0c-solutions-learning-checks.qmd

# Learning check solutions {#sec-lc-sol}


## Chapter 1 Solutions 

```{r, include=FALSE, purl=FALSE}
chap <- 1
lc <- 0
```

```{r message=FALSE}
library(dplyr)
library(ggplot2)
```

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  Repeat the above installing steps, but for the `dplyr`, `nycflights13`, and `knitr` packages. This will install the earlier mentioned `dplyr` package, the `nycflights13` package containing data on all domestic flights leaving a NYC airport in 2013, and the `knitr` package for writing reports in R. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** "Load" the `dplyr`, `nycflights13`, and `knitr` packages as well by repeating the above steps.

**Solution**: If the following code runs with no errors, you've succeeded!

```{r, eval=FALSE}
library(dplyr)
library(nycflights13)
library(knitr)
```


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What does any *ONE* row in this `flights` dataset refer to?

- A. Data on an airline 
- B. Data on a flight
- C. Data on an airport
- D. Data on multiple flights

**Solution**: This is data on a flight. Not a flight path! Example: 

* a flight path would be United 1545 to Houston
* a flight would be United 1545 to Houston at a specific date/time. For example: 2013/1/1 at 5:15am.


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What are some examples in this dataset of **categorical** variables?  What makes them different than **quantitative** variables?

**Solution**: Hint: Type `?flights` in the console to see what all the variables mean!

* Categorical:
    + `carrier` the company
    + `dest` the destination
    + `flight` the flight number. Even though this is a number, its simply a label. Example United 1545 is not  less than United 1714
* Quantitative:
    + `distance` the distance in miles
    + `time_hour` time


***



## Chapter 2 Solutions 

```{r, include=FALSE, purl=FALSE}
chap <- 2
lc <- 0
```

```{r, message=FALSE}
library(nycflights13)
library(ggplot2)
library(dplyr)
library(knitr)
library(moderndive)
```

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  Take a look at both the `flights` and `alaska_flights` data frames by running `View(flights)` and `View(alaska_flights)` in the console. In what respect do these data frames differ?

**Solution**: `flights` contains all flight data, while `alaska_flights` contains only data from Alaskan carrier "AS". We can see that flights has `r nrow(flights)` rows while `alaska_flights` has only `r nrow(alaska_flights)`


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  What are some practical reasons why `dep_delay` and `arr_delay` have a positive relationship?

**Solution**: The later a plane departs, typically the later it will arrive.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  What variables (not necessarily in the `flights` data frame) would you expect to have a negative correlation (i.e. a negative relationship) with `dep_delay`? Why? Remember that we are focusing on numerical variables here.

**Solution**: An example in the `weather` dataset is `visibility`, which measure visibility in miles. As visibility increases, we would expect departure delays to decrease. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why do you believe there is a cluster of points near (0, 0)? What does (0, 0) correspond to in terms of the Alaskan flights?

**Solution**: The point (0,0) means no delay in departure nor arrival. From the point of view of Alaska airlines, this means the flight was on time. It seems most flights are at least close to being on time.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What are some other features of the plot that stand out to you?

**Solution**: Different people will answer this one differently. One answer is most flights depart and arrive less than an hour late. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Create a new scatterplot using different variables in the `alaska_flights` data frame by modifying the example above.

**Solution**: Many possibilities for this one, see the plot below. Is there a pattern in departure delay depending on when the flight is scheduled to depart? Interestingly, there seems to be only two blocks of time where flights depart. 


```{r}
ggplot(data = alaska_flights, mapping = aes(x = dep_time, y = dep_delay)) +
  geom_point()
```

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  Why is setting the `alpha` argument value useful with scatterplots? What further information does it give you that a regular scatterplot cannot?

**Solution**: It thins out the points so we address overplotting. But more importantly it hints at the (statistical) **density** and **distribution** of the points: where are the points concentrated, where do they occur. We will see more about densities and distributions in Chapter 6 when we switch gears to statistical topics.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** After viewing the @fig-alpha above, give an approximate range of arrival delays and departure delays that occur the most frequently.  How has that region changed compared to when you observed the same plot without the `alpha = 0.2` set in @fig-noalpha?

**Solution**: The lower plot suggests that most Alaska flights from NYC depart between 12 minutes early and on time and arrive between 50 minutes early and on time.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Take a look at both the `weather` and `early_january_weather` data frames by running `View(weather)` and `View(early_january_weather)` in the console. In what respect do these data frames differ?

**Solution**: The rows of `early_january_weather` are a subset of `weather`. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** `View()` the `flights` data frame again. Why does the `time_hour` variable uniquely identify the hour of the measurement whereas the `hour` variable does not? 

**Solution**: Because to uniquely identify an hour, we need the `year`/`month`/`day`/`hour` sequence, whereas there are only 24 possible `hour`'s. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  Why should linegraphs be avoided when there is not a clear ordering of the horizontal axis?

**Solution**: Because lines suggest connectedness and ordering.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why are linegraphs frequently used when time is the explanatory variable?

**Solution**: Because time is sequential: subsequent observations are closely related to each other.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Plot a time series of a variable other than `temp` for Newark Airport in the first 15 days of January 2013.

**Solution**: Humidity is a good one to look at, since this very closely related to the cycles of a day.

```{r}
ggplot(data = early_january_weather, mapping = aes(x = time_hour, y = humid)) +
  geom_line()
```

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What does changing the number of bins from 30 to 40 tell us about the distribution of temperatures?

**Solution**: The distribution doesn't change much. But by refining the bin width, we see that the temperature data has a high degree of accuracy. What do I mean by accuracy? Looking at the `temp` variabile by `View(weather)`, we see that the precision of each temperature recording is 2 decimal places.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Would you classify the distribution of temperatures as symmetric or skewed?

**Solution**: It is rather symmetric, i.e. there are no __long tails__ on only one side of the distribution

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What would you guess is the "center" value in this distribution?  Why did you make that choice?

**Solution**: The center is around `r mean(weather$temp, na.rm=TRUE)`&deg;F. By running the `summary()` command, we see that the mean and median are very similar. In fact, when the distribution is symmetric the mean equals the median.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Is this data spread out greatly from the center or is it close?  Why?

**Solution**: This can only be answered relatively speaking! Let's pick things to be relative to Seattle, WA temperatures:

![](images/temp.png)

While, it appears that Seattle weather has a similar center of 55&deg;F, its
temperatures are almost entirely between 35&deg;F and 75&deg;F for a range of
about 40&deg;F. Seattle temperatures are much less spread out than New York i.e.
much more consistent over the year. New York on the other hand has much colder
days in the winter and much hotter days in the summer. Expressed differently,
the middle 50% of values, as delineated by the interquartile range is 30&deg;F:

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What other things do you notice about the faceted plot above?  How does a faceted plot help us see relationships between two variables?

**Solution**: 

* Certain months have much more consistent weather (August in particular), while others have crazy variability like January and October, representing changes in the seasons.
* Because we see `temp` recordings split by `month`, we are considering the relationship between these two variables. For example, for example for summer months, temperatures tend to be higher. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What do the numbers 1-12 correspond to in the plot above?  What about 25, 50, 75, 100?

**Solution**: 

* While month is technically a number between 1-12, we're viewing it as a categorical variable here. Specifically an **ordinal categorical** variable since there is a ordering to the categories
* 25, 50, 75, 100 are temperatures

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** For which types of data-sets would these types of faceted plots not work well in comparing relationships between variables? Give an example describing the nature of these variables and other important characteristics.

**Solution**: 

* We'd have 365 facets to look at. Way too many.
* We don't really care about day-to-day fluctuation in weather so much, but maybe more week-to-week variation. We'd like to focus on seasonal trends.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Does the `temp` variable in the `weather` data-set have a lot of variability?  Why do you say that?

**Solution**: Again, like in LC `r paste0("(LC ", chap, ".", (lc-4), ")")`, this is a relative question. I would say yes, because in New York City, you have 4 clear seasons with different weather. Whereas in Seattle WA and Portland OR, you have two seasons: summer and rain!


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What does the dot at the bottom of the plot for May correspond to?  Explain what might have occurred in May to produce this point.

**Solution**: It appears to be an outlier. Let's revisit the use of the `filter` command to hone in on it. We want all data points where the `month` is 5 and `temp<25`

``` {r}
weather %>% 
  filter(month==5 & temp < 25)
```

There appears to be only one hour and only at JFK that recorded 13.1 F (-10.5 C) in the month of May. This is probably a data entry mistake! Why wasn't the weather at least similar at EWR (Newark) and LGA (La Guardia)?

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Which months have the highest variability in temperature?  What reasons do you think this is?

**Solution**: We are now interested in the **spread** of the data. One measure some of you may have seen previously is the standard deviation. But in this plot we can read off the Interquartile Range (IQR):

* The distance from the 1st to the 3rd quartiles i.e. the length of the boxes
* You can also think of this as the spread of the **middle 50%** of the data

Just from eyeballing it, it seems

* November has the biggest IQR, i.e. the widest box, so has the most variation in temperature
* August has the smallest IQR, i.e. the narrowest box, so is the most consistent temperature-wise

Here's how we compute the exact IQR values for each month (we'll see this more in depth Chapter 5 of the text):

1. `group` the observations by `month` then
1. for each `group`, i.e. `month`, `summarize` it by applying the summary statistic function `IQR()`, while making sure to skip over missing data via `na.rm=TRUE` then
1. `arrange` the table in `desc`ending order of `IQR`

```{r, eval=FALSE}
weather %>%
  group_by(month) %>%
  summarize(IQR = IQR(temp, na.rm=TRUE)) %>%
  arrange(desc(IQR))
```
```{r, echo=FALSE}
weather %>%
  group_by(month) %>%
  summarize(IQR = IQR(temp, na.rm=TRUE)) %>%
  arrange(desc(IQR)) %>%
  kable()
```

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** We looked at the distribution of the numerical variable `temp` split by the numerical variable `month` that we converted to a categorical variable using the `factor()` function. Why would a boxplot of `temp` split by the numerical variable `pressure` similarly converted to a categorical variable using the `factor()` not be informative?

**Solution**: Because there are 12 unique values of `month` yielding only 12 boxes in our boxplot. There are many more unique values of `pressure` (`r weather$pressure %>% unique() %>% length()` unique values in fact), because values are to the first decimal place. This would lead to `r weather$pressure %>% unique() %>% length()` boxes, which is too many for people to digest. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Boxplots provide a simple way to identify outliers.  Why may outliers be easier to identify when looking at a boxplot instead of a faceted histogram?

**Solution**: In a histogram, the bin corresponding to where an outlier lies may not by high enough for us to see. In a boxplot, they are explicitly labelled separately. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why are histograms inappropriate for visualizing categorical variables?

**Solution**: Histograms are for numerical variables i.e. the horizontal part of each histogram bar represents an interval, whereas for a categorical variable each bar represents only one level of the categorical variable.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What is the difference between histograms and barplots?

**Solution**: See above.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** How many Envoy Air flights departed NYC in 2013?

**Solution**: Envoy Air is carrier code `MQ` and thus 26397 flights departed NYC in 2013.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What was the seventh highest airline in terms of departed flights from NYC in 2013? How could we better present the table to get this answer quickly?

**Solution**: What a pain! We'll see in Chapter 5 on Data Wrangling that applying `arrange(desc(n))` will sort this table in descending order of `n`!

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why should pie charts be avoided and replaced by barplots?

**Solution**: In our **opinion**, comparisons using horizontal lines are easier than comparing angles and areas of circles.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What is your opinion as to why pie charts continue to be used?

**Solution**: Legacy?


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What kinds of questions are not easily answered by looking at the above figure?

**Solution**: Because the red, green, and blue bars don't all start at 0 (only red does), it makes comparing counts hard.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What can you say, if anything, about the relationship between airline and airport in NYC in 2013 in regards to the number of departing flights?

**Solution**: The different airlines prefer different airports. For example, United is mostly a Newark carrier and JetBlue is a JFK carrier. If airlines didn't prefer airports, each color would be roughly one third of each bar.}


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why might the side-by-side (AKA dodged) barplot be preferable to a stacked barplot in this case?

**Solution**: We can easily compare the different aiports for a given carrier using a single comparison line i.e. things are lined up

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What are the disadvantages of using a side-by-side (AKA dodged) barplot, in general?

**Solution**: It is hard to get totals for each airline.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why is the faceted barplot preferred to the side-by-side and stacked barplots in this case?

**Solution**: Not that different than using side-by-side; depends on how you want to organize your presentation.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What information about the different carriers at different airports is more easily seen in the faceted barplot?

**Solution**: Now we can also compare the different carriers **within** a particular airport easily too. For example, we can read off who the top carrier for each airport is easily using a single horizontal line.

***

## Chapter 3 Solutions

```{r, include=FALSE, purl=FALSE}
chap <- 3
lc <- 0
```

```{r message=FALSE}
library(dplyr)
library(ggplot2)
library(nycflights13)
library(kableExtra)
``` 


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What's another way using the "not" operator `!` to filter only the rows that are not going to Burlington, VT nor Seattle, WA in the `flights` data frame? Test this out using the code above.

**Solution**: 

```{r, eval=FALSE, purl=FALSE}
# Original in book
not_BTV_SEA <- flights %>% 
  filter(!(dest == "BTV" | dest == "SEA"))

# Alternative way
not_BTV_SEA <- flights %>% 
  filter(!dest == "BTV" & !dest == "SEA")

# Yet another way
not_BTV_SEA <- flights %>% 
  filter(dest != "BTV" & dest != "SEA")
```


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Say a doctor is studying the effect of smoking on lung cancer for a large number of patients who have records measured at five year intervals. She notices that a large number of patients have missing data points because the patient has died, so she chooses to ignore these patients in her analysis. What is wrong with this doctor's approach?

**Solution**: The missing patients may have died of lung cancer! So to ignore them might seriously **bias** your results! It is very important to think of what the consequences on your analysis are of ignoring missing data! Ask yourself:

* There is a systematic reasons why certain values are missing? If so, you might be biasing your results!
* If there isn't, then it might be ok to "sweep missing values under the rug."


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Modify the above `summarize` function to create `summary_temp` to also use the `n()` summary function: `summarize(count = n())`. What does the returned value correspond to?

**Solution**: It corresponds to a count of the number of observations/rows:

```{r, purl=FALSE}
weather %>% 
  summarize(count = n())
```


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why doesn't the following code work?  Run the code line by line instead of all at once, and then look at the data.  In other words, run `summary_temp <- weather %>% summarize(mean = mean(temp, na.rm = TRUE))` first.

```{r eval=FALSE}
summary_temp <- weather %>%   
  summarize(mean = mean(temp, na.rm = TRUE)) %>% 
  summarize(std_dev = sd(temp, na.rm = TRUE))
```

**Solution**: Consider the output of only running the first two lines:

```{r, purl=FALSE}
weather %>%   
  summarize(mean = mean(temp, na.rm = TRUE))
```

Because after the first `summarize()`, the variable `temp` disappears as it has been collapsed to the value `mean`. So when we try to run the second `summarize()`, it can't find the variable `temp` to compute the standard deviation of. 


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Recall from Chapter -@sec-viz when we looked at plots of temperatures by months in NYC. What does the standard deviation column in the `summary_monthly_temp` data frame tell us about temperatures in New York City throughout the year?

**Solution**: 

```{r, echo=FALSE, purl=FALSE}
summary_temp_by_month <- weather %>% 
  group_by(month) %>% 
  summarize(
          mean = mean(temp, na.rm = TRUE),
          std_dev = sd(temp, na.rm = TRUE)
          )

kable(summary_temp_by_month) %>% 
  kable_styling(font_size = ifelse(knitr:::is_latex_output(), 10, 16),
                latex_options = c("HOLD_position"))
```

The standard deviation is a quantification of **spread** and **variability**. We
see that the period in November, December, and January has the most variation in
weather, so you can expect very different temperatures on different days.


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What code would be required to get the mean and standard deviation temperature for each day in 2013 for NYC?

**Solution**: 

```{r, purl=FALSE}
summary_temp_by_day <- weather %>% 
  group_by(year, month, day) %>% 
  summarize(
          mean = mean(temp, na.rm = TRUE),
          std_dev = sd(temp, na.rm = TRUE)
          )
summary_temp_by_day
```

Note: `group_by(day)` is not enough, because `day` is a value between 1-31. We need to `group_by(year, month, day)` 

```{r, purl=FALSE}
library(dplyr)
library(nycflights13)

summary_temp_by_month <- weather %>% 
  group_by(month) %>% 
  summarize(
          mean = mean(temp, na.rm = TRUE),
          std_dev = sd(temp, na.rm = TRUE)
          )
```


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Recreate `by_monthly_origin`, but instead of grouping via `group_by(origin, month)`, group variables in a different order `group_by(month, origin)`. What differs in the resulting dataset?

**Solution**: 

```{r, purl=FALSE}
by_monthly_origin <- flights %>% 
  group_by(month, origin) %>% 
  summarize(count = n())
```

```{r, eval=FALSE, purl=FALSE}
by_monthly_origin
```

```{r, echo=FALSE, purl=FALSE}
kable(by_monthly_origin) %>% 
  kable_styling(font_size = ifelse(knitr:::is_latex_output(), 10, 16),
                latex_options = c("HOLD_position"))
```

In `by_monthly_origin` the `month` column is now first and the rows are sorted by `month` instead of origin. If you compare the values of `count` in `by_origin_monthly` and `by_monthly_origin` using the `View()` function, you'll see that the values are actually the same, just presented in a different order. 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** How could we identify how many flights left each of the three airports for each `carrier`?

**Solution**: We could summarize the count from each airport using the `n()` function, which *counts rows*.

```{r, purl=FALSE}
count_flights_by_airport <- flights %>% 
  group_by(origin, carrier) %>% 
  summarize(count=n())
```

```{r, eval=FALSE, purl=FALSE}
count_flights_by_airport
```

```{r, echo=FALSE, purl=FALSE}
kable(count_flights_by_airport) %>% 
  kable_styling(font_size = ifelse(knitr:::is_latex_output(), 10, 16), 
                latex_options = c("HOLD_position"))
```

All remarkably similar! Note: the `n()` function counts rows, whereas the `sum(VARIABLE_NAME)` funciton sums all values of a certain numerical variable `VARIABLE_NAME`.


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** How does the `filter` operation differ from a `group_by` followed by a `summarize`?

**Solution**: 

* `filter` picks out rows from the original dataset without modifying them, whereas
* `group_by %>% summarize` computes summaries of numerical variables, and hence
reports new values.


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What do positive values of the `gain` variable in `flights` correspond to?  What about negative values?  And what about a zero value?

**Solution**: 

* Say a flight departed 20 minutes late, i.e. `dep_delay = 20`
* Then arrived 10 minutes late, i.e. `arr_delay = 10`.
* Then `gain = dep_delay - arr_delay = 20 - 10  = 10` is positive, so it "made up/gained time in the air."
* 0 means the departure and arrival time were the same, so no time was made up in  the air. We see in most cases that the `gain` is near 0 minutes.
* I never understood this. If the pilot says "we're going make up time in the air"
because of delay by flying faster, why don't you always just fly faster to begin
with?


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Could we create the `dep_delay` and `arr_delay` columns by simply subtracting `dep_time` from `sched_dep_time` and similarly for arrivals?  Try the code out and explain any differences between the result and what actually appears in `flights`.

**Solution**: No because you can't do direct arithmetic on times. The difference in time between 12:03 and 11:59 is 4 minutes, but `1203-1159 = 44`


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What can we say about the distribution of `gain`?  Describe it in a few sentences using the plot and the `gain_summary` data frame values.

**Solution**: Most of the time the gain is a little under zero, most of the time the gain is between -50 and 50 minutes. There are some extreme cases however!


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Looking at @fig-flights-key-diagram, when joining `flights` and `weather` (or, in other words, matching the hourly weather values with each flight), why do we need to join by all of `year`, `month`, `day`, `hour`, and `origin`, and not just `hour`?

**Solution**: Because `hour` is simply a value between 0 and 23; to identify a *specific* hour, we need to know which year, month, day and at which airport.


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What surprises you about the top 10 destinations from NYC in 2013?

**Solution**: This question is subjective! What surprises me is the high number of flights to Boston. Wouldn't it be easier and quicker to take the train?


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What are some advantages of data in normal forms?  What are some disadvantages?

**Solution**: When datasets are in normal form, we can easily `_join` them with other datasets! For example, we can join the `flights` data with the `planes` data.

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What are some ways to select all three of the `dest`, `air_time`, and `distance` variables from `flights`?  Give the code showing how to do this in at least three different ways.

**Solution**: 

```{r, purl=FALSE}
# The regular way:
flights %>% 
  select(dest, air_time, distance)

# Since they are sequential columns in the dataset
flights %>% 
  select(dest:distance)

# Not as effective, by removing everything else
flights %>% 
  select(-year, -month, -day, -dep_time, -sched_dep_time, -dep_delay, -arr_time,
         -sched_arr_time, -arr_delay, -carrier, -flight, -tailnum, -origin, 
         -hour, -minute, -time_hour)
```


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** How could one use `starts_with`, `ends_with`, and `contains` to select columns from the `flights` data frame?  Provide three different examples in total: one for `starts_with`, one for `ends_with`, and one for `contains`.

**Solution**: 

```{r, purl=FALSE}
# Anything that starts with "d"
flights %>% 
  select(starts_with("d"))
# Anything related to delays:
flights %>% 
  select(ends_with("delay"))
# Anything related to departures:
flights %>% 
  select(contains("dep"))
```


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Why might we want to use the `select()` function on a data frame?

**Solution**: To narrow down the data frame, to make it easier to look at. Using `View()` for example.


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Create a new data frame that shows the top 5 airports with the largest arrival delays from NYC in 2013.

**Solution**: 

```{r, purl=FALSE}
top_five <- flights %>% 
  group_by(dest) %>% 
  summarize(avg_delay = mean(arr_delay, na.rm = TRUE)) %>% 
  arrange(desc(avg_delay)) %>% 
  top_n(n = 5)
top_five
```


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Using the datasets included in the `nycflights13` package, compute the available seat miles for each airline sorted in descending order. After completing all the necessary data wrangling steps, the resulting data frame should have 16 rows (one for each airline) and 2 columns (airline name and available seat miles). Here are some hints:

1. **Crucial**: Unless you are very confident in what you are doing, it is worthwhile to not starting coding right away, but rather first sketch out on paper all the necessary data wrangling steps not using exact code, but rather high-level *pseudocode* that is informal yet detailed enough to articulate what you are doing. This way you won't confuse *what* you are trying to do (the algorithm) with *how* you are going to do it (writing `dplyr` code). 
1. Take a close look at all the datasets using the `View()` function: `flights`, `weather`, `planes`, `airports`, and `airlines` to identify which variables are necessary to compute available seat miles.
1. Figure @fig-flights-key-diagram showing how the various datasets can be joined will also be useful. 
1. Consider the data wrangling verbs in Table @tbl-wrangle-summary-table as your toolbox!


**Solution**: Here are some examples of student-written [pseudocode](https://twitter.com/rudeboybert/status/964181298691629056). Based on our own pseudocode, let's first display the entire solution.

```{r}
flights %>% 
  inner_join(planes, by = "tailnum") %>% 
  select(carrier, seats, distance) %>% 
  mutate(ASM = seats * distance) %>% 
  group_by(carrier) %>% 
  summarize(ASM = sum(ASM, na.rm = TRUE)) %>% 
  arrange(desc(ASM))
```

Let's now break this down step-by-step. To compute the available seat miles for a given flight, we need the `distance` variable from the `flights` data frame and the `seats` variable from the `planes` data frame, necessitating a join by the key variable `tailnum` as illustrated in @fig-flights-key-diagram. To keep the resulting data frame easy to view, we'll `select()` only these two variables and `carrier`:

```{r}
flights %>% 
  inner_join(planes, by = "tailnum") %>% 
  select(carrier, seats, distance)
```

Now for each flight we can compute the available seat miles `ASM` by multiplying the number of seats by the distance via a `mutate()`:

```{r}
flights %>% 
  inner_join(planes, by = "tailnum") %>% 
  select(carrier, seats, distance) %>% 
  # Added:
  mutate(ASM = seats * distance)
```

Next we want to sum the `ASM` for each carrier. We achieve this by first grouping by `carrier` and then summarizing using the `sum()` function:

```{r}
flights %>% 
  inner_join(planes, by = "tailnum") %>% 
  select(carrier, seats, distance) %>% 
  mutate(ASM = seats * distance) %>% 
  # Added:
  group_by(carrier) %>% 
  summarize(ASM = sum(ASM))
```

However, because for certain carriers certain flights have missing `NA` values, the resulting table also returns `NA`'s. We can eliminate these by adding a `na.rm = TRUE` argument to `sum()`, telling R that we want to remove the `NA`'s in the sum. We saw this in Section -@sec-summarize):

```{r}
flights %>% 
  inner_join(planes, by = "tailnum") %>% 
  select(carrier, seats, distance) %>% 
  mutate(ASM = seats * distance) %>% 
  group_by(carrier) %>% 
  # Modified:
  summarize(ASM = sum(ASM, na.rm = TRUE))
```

Finally, we `arrange()` the data in `desc()`ending order of `ASM`.

```{r}
flights %>% 
  inner_join(planes, by = "tailnum") %>% 
  select(carrier, seats, distance) %>% 
  mutate(ASM = seats * distance) %>% 
  group_by(carrier) %>% 
  summarize(ASM = sum(ASM, na.rm = TRUE)) %>% 
  # Added:
  arrange(desc(ASM))
```

While the above data frame is correct, the IATA `carrier` code is not always useful. For example, what carrier is `WN`? We can address this by joining with the `airlines` dataset using `carrier` is the key variable. While this step is not absolutely required, it goes a long way to making the table easier to make sense of. It is important to be empathetic with the ultimate consumers of your presented data!

```{r}
flights %>% 
  inner_join(planes, by = "tailnum") %>% 
  select(carrier, seats, distance) %>% 
  mutate(ASM = seats * distance) %>% 
  group_by(carrier) %>% 
  summarize(ASM = sum(ASM, na.rm = TRUE)) %>% 
  arrange(desc(ASM)) %>% 
  # Added:
  inner_join(airlines, by = "carrier")
```

***

## Chapter 4 Solutions

```{r, include=FALSE, purl=FALSE}
chap <- 4
lc <- 0
```

```{r message=FALSE}
library(dplyr)
library(ggplot2)
library(nycflights13)
library(tidyr)
library(readr)
library(fivethirtyeight)
``` 

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What are common characteristics of "tidy" datasets?

**Solution**: Rows correspond to observations, while columns correspond to variables.  

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** What makes "tidy" datasets useful for organizing data?

**Solution**: Tidy datasets are an organized way of viewing data. This format is required for the `ggplot2` and `dplyr` packages for data visualization and wrangling. 


**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`** Take a look the `airline_safety` data frame included in the `fivethirtyeight` data. Run the following:

```{r, eval=FALSE}
airline_safety
```

After reading the help file by running `?airline_safety`, we see that `airline_safety` is a data frame containing information on different airlines companies' safety records. This data was originally reported on the data journalism website FiveThirtyEight.com in Nate Silver's article ["Should Travelers Avoid Flying Airlines That Have Had Crashes in the Past?"](https://fivethirtyeight.com/features/should-travelers-avoid-flying-airlines-that-have-had-crashes-in-the-past/). Let's ignore the `incl_reg_subsidiaries` and `avail_seat_km_per_week` variables for simplicity:

```{r}
airline_safety_smaller <- airline_safety %>% 
  select(-c(incl_reg_subsidiaries, avail_seat_km_per_week))
airline_safety_smaller
```

This data frame is not in "tidy" format. How would you convert this data frame to be in "tidy" format, in particular so that it has a variable `incident_type_years` indicating the indicent type/year and a variable `count` of the counts?

**Solution**: Using the `gather()` function from the `tidyr` package:

```{r}
airline_safety_smaller_tidy <- airline_safety_smaller %>% 
  gather(key = incident_type_years, value = count, -airline)
airline_safety_smaller_tidy
```

If you look at the resulting `airline_safety_smaller_tidy` data frame in the spreadsheet viewer, you'll see that the variable `incident_type_years` has 6 possible values: `"incidents_85_99", "fatal_accidents_85_99", "fatalities_85_99", 
"incidents_00_14", "fatal_accidents_00_14", "fatalities_00_14"` corresponding to the 6 columns of `airline_safety_smaller` we tidied. 



**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  Convert the `dem_score` data frame into
a tidy data frame and assign the name of `dem_score_tidy` to the resulting long-formatted data frame.

**Solution**: Running the following in the console:

```{r}
dem_score <- read_csv("https://moderndive.com/data/dem_score.csv")

dem_score_tidy <- dem_score %>% 
  gather(key = year, value = democracy_score, - country)
```

Let's now compare the `dem_score` and `dem_score_tidy`. `dem_score` has democracy score information for each year in columns, whereas in `dem_score_tidy` there are explicit variables `year` and `democracy_score`. While both representations of the data contain the same information, we can only use `ggplot()` to create plots using the `dem_score_tidy` data frame.

```{r}
dem_score
dem_score_tidy
```

**`r paste0("(LC ", chap, ".", (lc <- lc + 1), ")")`**  Read in the life expectancy data stored at <https://moderndive.com/data/le_mess.csv> and convert it to a tidy data frame. 

**Solution**: The code is similar

```{r, eval=FALSE}
life_expectancy <- read_csv("https://moderndive.com/data/le_mess.csv")
life_expectancy_tidy <- life_expectancy %>% 
  gather(key = year, value = life_expectancy, -country)
```
```{r, echo=FALSE, purl=FALSE, message=FALSE, warning=FALSE}
life_expectancy <- read_csv('data/le_mess.csv')
life_expectancy_tidy <- life_expectancy %>% 
  gather(key = year, value = life_expectancy, -country)
```

We observe the same construct structure with respect to `year` in `life_expectancy` vs `life_expectancy_tidy` as we did in `dem_score` vs `dem_score_tidy`:

```{r, lc4-2solutions-4}
life_expectancy
life_expectancy_tidy
```

***

## Chapter 5 Solutions

```{r, include=FALSE, purl=FALSE}
chap <- 5
lc <- 0
```

```{r message=FALSE}
library(ggplot2)
library(dplyr)
library(moderndive)
library(gapminder)
library(skimr)
``` 


**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Conduct a new exploratory data analysis with the same outcome variable $y$ being `score` but with `age` as the new explanatory variable $x$. Remember, this involves three things:

a) Looking at the raw data values.
a) Computing summary statistics.
a) Creating data visualizations.

What can you say about the relationship between age and teaching scores based on this exploration?

**Solution**: 

```{r}
evals_ch5 <- evals %>%
  select(score, bty_avg, age)
```

- Looking at the raw data values: 
```{r}
glimpse(evals_ch5)
```

- Computing summary statistics:
```{r, eval=FALSE}
skim_with(numeric = list(hist = NULL), integer = list(hist = NULL))
```
```{r, include=FALSE}
skim_with(numeric = list(hist = NULL), integer = list(hist = NULL))
```

```{r, eval=FALSE}
evals_ch5 %>%
  select(score, age) %>%
  skim()
```

(Note that for formatting purposes, the inline histogram that is usually `printed with skim()` has been removed. This can be done by running `skim_with(numeric = list(hist = NULL), integer = list(hist = NULL))` prior to using the `skim()` function as well.)

- Creating data visualizations:
```{r}
ggplot(evals_ch5, aes(x = age, y = score)) +
  geom_point() +
  labs(
    x = "Age", y = "Teaching Score",
    title = "Scatterplot of relationship of teaching score and age"
  )
```

<!--
TODO: Albert needs to double check interpretation:
-->

Based on the scatterplot visualization, there seem to have a weak negative relationship between age and teaching score. As age increases, the teaching score see, to decrease slightly. 

**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Fit a new simple linear regression using `lm(score ~ age, data = evals_ch5)` where `age` is the new explanatory variable $x$. Get information about the "best-fitting" line from the regression table by applying the `get_regression_table()` function. How do the regression results match up with the results from your earlier exploratory data analysis?

**Solution**: 
```{r}
# Fit regression model:
score_age_model <- lm(score ~ age, data = evals_ch5)
# Get regression table:
get_regression_table(score_age_model)
```
$$
\begin{aligned}
\widehat{y} &= b_0 + b_1 \cdot x\\
\widehat{\text{score}} &= b_0 + b_{\text{age}} \cdot\text{age}\\
&= 4.462 - 0.006\cdot\text{age}
\end{aligned}
$$

<!--
TODO: Albert will verify Starry's interpretation:
-->

For every increase of 1 unit in `age`, there is an *associated* decrease of, *on average*, 0.006 units of `score`. It matches with the results from our earlier exploratory data analysis. 

<!-- **`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Generate a data frame of the residuals of the model where you used `age` as the explanatory $x$ variable. -->

<!-- **Solution**:  -->
<!-- ```{r} -->
<!-- score_age_regression_points <- get_regression_points(score_age_model) -->
<!-- score_age_regression_points -->
<!-- ``` -->

**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Conduct a new exploratory data analysis with the same explanatory variable $x$ being `continent` but with `gdpPercap` as the new outcome variable $y$. Remember, this involves three things:

1. Most crucially: Looking at the raw data values.
1. Computing summary statistics, such as means, medians, and interquartile ranges.
1. Creating data visualizations.

What can you say about the differences in GDP per capita between continents based on this exploration?

**Solution**: 

```{r}
gapminder2007 <- gapminder %>%
  filter(year == 2007) %>%
  select(country, lifeExp, continent, gdpPercap)
```

- Looking at the raw data values: 
```{r}
glimpse(gapminder2007)
```

- Computing summary statistics, such as means, medians, and interquartile ranges:
```{r eval=FALSE}
gapminder2007 %>%
  select(gdpPercap, continent) %>%
  skim()
```

- Creating data visualizations:
```{r}
ggplot(gapminder2007, aes(x = continent, y = gdpPercap)) +
  geom_boxplot() +
  labs(
    x = "Continent", y = "GPD per capita",
    title = "GDP by continent"
  )
```

<!--
TODO: Albert needs to double check interpretation:
-->

Based on this exploration, it seems that GDP's are very different among different continents, which means that continent might be a statistically significant predictor for an area's GDP. 

**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Fit a new linear regression using `lm(gdpPercap ~ continent, data = gapminder2007)` where `gdpPercap` is the new outcome variable $y$. Get information about the "best-fitting" line from the regression table by applying the `get_regression_table()` function. How do the regression results match up with the results from your previous exploratory data analysis?

**Solution**: 
```{r}
# Fit regression model:
gdp_model <- lm(gdpPercap ~ continent, data = gapminder2007)
# Get regression table:
get_regression_table(gdp_model)
```
$$
\begin{aligned}
\widehat{y} = \widehat{\text{gdpPercap}} &= b_0 + b_{\text{Amer}}\cdot\mathbb{1}_{\mbox{Amer}}(x) + b_{\text{Asia}}\cdot\mathbb{1}_{\mbox{Asia}}(x) + \\
& \qquad b_{\text{Euro}}\cdot\mathbb{1}_{\mbox{Euro}}(x) + b_{\text{Ocean}}\cdot\mathbb{1}_{\mbox{Ocean}}(x)\\
&= 3089 + 7914\cdot\mathbb{1}_{\mbox{Amer}}(x) + 9384\cdot\mathbb{1}_{\mbox{Asia}}(x) + \\
& \qquad 21965\cdot\mathbb{1}_{\mbox{Euro}}(x) + 26721\cdot\mathbb{1}_{\mbox{Ocean}}(x)
\end{aligned}
$$

<!--
TODO: Albert will double check Starry's interpretation:
-->

In our previous exploratory data analysis, it seemed that continent is a statistically significant predictor for an area's GDP. Here, by fit a new linear regression using `lm(gdpPercap ~ continent, data = gapminder2007)` where `gdpPercap` is the new outcome variable $y$, we are able to write an equation to predict `gdpPercap` using the continent as statistically significant predictors. Therefore, the regression results matches with the results from your previous exploratory data analysis. 

**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Using either the sorting functionality of RStudio's spreadsheet viewer or using the data wrangling tools you learned in Chapter \@ref(wrangling), identify the five countries with the five smallest (most negative) residuals? What do these negative residuals say about their life expectancy relative to their continents?

**Solution**: 
Using the sorting functionality of RStudio's spreadsheet viewer, we can identify that the five countries with the five smallest (most negative) residuals are: Afghanistan, Swaziland, Mozambique, Haiti, and Zambia. 

These negative residuals indicate that these data points have the biggest negative deviations from their group means. This means that these five countries' average life expectancies are the lowest comparing to their respective continents' average life expectancies. For example, the residual for Afghanistan is $-26.900$ and it is the smallest residual. This means that the average life expectancy of Afghanistan is $26.900$ years lower than the average life expectancy of its continent, Asia. 

**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Repeat this process, but identify the five countries with the five largest (most positive) residuals. What do these positive residuals say about their life expectancy relative to their continents?

**Solution**: 
Using either the sorting functionality of RStudio's spreadsheet viewer, we can identify that the five countries with the five largest (most positive) residuals are: Reunion, Libya, Tunisia, Mauritius, and Algeria. 

These positive residuals indicate that the data points are above the regression line with the longest distance. This means that these five countries' average life expectancies are the highest comparing to their respective continents' average life expectancies. For example, the residual for Reunion is $21.636$ and it is the largest residual. This means that the average life expectancy of Reunion is $21.636$ years higher than the average life expectancy of its continent, Africa. 

<!-- **`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Note in the following plot there are 3 points marked with dots along with: -->

<!-- * The "best" fitting solid regression line `r if_else(is_latex_output(), "", "in blue")` -->
<!-- * An arbitrarily chosen dotted `r if_else(is_latex_output(), "", "red")` line  -->
<!-- * Another arbitrarily chosen dashed `r if_else(is_latex_output(), "", "green")` line -->

<!-- ```{r, fig.cap="Regression line and two others.", out.width="80%", echo=FALSE} -->
<!-- example <- tibble( -->
<!--   x = c(0, 0.5, 1), -->
<!--   y = c(2, 1, 3) -->
<!-- ) -->

<!-- ggplot(example, aes(x = x, y = y)) + -->
<!--   geom_smooth(method = "lm", se = FALSE, fullrange = TRUE) + -->
<!--   geom_hline(yintercept = 2.5, col = "red", linetype = "dotted", size = 1) + -->
<!--   geom_abline( -->
<!--     intercept = 2, slope = -1, col = "forestgreen", -->
<!--     linetype = "dashed", size = 1 -->
<!--   ) + -->
<!--   geom_point(size = 4) -->
<!-- ``` -->
<!-- Compute the sum of squared residuals by hand for each line and show that of these three lines, the regression line `r if_else(is_latex_output(), "", "in blue")` has the smallest value. -->

<!-- **Solution**:   -->

<!-- * The "best" fitting solid regression line `r if_else(is_latex_output(), "", "in blue")`:  -->

<!-- $$ -->
<!-- \sum_{i=1}^{n}(y_i - \widehat{y}_i)^2 = (2.0-1.5)^2+(1.0-2.0)^2+(3.0-2.5)^2 = 1.5 -->
<!-- $$ -->

<!-- * An arbitrarily chosen dotted `r if_else(is_latex_output(), "", "red")` line:  -->

<!-- $$ -->
<!-- \sum_{i=1}^{n}(y_i - \widehat{y}_i)^2 = (2.0-2.5)^2+(1.00-2.5)^2+(3.0-2.5)^2 = 2.75 -->
<!-- $$ -->

<!-- * Another arbitrarily chosen dashed `r if_else(is_latex_output(), "", "green")` line: -->

<!-- $$ -->
<!-- \sum_{i=1}^{n}(y_i - \widehat{y}_i)^2 = (2.0-2.0)^2+(1.0-1.5)^2+(3.0-1.0)^2 = 4.25 -->
<!-- $$ -->

<!-- As calculated, $1.5 < 2.75 < 4.25$. Therefore, we show that the regression line in blue has the smallest value of the residual sum of squares. -->



***

## Chapter 6 Solutions

```{r, include=FALSE, purl=FALSE}
chap <- 6
lc <- 0
```

```{r message=FALSE}
library(tidyverse)
library(moderndive)
library(skimr)
library(ISLR2)
``` 

<!-- **`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Compute the observed values, fitted values, and residuals not for the interaction model as we just did, but rather for the parallel slopes model we saved in `score_model_parallel_slopes`. -->

<!-- **Solution**:  -->
<!-- ```{r} -->
<!-- regression_points_parallel <- get_regression_points(score_model_parallel_slopes) -->
<!-- regression_points_parallel -->
<!-- ``` -->

**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Conduct a new exploratory data analysis with the same outcome variable $y$ being `debt` but with `credit_rating` and `age` as the new explanatory variables $x_1$ and $x_2$. Remember, this involves three things:

1. Most crucially: Looking at the raw data values.
1. Computing summary statistics, such as means, medians, and interquartile ranges.
1. Creating data visualizations.

What can you say about the relationship between a credit card holder's debt and their credit rating and age?

**Solution**: 

```{r}
credit_ch6 <- Credit %>%
  as_tibble() %>% 
  select(debt = Balance, credit_limit = Limit, 
         income = Income, credit_rating = Rating, age = Age)
```

* Most crucially: Looking at the raw data values.
```{r}
credit_ch6 %>%
  select(debt, credit_rating, age) %>%
  head()
```

* Computing summary statistics, such as means, medians, and interquartile ranges.
```{r, eval=FALSE}
skim_with(numeric = list(hist = NULL), integer = list(hist = NULL))
```
```{r, include=FALSE}
skim_with(numeric = list(hist = NULL), integer = list(hist = NULL))
```
```{r, eval=FALSE}
credit_ch6 %>%
  select(debt, credit_rating, age) %>%
  skim()
```
* Creating data visualizations.
```{r}
ggplot(credit_ch6, aes(x = credit_rating, y = debt)) +
  geom_point() +
  labs(
    x = "Credit rating", y = "Credit card debt (in $)",
    title = "Debt and credit rating"
  ) +
  geom_smooth(method = "lm", se = FALSE)

ggplot(credit_ch6, aes(x = age, y = debt)) +
  geom_point() +
  labs(
    x = "Age (in year)", y = "Credit card debt (in $)",
    title = "Debt and age"
  ) +
  geom_smooth(method = "lm", se = FALSE)
```
It seems that there is a positive relationship between one's credit rating and their debt, and very little relationship between one's age and their debt. 

**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`** Fit a new simple linear regression using `lm(debt ~ credit_rating + age, data = credit_ch6)` where `credit_rating` and `age` are the new numerical explanatory variables $x_1$ and $x_2$. Get information about the "best-fitting" regression plane from the regression table by applying the `get_regression_table()` function. How do the regression results match up with the results from your previous exploratory data analysis? 

**Solution**: 

```{r}
# Fit regression model:
debt_model_2 <- lm(debt ~ credit_rating + age, data = credit_ch6)
# Get regression table:
get_regression_table(debt_model_2)
```
The coefficients for both new numerical explanatory variables $x_1$ and $x_2$, `credit_rating` and `age`, are $2.59$ and $-2.35$ respectively, which means that `debt` and `credit_rating` are positively correlated, and `debt` and `age` are negatively correlated. This matches up with the results from your previous exploratory data analysis. 


**`r paste0("(LC", chap, ".", (lc <- lc + 1), ")")`**
Compute the observed values, fitted values, and residuals not for the interaction model as we just did, but rather for the parallel slopes model we saved in `score_model_parallel_slopes`.

**Solution**: 

```{r}
evals_ch6 <- evals %>%
  select(ID, score, age, gender)

score_model_parallel_slopes <- lm(score ~ age + gender, data = evals_ch6)

score_model_parallel_data <- evals_ch6 %>% 
  select(score, age, gender) %>% 
  mutate(score_hat = fitted(score_model_parallel_slopes),
         residual = residuals(score_model_parallel_slopes)) %>% 
  rownames_to_column("ID")

score_model_parallel_data
```