A1 — Describe an algorithm
Specify input and desired output → identify missing answers → select valid answers → compute sum and sample size → divide sum by valid count → document the exclusion rule.
This answer key is public
All exercises and numbers below are original, fictional teaching examples. They are not solutions to UCL's original assessments. Publishing the key is intentional; students should attempt the starter file before checking it.
Worked R program
# ferdo R Lab — Lesson 0 · instructor worked solution (synthetic data)
scores <- c(12, 10, 14, NA, 9)
length(scores) # 5
mean(scores) # NA
mean(scores, na.rm = TRUE) # 11.25
sum(!is.na(scores)) # 4
id <- c("A", "B", "C", "D", "E")
before <- c(12, 10, 14, NA, 9)
after <- c(14, 13, 13, 15, 12)
delta <- after - before
print(data.frame(id, before, after, delta))
valid_pairs <- sum(!is.na(delta))
increased <- sum(delta > 0, na.rm = TRUE)
decreased <- sum(delta < 0, na.rm = TRUE)
paired_mean_change <- mean(delta, na.rm = TRUE)
pre_mean <- mean(before, na.rm = TRUE)
post_mean <- mean(after)
stopifnot(valid_pairs == 4, increased == 3, decreased == 1,
paired_mean_change == 1.75,
pre_mean == 11.25, post_mean == 13.4)
Download instructor .R solution
Uses only base R; no additional packages or private data.
Expected numerical answers
| Calculation | Answer | Reason |
|---|---|---|
| Number of values | 5 | `length()` counts the missing element too. |
| Mean without removing NA | NA | R propagates missingness under the default rule. |
| Mean of observed pretest values | 11.25 | (12+10+14+9)/4 |
| Valid matched participants | 4 | D has no pretest; D's change is unknown. |
| Individual change values | 2, 3, −1, NA, 3 | Computed as posttest minus pretest for each person. |
| Positive / negative changes | 3 / 1 | Only matched records can be classified. |
| Mean of matched changes | 1.75 | (2+3−1+3)/4 |
| Mean of all posttest values | 13.40 | (14+13+13+15+12)/5 |
| Posttest mean − observed pretest mean | 2.15 | Different denominators; not the same as mean matched change. |
Six exercises with model answers
A2 — Predict an output
`mean(c(2,4,6))` = 4.
B1 — Count valid pretest records
`sum(!is.na(before))` = 4.
B2 — Determine change direction
`sum(delta>0,na.rm=TRUE)` = 3; `sum(delta<0,na.rm=TRUE)` = 1.
C1 — Fix a missing mean
`mean(before)` returns NA; `mean(before, na.rm=TRUE)` returns 11.25. This excludes missingness from the calculation but does not solve attrition bias.
C2 — Challenge causality
A difference alone cannot establish a treatment effect: consider missing observations, regression to the mean, selection and control-group/counterfactual evidence.
Why 1.75 ≠ 2.15
The mean of paired individual changes uses four people because D has no pretest; the mean of all posttest scores includes five people. Subtracting the mean of four observed pretests from the mean of five posttests compares different sets of records. Even correct arithmetic can be a poor match for the research question.
Neither figure, by itself, establishes causation.
Common conceptual mistakes
- `length(x)` includes NA; `sum(!is.na(x))` counts observed values.
- `na.rm=TRUE` omits missing items from that operation; it does not replace them with zero.
- `<-` assigns, whereas `==` compares for equality.
- R is a language; RStudio is a development environment.
- Code that runs without error is not necessarily an appropriate statistical analysis.
- Screening measures like AQ-10 are not clinical diagnoses.
Exit-ticket model answers
- Programming specifies a testable sequence to solve a problem; typing syntax is only one part.
- No. R executes code; RStudio makes writing and running it easier.
- Define the question, input and desired output, then plan the transformations and checks.
- Check whether values are truly missing and choose a defensible rule for handling NA.
- An error-free command can use incorrect labels, omit key observations or support an invalid interpretation.