Let’s begin by loading the packages we’ll need to get started

library(tidyverse)
## ── Attaching packages ──────────────────────────────────────────────────────────────────────────────────── tidyverse 1.2.1 ──
## ✔ ggplot2 3.2.1     ✔ purrr   0.3.3
## ✔ tibble  2.1.3     ✔ dplyr   0.8.3
## ✔ tidyr   1.0.0     ✔ stringr 1.4.0
## ✔ readr   1.3.1     ✔ forcats 0.4.0
## ── Conflicts ─────────────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()
library(knitr)

options(scipen=4)

We’ll begin by doing all the same data processing as in previous lectures

# Load data from MASS into a tibble
# Rename variables
# Recode categorical variables
birthwt <- as_tibble(MASS::birthwt) %>%
  rename(birthwt.below.2500 = low, 
         mother.age = age,
         mother.weight = lwt,
         mother.smokes = smoke,
         previous.prem.labor = ptl,
         hypertension = ht,
         uterine.irr = ui,
         physician.visits = ftv,
         birthwt.grams = bwt)  %>%
  mutate(race = recode_factor(race, `1` = "white", `2` = "black", `3` = "other")) %>%
  mutate_at(c("mother.smokes", "hypertension", "uterine.irr", "birthwt.below.2500"),
            ~ recode_factor(.x, `0` = "no", `1` = "yes"))

Assessing significance of factors and interactions in regression

Factors in linear regression

Interpreting coefficients of factor variables

In the case of quantitative predictors, we’re more or less comfortable with the interpretation of the linear model coefficient as a “slope” or a “unit increase in outcome per unit increase in the covariate”. This isn’t the right interpretation for factor variables. In particular, the notion of a slope or unit change no longer makes sense when talking about a categorical variable. E.g., what does it even mean to say “unit increase in major” when studying the effect of college major on future earnings?

To understand what the coefficients really mean, let’s go back to the birthwt data and try regressing birthweight on mother’s race and mother’s age.

# Fit regression model
birthwt.lm <- lm(birthwt.grams ~ race + mother.age, data = birthwt)

# Regression model summary
summary(birthwt.lm)
## 
## Call:
## lm(formula = birthwt.grams ~ race + mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2131.57  -488.02    -1.16   521.87  1757.07 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2949.979    255.352  11.553   <2e-16 ***
## raceblack   -365.715    160.636  -2.277   0.0240 *  
## raceother   -285.466    115.531  -2.471   0.0144 *  
## mother.age     6.288     10.073   0.624   0.5332    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 715.7 on 185 degrees of freedom
## Multiple R-squared:  0.05217,    Adjusted R-squared:  0.0368 
## F-statistic: 3.394 on 3 and 185 DF,  p-value: 0.01909

Note that there are two coefficients estimated for the race variable (raceother and racewhite). What’s happening here?

When you put a factor variable into a regression, you’re allowing a different intercept at every level of the factor. In the present example, you’re saying that you want to model birthwt.grams as

Baby’s birthweight = Intercept(based on mother’s race) + \(\beta\) * mother’s age


We can rewrite this more succinctly as: \[ y = \text{Intercept}_{race} + \beta \times \text{age} \]

Essentially you’re saying that your data is broken down into 3 racial groups, and you want to model your data as having the same slope governing how birthweight changes with mother’s age, but potentially different intercepts. Here’s a picture of what’s happening.

# Calculate race-specific intercepts
intercepts <- c(coef(birthwt.lm)["(Intercept)"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceother"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["racewhite"])

lines.df <- data.frame(intercepts = intercepts,
                       slopes = rep(coef(birthwt.lm)["mother.age"], 3),
                       race = levels(birthwt$race))

qplot(x = mother.age, y = birthwt.grams, color = race, data = birthwt) + 
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)
## Warning: Removed 1 rows containing missing values (geom_abline).

How do we interpret the 2 race coefficients? For categorical variables, the interpretation is relative to the given baseline. The baseline is just whatever level comes first (here, “black”). E.g., the estimate of raceother means that the estimated intercept is -285.4657796 higher among “other” race mothers compared to black mothers. Similarly, the estimated intercept is NA higher for white mothers than black mothers.

Another way of putting it: Among mothers of the same age, babies of white mothers are born on average weighing NAg more than babies of black mothers.

Why is one of the levels missing in the regression?

As you’ve already noticed, there is no coefficient called “raceblack” in the estimated model. This is because this coefficient gets absorbed into the overall (Intercept) term.

Let’s peek under the hood. Using the model.matrix() function on our linear model object, we can get the data matrix that underlies our regression. Here are the first 20 rows.

head(model.matrix(birthwt.lm), 20)
##    (Intercept) raceblack raceother mother.age
## 1            1         1         0         19
## 2            1         0         1         33
## 3            1         0         0         20
## 4            1         0         0         21
## 5            1         0         0         18
## 6            1         0         1         21
## 7            1         0         0         22
## 8            1         0         1         17
## 9            1         0         0         29
## 10           1         0         0         26
## 11           1         0         1         19
## 12           1         0         1         19
## 13           1         0         1         22
## 14           1         0         1         30
## 15           1         0         0         18
## 16           1         0         0         18
## 17           1         1         0         15
## 18           1         0         0         25
## 19           1         0         1         20
## 20           1         0         0         28

Even though we think of the regression birthwt.grams ~ race + mother.age as being a regression on two variables (and an intercept), it’s actually a regression on 3 variables (and an intercept). This is because the race variable gets represented as two dummy variables: one for race == other and the other for race == white.

Why isn’t there a column for representing the indicator of race == black? This gets back to our colinearity issue. By definition, we have that

raceblack + raceother + racewhite = 1 = (Intercept)


This is because for every observation, one and only one of the race dummy variables will equal 1. Thus the group of 4 variables {raceblack, raceother, racewhite, (Intercept)} is perfectly colinear, and we can’t include all 4 of them in the model. The default behavior in R is to remove the dummy corresponding to the first level of the factor (here, raceblack), and to keep the rest.

Interaction terms

Let’s go back to the regression line plot we generated above.

qplot(x = mother.age, y = birthwt.grams, color = race, data = birthwt) + 
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)
## Warning: Removed 1 rows containing missing values (geom_abline).

We have seen similar plots before by using the geom_smooth or stat_smooth commands in ggplot. Compare the plot above to the following.

qplot(x = mother.age, y = birthwt.grams, color = race, data = birthwt) +
  stat_smooth(method = "lm", se = FALSE, fullrange = TRUE)

In this case we have not only race-specific intercepts, but also race-specific slopes. The plot above corresponds to the model:

Baby’s birthweight = Intercept(based on mother’s race) + \(\beta\)(based on mother’s race) * mother’s age


We can rewrite this more succinctly as: \[ y = \text{Intercept}_{race} + \beta_{race}\times \text{age} \]

To specify this interaction model in R, we use the following syntax

birthwt.lm.interact <- lm(birthwt.grams ~ race * mother.age, data = birthwt)

summary(birthwt.lm.interact)
## 
## Call:
## lm(formula = birthwt.grams ~ race * mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2182.35  -474.23    13.48   523.86  1496.51 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           2583.54     321.52   8.035 1.11e-13 ***
## raceblack             1022.79     694.21   1.473   0.1424    
## raceother              326.05     545.30   0.598   0.5506    
## mother.age              21.37      12.89   1.658   0.0991 .  
## raceblack:mother.age   -62.54      30.67  -2.039   0.0429 *  
## raceother:mother.age   -26.03      23.20  -1.122   0.2633    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 710.7 on 183 degrees of freedom
## Multiple R-squared:  0.07541,    Adjusted R-squared:  0.05015 
## F-statistic: 2.985 on 5 and 183 DF,  p-value: 0.01291

We now have new terms appearing. Terms like racewhite:mother.age are deviations from the baseline slope (the coefficient of mother.age in the model) in the same way that terms like racewhite are deviations from the baseline intercept. This models says that:

On average among black mothers, every additional year of age is associated with a 21.4g decrease in the birthweight of the baby.

To get the slope for white mothers, we need to add the interaction term to the baseline.

\[ \begin{aligned} \beta_{racewhite} &= \beta_{raceblack} + \beta_{racewhite:mother.age} \\ &= \text{mother.age} + \text{racewhite:mother.age} \\ &= 21.4 + NA \\ &= NA \end{aligned} \]

This slope estimate is positive, which agrees with the regression plot above.

Is a categorical variable in a regression statistically significant?

Last class we considered modelling birthweight as a linear function of mother’s age, allowing for a race-specific intercept for each of the three race categories. This model is fit again below.

birthwt.lm <- lm(birthwt.grams ~ race + mother.age, data = birthwt)

Here’s a visualization of the model fit that we wound up with. Note that while there are 3 lines shown, this is a visualization of just one model: birthwt.grams ~ race + mother.age. This model produces 3 lines because the coefficients of the race variable result in different intercepts.

# Calculate race-specific intercepts
intercepts <- c(coef(birthwt.lm)["(Intercept)"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceother"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["racewhite"])

lines.df <- data.frame(intercepts = intercepts,
                       slopes = rep(coef(birthwt.lm)["mother.age"], 3),
                       race = levels(birthwt$race))

qplot(x = mother.age, y = birthwt.grams, color = race, data = birthwt) + 
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)
## Warning: Removed 1 rows containing missing values (geom_abline).

At this stage we may be interested in assessing whether the race variable is statistically significant. i.e., Does including the race variable significantly improve the fit of our model, or is the simpler model birthwt.grams ~ mother.age just as good?

Essentially, we want to know if the race-specific intercepts capture significantly more variation in the outcome (birthweight) than the single intercept model, or if allowing for different intercepts isn’t doing much more than capturing random fluctuations in the data.

Here’s a picture of the two models we’re comparing:

library(gridExtra)
## 
## Attaching package: 'gridExtra'
## The following object is masked from 'package:dplyr':
## 
##     combine
plot.complex <- qplot(x = mother.age, y = birthwt.grams, 
                      color = race, data = birthwt) + 
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

# Single intercept model (birthwt.grams ~ mother.age)
p <- ggplot(birthwt, aes(x = mother.age, y = birthwt.grams))
plot.simple <- p + geom_point(aes(colour = race)) + stat_smooth(method = "lm")

grid.arrange(plot.complex, plot.simple, ncol = 2)
## Warning: Removed 1 rows containing missing values (geom_abline).

To test this hypothesis, we use the anova function (not to be confused with the aov function). This function compares two nested models, accounting for their residual sums of squares (how well they fit the data) and their complexity (how many more variables are in the larger model) to assess statistical significance.

# Fit the simpler model with mother.age as the only predictor
birthwt.lm.simple <- lm(birthwt.grams ~ mother.age, data = birthwt)

# Compare to more complex model
anova(birthwt.lm.simple, birthwt.lm)
## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ mother.age
## Model 2: birthwt.grams ~ race + mother.age
##   Res.Df      RSS Df Sum of Sq      F  Pr(>F)  
## 1    187 99154173                              
## 2    185 94754346  2   4399826 4.2951 0.01502 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

This output tells us that the race variable is statistically significant: It is unlikely that the improvement in fit when the add the race variable is simply due to random fluctuations in the data. Thus it is important to consider race when modeling how birthweight depends on the mother’s age.

Is an interaction term significant?

Assessing significance of interaction terms operates on the same principle. We once again ask whether the improvement in model fit is worth the increased complexity of our model. For instance, consider the example we saw last class, where we allowed for a race-specific slope in addition to the race-specific intercept from before.

birthwt.lm.interact <- lm(birthwt.grams ~ race * mother.age, data = birthwt)
summary(birthwt.lm.interact)
## 
## Call:
## lm(formula = birthwt.grams ~ race * mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2182.35  -474.23    13.48   523.86  1496.51 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           2583.54     321.52   8.035 1.11e-13 ***
## raceblack             1022.79     694.21   1.473   0.1424    
## raceother              326.05     545.30   0.598   0.5506    
## mother.age              21.37      12.89   1.658   0.0991 .  
## raceblack:mother.age   -62.54      30.67  -2.039   0.0429 *  
## raceother:mother.age   -26.03      23.20  -1.122   0.2633    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 710.7 on 183 degrees of freedom
## Multiple R-squared:  0.07541,    Adjusted R-squared:  0.05015 
## F-statistic: 2.985 on 5 and 183 DF,  p-value: 0.01291

Here’s a side-by-side visual comparison of the race + mother.age model and the race + mother.age + race*mother.age interaction model.

plot.interact <- qplot(x = mother.age, y = birthwt.grams, color = race, data = birthwt) + stat_smooth(method = "lm", se = FALSE, fullrange = TRUE)
grid.arrange(plot.interact, plot.complex, ncol = 2)
## Warning: Removed 1 rows containing missing values (geom_abline).

So, do the lines with different slopes fit the data significantly better than the common slope model? Let’s compare the two with the anova() function.

anova(birthwt.lm, birthwt.lm.interact)
## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ race + mother.age
## Model 2: birthwt.grams ~ race * mother.age
##   Res.Df      RSS Df Sum of Sq      F Pr(>F)
## 1    185 94754346                           
## 2    183 92431148  2   2323199 2.2998 0.1032

This p-value turns out to not be statistically significant. So even though the estimated slopes in the interaction model look very different, our estimates are quite variable, so we don’t have enough evidence to conclude that the interaction term (different slopes) is providing significant additional explanatory power over the simpler race + mother.age model.

Is my complex model signficantly better than a simpler one?

The testing strategy above applies to any two nested models. Here’s an example where we add in a few more variables and see how it compares to the race + mother.age model from earlier.

birthwt.lm.complex <- lm(birthwt.grams ~ mother.smokes + physician.visits + race + mother.age, data = birthwt)

summary(birthwt.lm.complex)
## 
## Call:
## lm(formula = birthwt.grams ~ mother.smokes + physician.visits + 
##     race + mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2335.06  -455.16    31.74   499.29  1623.57 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      3282.407    261.326  12.561  < 2e-16 ***
## mother.smokesyes -424.651    110.371  -3.847 0.000165 ***
## physician.visits   14.391     48.953   0.294 0.769102    
## raceblack        -444.340    156.586  -2.838 0.005057 ** 
## raceother        -445.161    119.666  -3.720 0.000265 ***
## mother.age          1.547      9.996   0.155 0.877155    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 691.7 on 183 degrees of freedom
## Multiple R-squared:  0.1241, Adjusted R-squared:  0.1001 
## F-statistic: 5.184 on 5 and 183 DF,  p-value: 0.000179

Let’s compare to our earlier model:

anova(birthwt.lm, birthwt.lm.complex)
## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ race + mother.age
## Model 2: birthwt.grams ~ mother.smokes + physician.visits + race + mother.age
##   Res.Df      RSS Df Sum of Sq      F    Pr(>F)    
## 1    185 94754346                                  
## 2    183 87567280  2   7187067 7.5098 0.0007336 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Highly significant! This is probably due to the fact that mother’s smoking status has a tremendously high association with birthweight.

Summarizing stratified regressions

Gapminder life expectancy data

gapminder <- read_delim("http://www.andrew.cmu.edu/user/achoulde/94842/data/gapminder_five_year.txt",
                        delim = "\t") # Load data
## Parsed with column specification:
## cols(
##   country = col_character(),
##   year = col_double(),
##   pop = col_double(),
##   continent = col_character(),
##   lifeExp = col_double(),
##   gdpPercap = col_double()
## )
gapminder # A look at the data
## # A tibble: 1,704 x 6
##    country      year      pop continent lifeExp gdpPercap
##    <chr>       <dbl>    <dbl> <chr>       <dbl>     <dbl>
##  1 Afghanistan  1952  8425333 Asia         28.8      779.
##  2 Afghanistan  1957  9240934 Asia         30.3      821.
##  3 Afghanistan  1962 10267083 Asia         32.0      853.
##  4 Afghanistan  1967 11537966 Asia         34.0      836.
##  5 Afghanistan  1972 13079460 Asia         36.1      740.
##  6 Afghanistan  1977 14880372 Asia         38.4      786.
##  7 Afghanistan  1982 12881816 Asia         39.9      978.
##  8 Afghanistan  1987 13867957 Asia         40.8      852.
##  9 Afghanistan  1992 16317921 Asia         41.7      649.
## 10 Afghanistan  1997 22227415 Asia         41.8      635.
## # … with 1,694 more rows

Example: Maximum life expectancy for each continent

Before diving into a regression, let’s look at some data summaries that use some familiar functions. Here’s how we can form a table showing the maximum life expectancy on each continent each year, and the country that attained that maximum.

gapminder %>%
  group_by(continent, year) %>%
  summarize(max.life.exp = max(lifeExp),
            country = country[which.max(lifeExp)])
##   max.life.exp country
## 1       82.603   Japan

Example (more involved): Fitting a linear model for each country

We’re now going to go through an example where we get the life expectancy in 1952 and the rate of change in life expectancy over time for each country. The rate of change will be obtained by regressing lifeExp on year.

(1) Figure out how to extract the desired information.

Let’s start with the data for a single country.

country.name <- "Ireland"  # Pick a country
gapminder.sub <- filter(gapminder, country == country.name)  # Pull data for this country
gapminder.sub
## # A tibble: 12 x 6
##    country  year     pop continent lifeExp gdpPercap
##    <chr>   <dbl>   <dbl> <chr>       <dbl>     <dbl>
##  1 Ireland  1952 2952156 Europe       66.9     5210.
##  2 Ireland  1957 2878220 Europe       68.9     5599.
##  3 Ireland  1962 2830000 Europe       70.3     6632.
##  4 Ireland  1967 2900100 Europe       71.1     7656.
##  5 Ireland  1972 3024400 Europe       71.3     9531.
##  6 Ireland  1977 3271900 Europe       72.0    11151.
##  7 Ireland  1982 3480000 Europe       73.1    12618.
##  8 Ireland  1987 3539900 Europe       74.4    13873.
##  9 Ireland  1992 3557761 Europe       75.5    17559.
## 10 Ireland  1997 3667233 Europe       76.1    24522.
## 11 Ireland  2002 3879155 Europe       77.8    34077.
## 12 Ireland  2007 4109086 Europe       78.9    40676.
# Scatterplot of life exp vs year
# with a regression line overlaid
qplot(year, lifeExp, data = gapminder.sub, main = paste("Life expectancy in", country.name)) +
  geom_smooth(method = "lm")

We can confirm that it’s a pretty good model for other countries as well, though not for all of them

ggplot(data = gapminder, aes(x = year, y = lifeExp)) +
  facet_wrap( ~ country) +
  geom_point() +
  geom_smooth(method = "lm")

Now let’s fit a regression and extract the slope.

life.exp.lm <- lm(lifeExp ~ year, data = gapminder.sub) # Fit model
coef(life.exp.lm) # Get coefficients
##  (Intercept)         year 
## -321.1399594    0.1991196
coef(life.exp.lm)["year"]  # The slope that we wanted
##      year 
## 0.1991196

Here’s how we can do everything in one line:

coef(lm(lifeExp ~ year, data = gapminder.sub))["year"]
##      year 
## 0.1991196
(2) Extract information for each country

Here we’ll extract the slope in the way we practiced above, and we’ll also extract the “origin” life expectancy: the given country’s life expectancy in 1952, the first year of our data. For the purpose of plotting we’ll also want the continent information, so we’ll capture that in this call too.

progress.df <- gapminder %>%
  group_by(country) %>%
  summarize(continent = continent[1],
            origin = lifeExp[year == 1952],
            slope =  lm(lifeExp ~ year)$coef["year"]) 

progress.df
## # A tibble: 142 x 4
##    country     continent origin slope
##    <chr>       <chr>      <dbl> <dbl>
##  1 Afghanistan Asia        28.8 0.275
##  2 Albania     Europe      55.2 0.335
##  3 Algeria     Africa      43.1 0.569
##  4 Angola      Africa      30.0 0.209
##  5 Argentina   Americas    62.5 0.232
##  6 Australia   Oceania     69.1 0.228
##  7 Austria     Europe      66.8 0.242
##  8 Bahrain     Asia        50.9 0.468
##  9 Bangladesh  Asia        37.5 0.498
## 10 Belgium     Europe      68   0.209
## # … with 132 more rows

What can we learn from this output?

Let’s summarize our findings by creating a bar chart for the origin and slope, with the bars colored by continent. We’ll do this in ggplot.

Plotting origins coloured by continent

This code is analogous to that presented at the end of Lecture 6

# Reorder country factor by origin
# Construct bar chart
progress.df %>%
  mutate(country = reorder(country, origin)) %>%
  ggplot(aes(x = country, y = origin, fill = continent)) +
  geom_bar(stat = "identity") +
  theme(axis.text.x = element_text(angle = 60, vjust = 1, hjust = 1))

Plotting slopes coloured by continent

# Reorder country factor by slope
# Construct bar chart
progress.df %>%
  mutate(country = reorder(country, slope)) %>%
  ggplot(aes(x = country, y = slope, fill = continent)) +
  geom_bar(stat = "identity") +
  theme(axis.text.x = element_text(angle = 60, vjust = 1, hjust = 1))

These are very interesting plots. What can you tell from looking at them?

Looking at per capita GDP by year

Let’s start by looking at some plots of how GDP per capita varied by year

# Use qplot from ggplot2 to generate plots
qplot(year, gdpPercap, facets = ~ country, data = gapminder, colour = continent) +
  theme(axis.text.x = element_text(angle = 90, hjust = 1))

What if we want to rearrange the plots by continent? This can be done by changing the order of the country level.

# First step: reorder the countries by continent
# Produce a data frame that has the countries ordered alphabetically within continent
# Arrange sorts the data according to the variable(s) provided
country.df <- gapminder %>%
  group_by(country) %>%
  summarize(continent = continent[1]) %>%
  arrange(continent)

gapminder.ordered <- gapminder %>%
  mutate(country = factor(country, levels = country.df$country))

# Let's make sure that things are now ordered correctly...
levels(gapminder.ordered$country)
## character(0)
# Use qplot from ggplot2 to generate plots
qplot(year, gdpPercap, facets = ~ country, data = gapminder.ordered, colour = continent) + 
  theme(axis.text.x = element_text(angle = 90, hjust = 1)) + 
  stat_smooth(method = "lm")