Today we’ll be applying the methods we’ve learned so far.
The dataset we’ll be using comes from the Armed Conflict Location and Event Data database. It contains data on armed conflict events and fatalities in Palestine between January 2020 to January 2025. I’ve cleaned the data somewhat so that it contains the following variables:
Admin1: “West Bank” or “Gaza Strip”Admin2: The city or townMonth: 1 to 12YearEvents: The number of armed conflict events in the
given month and year, per ACLED’s definition.Fatalities: The total number of fatalities across all
events in the given month and year, per ACLED’s estimate.Download the data (filename acledpalestine.csv) here.
Import the data and filter out only the rows of data from the West Bank.
We will only be looking at armed conflict events in the West Bank.
Create a new numeric binary variable called prepost that
takes on value \(1\) for time periods
post-October 2023 (including October 2023) and \(0\) for time periods pre-October 2023.
# [Your Code Here]
df <- read.csv("acledpalestine.csv")
df <- df %>%
filter(Admin1 == "West Bank") %>%
mutate(prepost = ifelse((Month >= 10 & Year == 2023) | (Year > 2023), 1, 0))What is the average number of monthly fatalities in the West Bank prior to October 2023? What is the average number after (and including) October 2023? Then compute the difference-in-means estimator.
# [Your Code Here]
mean(df[df$prepost == 1, ]$Fatalities)## [1] 5.585227mean(df[df$prepost == 0, ]$Fatalities)## [1] 1.052525diff_prepost <- mean(df[df$prepost == 1, ]$Fatalities) - mean(df[df$prepost == 0, ]$Fatalities)
df %>%
group_by(prepost) %>%
summarize(mean_fatalities = mean(Fatalities))## # A tibble: 2 × 2
## prepost mean_fatalities
## <dbl> <dbl>
## 1 0 1.05
## 2 1 5.59For a sample of \(n_X\) independent and identically distributed (IID) random variables \(X_1, \cdots, X_{n_X}\) and another sample of \(n_Y\) IID random variables \(Y_1, \cdots, Y_{n_Y}\), the standard error for the difference-in-means estimator when variances are unequal1 is given by:
\[ SE = \sqrt{\frac{s^2_X}{n_X}+\frac{s^2_Y}{n_Y}} \]
where \(s_X^2\) and \(s_Y^2\) are the sample variances
Compute the standard error of the difference-in-means estimator.
sd_post <- sd(df[df$prepost == 1, ]$Fatalities)
sd_pre <- sd(df[df$prepost == 0, ]$Fatalities)
n_post <- sum(df$prepost == 1)
n_pre <- sum(df$prepost == 0)
se_df <- sqrt((sd_post^2 / n_post) + (sd_pre^2 / n_pre))
se_df## [1] 0.6093912The 95% confidence interval (CI) for the difference in means estimator \(\hat{\delta}\) (where the variance of the two groups is not equal) is given by:
\[ CI \approx \left( \hat{\delta} - 1.96 \times SE \; , \; \hat{\delta} + 1.96 \times SE \right) \] 1.96 is the (approximate) critical value for a 95% confidence level under the standard normal distribution, but you can use 2.576 for 99% CI and 1.645 for 90% CI.
Calculate the 95% confidence intervals for the difference-in-means estimate.
ci_df <- c(diff_prepost - se_df*1.96, diff_prepost + se_df*1.96)
ci_df## [1] 3.338295 5.727109We want to know if the difference in means is statistically significant. To do so, we first compute the t-statistic, which for the two-sample t-test is \[ t = \frac{\hat{\delta}}{SE} \] Under the null hypothesis (\(\hat{\delta} = 0\)), the t-statistic should follow a t-distribution. The t-distribution has degrees of freedom \(\texttt{df} = n_X + n_Y - 2\) if variances are equal or uses Welch’s approximation of the degrees of freedom2 The t-distribution is very similar to the normal distribution but has fatter tails to capture the greater variability from using an estimate of the standard deviation rather than the population sample deviation (which is unknown).
Thus, given some t-statistic we can calculate the p-value: the
probability that we get this t-statistic when the null is true (i.e.,
when there is no difference in means). We do so by using
pt(q, df), which calculates the area under the curve up to
q for the \(t_{\texttt{df}}\) distribution.
You can conduct a t-test in R by using
t.test(x_values, y_values, "two.sided", mu=0, var.equal = TRUE)
t.test(df[df$prepost == 1, ]$Fatalities, df[df$prepost == 0, ]$Fatalities, alternative = "two.sided", mu=0, var.equal = FALSE)##
## Welch Two Sample t-test
##
## data: df[df$prepost == 1, ]$Fatalities and df[df$prepost == 0, ]$Fatalities
## t = 7.4381, df = 185.53, p-value = 3.673e-12
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 3.330475 5.734929
## sample estimates:
## mean of x mean of y
## 5.585227 1.052525When variances are equal, it is better to use the pooled standard error, \(\sqrt{\frac{s_{\text{total}}^2}{\frac{1}{n_X} + \frac{1}{n_Y}}}\)↩︎
You don’t need to know the formula for this. R can calculate it for you. If you’re interested, it’s \(df= \frac{ \left(\dfrac{s_{1}^{2}}{n_1}+\dfrac{s_{2}^{2}}{n_2}\right)^{\!2} }{ \dfrac{(s_{1}^{2}/n^{}_1)^2}{n_1-1}+ \dfrac{(s_{2}^{2}/n^{}_2)^2}{n_2-1} }\).↩︎