# Session 1 Homework: Foundations of Policy Evaluation

Estimated workload: 45-60 minutes, including 15-25 minutes in R.

Submit short answers in complete sentences. Use the notation from the lecture: treatment `D`, outcome `Y`, potential outcomes `Y(1)` and `Y(0)`, and assignment `Z` only when a lottery or offer is discussed.

## Exercise 1. From a Policy Question to a Credible Comparison

A city offers free after-school tutoring to eligible pupils. At the end of the year, pupils who attended tutoring have higher test scores than pupils who did not attend. The city is considering whether to expand the programme next year, but participation this year was voluntary.

Answer the following questions in short paragraphs.

1. Rewrite the city's question as an evaluation question. Name the unit, treatment `D`, outcome `Y`, target population, time horizon, and the estimand that best matches the expansion decision.

2. For pupils who attended tutoring, what is the missing counterfactual? Explain why comparing attendees with non-attendees may fail to recover it.

3. Give one plausible selection story that would make the observed attendee/non-attendee difference too optimistic, and one story that would make it too pessimistic.

4. Suppose next year the city has too few tutoring places and assigns offers by lottery. Define `Z`, `D`, and `Y`. Which comparison would estimate the effect of being offered tutoring? What would still be needed to learn the effect of actually attending tutoring?

## Exercise 2. A Short R Comparison

This exercise uses a tiny artificial dataset. The goal is to reproduce one empirical idea from the lecture: a treated-control comparison is easy to compute, but its causal interpretation requires an argument.

The companion R file is `r_exercises/week_01_foundations.R`.

Copy the code below into R or RStudio and run it line by line. Replace the `___` blanks before running the last four lines.

```r
dat <- data.frame(
  pupil = c("A", "B", "C", "D", "E", "F", "G", "H"),
  D = c(1, 1, 1, 1, 0, 0, 0, 0),
  X = c(62, 70, 58, 75, 65, 69, 60, 72),
  Y = c(73, 78, 69, 82, 68, 71, 64, 75)
)

head(dat)

treated_mean <- mean(dat$Y[dat$D == 1])
control_mean <- mean(dat$Y[dat$D == ___])
naive_difference <- treated_mean - ___

model <- lm(Y ~ ___, data = dat)
summary(model)
```

Answer the following questions.

1. Report the mean outcome for pupils with `D = 1`, the mean outcome for pupils with `D = 0`, and the naive difference.

2. In the regression `lm(Y ~ D, data = dat)`, what does the coefficient on `D` measure? Compare it with the naive difference.

3. Can this coefficient be interpreted as a causal effect of tutoring? State the key missing counterfactual or assumption that would be needed.