Indian Statistical Institute admission tests (BStat, MStat, JRF in Statistics) · India · Indian Statistical Institute
Original practice questions written to the published syllabus · not past papersModels for counts, proportions and categories: contingency tables, the exponential-family GLM with its link functions, logistic and Poisson regression, log-linear models, overdispersion, ordinal outcomes and generalised linear mixed models, applied to germination, mortality, pest counts and adoption data.
In the exams: AICE-JRF Ag Stat Unit 2 'contingency tables, log linear models' (heavy); GATE DA logistic regression (present); light in UPSC ISS
How our questions for Indian Statistical Institute admission tests spread across the units of Categorical Data Analysis & Generalized Linear Models. Difficulty: 15 easy, 45 medium, 13 hard.
| Link functions, deviance and iteratively reweighted least squares | 8 | |
| Poisson regression for counts | 8 | |
| Ordinal and multinomial logistic models | 8 | |
| Contingency tables: association, odds ratios and tests | 7 | |
| Zero-inflated and hurdle models | 7 | |
| Model checking: residuals and goodness of fit | 7 | |
| Exponential family and the generalized linear model | 6 | |
| Overdispersion: quasi-likelihood and negative binomial models | 6 | |
| Generalized linear mixed models | 6 | |
| Logistic regression for binary and binomial data | 5 | |
| Log-linear models for multi-way tables | 5 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
For the 2 × 2 table with cells 80, 20 (treated) and 60, 40 (untreated), what is the 95% confidence interval for the odds ratio?
B. 1.42 to 5.02
SE(log θ̂) = √(1/80 + 1/20 + 1/60 + 1/40) = 0.323; exp(log 2.67 ± 1.96 × 0.323) = 1.42 to 5.02. Adding ±1.96 SE directly to θ̂ ignores the skewness of the odds ratio; using the variance instead of the SE gives too narrow an interval.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
What is the variance function V(μ) of the Poisson family?
C. μ
Poisson Var(y) = μ. The normal has V = 1, the gamma μ², the binomial proportion μ(1 − μ).
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
For a logit link, by roughly what factor do probit coefficients differ from logit coefficients fitted to the same data?
C. Probit coefficients are smaller, about 1/1.6 to 1/1.8 of the logit ones
The logistic distribution has SD π/√3 ≈ 1.81 against 1 for the standard normal, so logit coefficients are larger by about 1.6-1.8. The exact ratio depends on the data.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
A Poisson GLM with log link includes an intercept. What property do the fitted values have?
A. Σμ̂ᵢ = Σyᵢ
The likelihood equation for the intercept is Σ(yᵢ − μ̂ᵢ) = 0, so fitted and observed totals agree.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
How can the proportional-odds assumption be checked?
B. Score test, or slopes of cut-point logits
Separate cumulative binary logits should give similar slopes if the assumption holds; the score test formalises this. Partial proportional-odds models relax it.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
For a zero-inflated Poisson with π = 0.3 and λ = 2, what are the mean and variance?
B. Mean 1.4, variance 2.24
E(Y) = (1 − π)λ = 1.4; Var(Y) = (1 − π)λ(1 + πλ) = 1.4 × 1.6 = 2.24. Zero inflation alone makes the counts overdispersed.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
In a hurdle model for pest counts, where do zeros come from?
C. Only from the binary part
The count part is zero-truncated, so every zero comes from the binary zero-versus-positive model. In zero-inflated models zeros come from both sources.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
In a Poisson GLMM with log link and a normal random intercept, how do conditional and marginal slopes compare?
B. They are equal; only the intercept differs
E(y) = exp(x′β)E(e^b) = exp(x′β + σ²/2), so only the intercept shifts by σ²/2; slopes are the same on both scales.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
How can an observation-level random effect help in a Poisson GLMM?
A. It absorbs overdispersion
A normal random effect for each observation makes the counts a Poisson-lognormal mixture, with variance exceeding the mean.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
How is leverage defined for a GLM?
A. From the weighted hat matrix W^(1/2)X(X′WX)⁻¹X′W^(1/2)
IRLS weights enter the hat matrix, so leverage depends on fitted means as well as the design; Cook's distance combines leverage and residual size.
Further reading: McCullagh and Nelder, Generalized Linear Models; Agresti, Categorical Data Analysis; Dobson and Barnett, An Introduction to Generalized Linear Models
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