The arithmetic is genuinely annoying to do by hand.
Start by clearly stating the formula for Welch's t-statistic and Satterthwaite's degrees of freedom, then compute each component step-by-step with the given numbers. After obtaining the t-statistic and df, use a t-distribution to find the two-sided p-value, and finally explain why Welch's is preferred over the pooled t-test in this context.
Pro tip: Mention that Welch's t-test is the default in many statistical software packages (e.g., R's t.test) because it maintains Type I error rates even when variances are unequal, and it performs well even when sample sizes are unequal, as in this case.
Clearly define the null and alternative hypotheses (two-sided) and state that you will use Welch's t-test due to unequal variances and sample sizes.
Calculate the standard error using the formula SE = sqrt(s1^2/n1 + s2^2/n2), then compute t = (mean1 - mean2) / SE.
Use the formula df = (s1^2/n1 + s2^2/n2)^2 / [ (s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1) ] and compute the value.
Using the t-statistic and df, find the two-sided p-value from a t-distribution (e.g., using a t-table or approximation) and round to three decimal places.
Explain that Welch's t-test does not assume equal variances, is robust to unequal sample sizes, and controls Type I error better when variances differ.
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Part one is fine, just scale the rate to λ=3 for the 1.5-hour window and use the complement.
First, compute the probability using the Poisson distribution with mean λt = 2 * 1.5 = 3. Then, for the conditional probability, recognize that given N(2)=1, the event time is uniformly distributed on [0,2] due to the memoryless property of the Poisson process. Finally, calculate the probability that the event occurred within the first 0.5 hours as 0.5/2 = 0.25, and explicitly name the property as the memoryless property (or independent increments).
Pro tip: When explaining the conditional part, emphasize that the uniform distribution of the single event time is a direct consequence of the memoryless property, which is a key concept in Poisson processes. This shows deep understanding beyond just plugging into formulas.
Recognize that N(t) follows a Poisson distribution with mean λt. For the first part, λ=2 and t=1.5, so mean = 3.
Calculate 1 - P[N(1.5) ≤ 2] using the Poisson PMF: P(N=k) = e^{-3} * 3^k / k!. Sum for k=0,1,2 and subtract from 1.
Given N(2)=1, the time of the event is uniformly distributed on [0,2] due to the memoryless property. So the conditional probability is the length of the interval [0,0.5] divided by the total interval [0,2].
Compute P(event in first 0.5 hours | N(2)=1) = 0.5 / 2 = 0.25.
Explicitly state that the property used is the memoryless property of the Poisson process (or the property that conditional on the number of events, the event times are uniformly distributed).
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This is the question I spent the most time on and still felt shaky leaving it.
Start by framing the problem: with 200 predictors, n=500, and block correlations ~0.9, you need a regularized linear model that balances prediction and interpretability. Compare LASSO, Ridge, and Elastic Net, arguing that Elastic Net is likely best because it handles correlated groups while still performing some selection. Then explain nested CV for tuning λ and α, and discuss how high within-group correlations affect selection stability and prediction error.
Pro tip: Emphasize that with such high correlations, LASSO's selection is unstable and may arbitrarily pick one predictor per group, while Elastic Net's L2 component encourages grouping and stability. Also mention that Ridge often has the best prediction error but no sparsity, so Elastic Net offers a practical compromise.
Explain that Ridge minimizes prediction error but keeps all predictors, LASSO gives sparse models but unstable selection under high correlation, and Elastic Net combines both penalties to balance the two goals.
Argue that Elastic Net is preferable because its L2 component handles correlated groups by shrinking coefficients together, while the L1 component still yields some sparsity for interpretability.
Outline an outer loop for unbiased performance estimation and an inner loop for hyperparameter tuning (grid search over λ and α) using cross-validation, ensuring no data leakage.
Explain that correlations ~0.9 cause LASSO to arbitrarily select one variable per group, leading to unstable selection across resamples; Elastic Net mitigates this by grouping correlated variables, improving stability at the cost of slightly less sparsity.
Suggest using Elastic Net with nested CV, and if interpretability is paramount, consider post-selection inference or grouping methods; if prediction is paramount, Ridge might be sufficient.
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Welch's t-test doesn't apply here because the pairing structure matters and you'd be throwing away information.
Clarify that the paired structure means you should analyze within-subject differences, not use Welch's t-test (which is for independent groups). Assess whether the mild non-normality and unequal variances of the original measurements affect the distribution of the differences; with n=25, the paired t-test is likely robust if the differences are not severely skewed or heavy-tailed. If the differences show strong non-normality or outliers, use a nonparametric alternative like the Wilcoxon signed-rank test, and report an appropriate effect size (Cohen's dz for the t-test, r for Wilcoxon).
Pro tip: Emphasize that the paired t-test is on the differences, so unequal variances of the two measurements are irrelevant; what matters is the distribution of the differences. Also, mention that with n=25, the Central Limit Theorem provides reasonable robustness unless non-normality is extreme.
Recognize that the data are paired (repeated measures on the same subject), so the analysis should focus on within-subject differences. Welch's t-test is for independent samples and is therefore inappropriate.
The paired t-test assumes that the differences are approximately normally distributed. Unequal variances of the two measurements do not violate this assumption. With n=25, mild non-normality is usually acceptable due to the robustness of the t-test.
If the differences are severely non-normal (e.g., strong skewness, outliers), consider a nonparametric alternative such as the Wilcoxon signed-rank test. Otherwise, proceed with the paired t-test.
State your decision: if differences are roughly symmetric without extreme outliers, use the paired t-test; if not, use Wilcoxon signed-rank. Justify based on robustness and the specific data characteristics.
For the paired t-test, report Cohen's dz (mean difference divided by standard deviation of differences). For Wilcoxon, report the matched-pairs rank-biserial correlation or r = Z/√N.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.