| Question | Formula | Condition / interpretation |
|---|---|---|
| Discrete expectation | E(X)=Σ x p(x) Var(X)=E(X²)−E(X)² | Sum over support; required moments must exist. |
| Binomial count | P(X=k)=choose(n,k)pᵏ(1−p)ⁿ⁻ᵏ E(X)=np; Var(X)=np(1−p) | Fixed n, independent trials, common p. |
| Standardise a normal variable | Z=(X−μ)/σ | X normal; σ is the SD, not the variance. |
| SE of a mean | SE(X̄)=σ/√n; estimated SE=s/√n | Independent identically distributed observations, finite variance. |
| Known-SD mean interval | x̄ ± z1−α/2 σ/√n | Exact under iid normal sampling and known σ; otherwise possible large-sample approximation. |
| Unknown-SD mean interval | x̄ ± tn−1,1−α/2 s/√n | Exact under iid normal sampling. For paired data, use the differences. |
| Approximate proportion interval | p̂ ± z1−α/2 √[p̂(1−p̂)/n] | Wald approximation; problematic near boundaries or with small n. Wilson/exact alternatives are discussed. |
| Mean test | t=(x̄−μ₀)/(s/√n) | t with n−1 df under the iid normal null. |
| Welch comparison | t=[x̄₁−x̄₂−Δ₀]/√[s₁²/n₁+s₂²/n₂] | Independent groups; approximate Welch df. Does not require equal variances. |
| Prediction of one future observation | x̄ ± tn−1,1−α/2 s√(1+1/n) | Independent future draw from the same normal population. |
| Pearson chi-square | X²=Σ(O−E)²/E Eij=rowi×columnj/N | Independent count observations and adequate expected counts. Table df=(r−1)(c−1). |
| Goodness of fit | Ej=npj; df=k−1 | k categories with prespecified probabilities and no estimated parameters. Fitting parameters requires further care. |
Use these as separate practice prompts. They do not specify the final exam's questions, marks or permitted aids.
Four of ten applicant files concern vocational graduates. Three distinct files are selected at random. Find the probability that exactly two are vocational.
Worked answer: Hypergeometric sampling without replacement. P=choose(4,2)choose(6,1)/choose(10,3)=36/120=0.30. A binomial model would ignore the changing success probability.
A population has SD 12. Compare the SE of the sample mean for n=36 and n=144 independent observations.
Worked answer: 12/√36=2 and 12/√144=1. Four times as many observations halves the SE. This does not remove selection bias.
Synthetic earnings: n=100, mean €1800 and known population SD €400. Use z=1.96 for a 95% mean interval.
Worked answer: SE=40; margin=78.4; interval [€1721.60, €1878.40]. Under the normal independent-sampling model, 95% of repeated intervals cover the fixed mean. It is not an interval containing 95% of individual earnings.
Synthetic search duration: n=25, mean 5.4 months, SD 2. Test mean 6 against a prespecified lower alternative. The lower 5% t critical value with 24 df is −1.711.
Worked answer: t=(5.4−6)/(2/5)=−1.5. It is not below −1.711, so fail to reject at 5%. Do not conclude the mean equals 6; question the normal model for possibly skewed durations.
Scores are observed before and after training for the same 30 workers. A significant positive mean change is found. Which test fits the pairing, and what causal claim is justified?
Worked answer: A paired t test is a one-sample t test of after-minus-before differences, exact for independent normal differences across workers. Positive change alone does not establish training impact; time trends or other changes may explain it.
In a three-by-two table, one row total is 60, one column total is 80 and N=200. Find that cell's expected count under independence and the table's df.
Worked answer: E=60×80/200=24; df=(3−1)(2−1)=2. Inspect all expected counts and the independence of observations before using the chi-square approximation.
For each question: can you name the target, justify the model, show the calculation, interpret the result and state a limitation? The website records practice completion when each module reaches at least 4/5. These formative exercises do not contribute to the examination grade.