; the region at or above 25 percent is shaded blue and contains all but one justice.

What if we want to know the frequency that support for the liberal position exceeds 50% among justices nominated by Republicans? All we’ve got now is to do about the same thing with part of our sample — a subsample. The code below calculates the frequency. It’s the row-picking from before done to a single column rather than the whole table: Y[X==0] keeps the entries of Y where X is 0, the way EMdata[EMdata$party == 0, ] kept the rows.

Y = EMdata$CLlib
X = EMdata$party
mean(Y[X==0] >= 50)
[1] 0.3333333

And this code draws a plot to help us interpret it.

CLindicator50.repub.plot = ggplot(EMdata) +  
  geom_point(aes(x=party, y=CLlib), 
             position=position_jitter(w=.1, h=0), alpha=.4) + 
  stat_summary(aes(x=party, y=CLlib), geom="pointrange", 
               fun.data=mean_sd, fun.args = list(mult=2)) +  
  annotate("rect", xmin = -Inf, xmax = .5, ymax = Inf, ymin = 50, alpha = .1, fill = "blue") +
  xlab("Nominating Party") + 
  ylab("%Support for Liberal Position on CL Cases")

CLindicator50.repub.plot

Liberal civil-liberties voting by nominating party; the Republican region at or above 50 percent is shaded blue.

Exercise 6.5  

Exercise

What if we want to know how often support is greater than or equal to 50% among justices nominated by Democrats? Calculate this frequency. Then, by adapting the R code above, draw a plot to help you interpret it. Do you think this frequency is a reasonable summary of the way Democrat-nominated justices vote in civil liberties cases? With reference to the plot, explain why or why not. What about frequency 33% as a summary of the way Republican-nominated justices vote?


  1. Usually we don’t really need to think at this level of subtlety to have the right intuition. Thinking ‘the usual deviation’ is enough. But it’s good to know what’s going on under the hood in case you do need it.↩︎

  2. You may be slightly off. In particular, you may be off by a factor of \(\sqrt{26/27}\). That’s ok. There are two conventions for calculating the sample standard deviation—one involves division by \(n\) and the other \(n-1\). That’s a factor of 26/27 inside the square root, so it’s the square root of that factor in the standard deviation itself.↩︎