---
title: "Homework 0: Basic Descriptive Statistics"
tlda-classroom-server: "https://pic.cormorant-matrix.ts.net"
tlda-classroom-assignment: "week0-homework"
tlda-answer-baseline: "week0-homework.qmd.support/baseline.txt"
filters:
  - "week0-homework.qmd.support/answer-placement-warning.lua"
---

{{< include shared-code.qmd >}}

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.

```{r}
#| echo: false
#| fig-alt: "One thousand purple observations shown with their histogram, a solid blue mean line, and dashed lines one standard deviation below and above the mean."
Y = rpois(1000,20)/20
ggplot(data.frame(x=1:length(Y), y=Y)) + 
  geom_point(aes(y=x,x=y), alpha=.2, size=1/2, color='purple') + 
  geom_vline(aes(xintercept=mean(y)), color='blue', alpha=.5) + 
  geom_vline(aes(xintercept=mean(y)-sd(y)), color='blue', alpha=.5, linetype='dashed') + 
  geom_vline(aes(xintercept=mean(y)+sd(y)), color='blue', alpha=.5, linetype='dashed') +
  geom_histogram(aes(x=y, y=300*after_stat(density)), bins=20, alpha=.15) + 
  xlab('') + ylab('') + theme(axis.text.y=element_blank(), axis.text.x=element_blank())
```


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.^[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.]


In the figure above, we visualize a list of n=`r length(Y)` 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

::: {#exr-calculations-1 .callout-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  

:::

::: {#ans-exr-calculations-1 .callout-answer}

*(your answer here)*

:::


::: {#exr-calculations-2 .callout-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 

:::

::: {#ans-exr-calculations-2 .callout-answer}

*(your answer here)*

:::

::: {#exr-calculations-3 .callout-exercise}
Repeat this exercise for a new pair of lists. 

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

:::

::: {#ans-exr-calculations-3 .callout-answer}

*(your answer here)*

:::

::: {#exr-calculations-4 .callout-exercise}
Repeat it again. 

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

:::

::: {#ans-exr-calculations-4 .callout-answer}

*(your answer here)*

:::


## Properties

::: {#exr-properties-1 .callout-exercise}
Can a standard deviation ever be negative? Explain.
:::

::: {#ans-exr-properties-1 .callout-answer}

*(your answer here)*

:::

::: {#exr-properties-2 .callout-exercise}
For a set of non-negative numbers, can the standard deviation ever be larger than the average? Explain.
:::

::: {#ans-exr-properties-2 .callout-answer}

*(your answer here)*

:::


## Visualization

```{r}
#| echo: false
set.seed(220220)
sample.1 = rbeta(10000, 5, 2)
sample.2 = rbeta(10000, 2, 5)
sample.3 = rbeta(10000, 5, 5)
```

Consider the three histograms below.

```{r}
#| echo: false
#| warning: false
#| layout-nrow: 3
#| fig-cap:
#|  - "Histogram 1"
#|  - "Histogram 2"
#|  - "Histogram 3"
#| fig-alt:
#| - "Left-skewed beta-sample histogram concentrated near .7."
#| - "Right-skewed beta-sample histogram concentrated near .3."
#| - "Symmetric beta-sample histogram centered near .5."

ggplot() + geom_histogram(aes(x=sample.1, y=after_stat(density)), bins=100, alpha=.3, color='black') + xlim(-.25,1.25) + xlab('')
ggplot() + geom_histogram(aes(x=sample.2, y=after_stat(density)), bins=100, alpha=.3, color='black') + xlim(-.25,1.25) + xlab('')
ggplot() + geom_histogram(aes(x=sample.3, y=after_stat(density)), bins=100, alpha=.3, color='black') + xlim(-.25,1.25) + xlab('')
```

::: {#exr-visual-1 .callout-exercise}
The means of the samples we've histogrammed are approximately `r round(mean(sample.2),1)`,
`r round(mean(sample.1),1)`, `r round(mean(sample.3),1)`. Which histogram corresponds to which mean?
:::

::: {#ans-exr-visual-1 .callout-answer}

*(your answer here)*

:::


::: {#exr-visual-2 .callout-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.
:::

::: {#ans-exr-visual-2 .callout-answer}

*(your answer here)*

:::

::: {#exr-sketch-by-hand .callout-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.
:::


::: {#ans-exr-sketch-by-hand .callout-answer}

*(paste your photo here. be sure to identify which has the larger standard deviation)*

:::

# Supreme Court Justices

## The Data
Start **R** and run this block to get the dataset we'll be working with.

```{r}
#| echo: true
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.

```{r}
#| echo: true
#| fig-alt: "Histogram of Supreme Court justices' percentages of liberal civil-liberties votes, with the mean marked by a blue vertical line."

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
```

::: {.callout-note title="How 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.
:::

::: { #exr-justices-calc .callout-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.^[You may be slightly off. In particular, you may be off by a factor of $\sqrt{`r n=nrow(EMdata); n-1`/`r n`}$. 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 `r n-1`/`r n` inside the square root, so it's the square root of that factor in the standard deviation itself.]
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`.
:::


::: {#ans-exr-justices-calc .callout-answer}

```{r}
my.sd = function(x) {
                                                                              # <1>
}

CL.histogram                                                                  # <2>
```
1. Your calculation goes here.
2. Add your lines to this plot.

*(your interpretation here)*

:::

::: {#exr-justices-histogram-ur .callout-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`.
:::


::: {#ans-exr-justices-histogram-ur .callout-answer}

```{r}
# your plot goes here
```

*(your answer here)*

:::

## 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.

::: {#exr-justices-histogram-clib-party .callout-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.
:::


::: {#ans-exr-justices-histogram-clib-party .callout-answer}

```{r}
# your two plots and two means go here
```

*(your interpretation here)*

:::


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.

```{r}
#| echo: true
#| fig-alt: "Liberal civil-liberties voting by nominating party, with individual justices jittered and each party mean plus or minus two standard deviations."

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
```

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.

::: {#exr-justices-scatter .callout-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?
:::


::: {#ans-exr-justices-scatter .callout-answer}

```{r}
# your means, standard deviations, and interval endpoints go here
```

*(your interpretation here)*

:::


# 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** {#sec-freq-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.
```{r}
#| echo: true
Y = c(1,2,3,4,5)                              #<1>
ge.3 = function(x) { ifelse(x >= 3, 1, 0) }   #<2>
freq.X.ge.3 = mean(ge.3(Y))                   #<3>
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.

And here's what we'd usually write in practice. It's more compact.
```{r}
#| echo: true
mean(Y >= 3)
```
## 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.

```{r}
#| echo: true 
#| fig-alt: "Five values from one through five in a vertical column, with the region at or above three shaded blue to show the indicator function."

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

ggplot(freq.data) +
  geom_point(aes(x=X, y = Y)) +
  annotate("rect", xmin = -Inf, xmax = Inf, ymax = Inf, ymin = 3,                            #<2>
           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. 



## 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.

::: {#exr-indicators-def .callout-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 $1_{\le 0}(Y_1) \ldots 1_{\le 0}(Y_5)$?
:::

::: {#ans-exr-indicators-def .callout-answer}

*(your answer here)*

:::

::: {#exr-indicators-calc .callout-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. 
:::

::: {#ans-exr-indicators-calc .callout-answer}

*(your answer here)*

:::


::: {#exr-indicators-viz .callout-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.
:::


::: {#ans-exr-indicators-viz .callout-answer}

*(your answer here)*

:::

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
$$
1_{\ge 25}(Y_i) \qfor 
1_{\ge 25}(y) = \begin{cases}
1 & \text{ if } y \ge 25 \\
0 & \text{ otherwise.}
\end{cases}
$$


::: {#exr-freq-CLlib .callout-exercise}
Calculate the frequency that support for the liberal position is greater than or equal to 25%.  What is it?
:::

::: {#ans-exr-freq-CLlib .callout-answer}

*(your answer here)*

:::

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

```{r}
#| echo: true
#| fig-alt: "Liberal civil-liberties voting by nominating party; the region at or above 25 percent is shaded blue and contains all but one justice."

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
```


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.

```{r}
#| echo: true
Y = EMdata$CLlib
X = EMdata$party
mean(Y[X==0] >= 50)
```

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

```{r}
#| echo: true
#| fig-alt: "Liberal civil-liberties voting by nominating party; the Republican region at or above 50 percent is shaded blue."
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
```



::: {#exr-indicators-CLlib-viz .callout-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 
`r round(100*mean(Y[X==0] >= 50))`% as a summary of the way Republican-nominated justices vote?
:::


::: {#ans-exr-indicators-CLlib-viz .callout-answer}

*(your answer here)*

:::
