ICAR AICE-JRF/SRF (PhD) — Subject 59 Forestry · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersDesigning and analysing forestry research: field experiments with long-lived trees, advanced designs, mixed models, long-term plots and scientific writing.
In the exams: ICAR AICE-JRF 59 (PhD level across units); ICAR AICE-JRF Part A; ASRB NET and ARS recruitment
How our questions for ICAR AICE-JRF/SRF spread across the units of Forestry Research Methods & Experimental Design. Difficulty: 0 easy, 26 medium, 18 hard.
| Factorial, split-plot and incomplete block designs | 6 | |
| Mixed-effects models and repeated measures | 6 | |
| Multivariate methods: PCA, cluster analysis, ordination | 6 | |
| Meta-analysis and systematic reviews | 6 | |
| Scientific method and research problem formulation | 5 | |
| Field experiments with trees: plot size, borders, replication | 4 | |
| Research ethics, data management and reproducible analysis | 4 | |
| Scientific writing, peer review and research proposals | 4 | |
| Long-term experiments and permanent sample plots | 3 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Why can a well-designed manipulative experiment support a causal claim more strongly than an observational study across sites?
B. Randomisation breaks links between treatment and unmeasured confounders
Random assignment makes treatment independent of unmeasured variables on average. Observational studies can be large and well analysed, but confounding remains possible.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
Two treatments are to be compared with σ = 4 cm, smallest important difference δ = 3 cm, α = 0.05 (z = 1.96) and power 0.80 (z = 0.84). Using n ≈ 2(z₁ + z₂)²σ² ÷ δ², how many replicates per treatment are needed?
B. 28
n ≈ 2 × (1.96 + 0.84)² × 16 ÷ 9 = 2 × 7.84 × 1.78 ≈ 27.9, rounded up to 28. 14 omits the factor of 2; 56 counts both treatments together.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
Which practice best reduces the risk that a research question is unanswerable with the planned study?
B. Pilot data and a power analysis before committing to the design
A pilot study estimates variance and feasibility; power analysis tests whether the design can detect the effect. Deciding analyses after seeing the data invites p-hacking.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
Alpha designs are popular in tree breeding trials mainly because they are:
C. Resolvable incomplete block designs flexible in size
Alpha (alpha-lattice) designs group incomplete blocks into complete replicates and exist for many entry numbers and block sizes (Patterson and Williams 1976).
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
Two models differ only in their fixed effects. To compare them by AIC or a likelihood ratio test you should fit them by:
A. Maximum likelihood
REML likelihoods depend on the fixed-effects structure and are not comparable across different fixed effects; ML must be used.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
Best linear unbiased predictions (BLUPs) of family effects in a breeding trial:
D. Shrink extreme family means towards the mean
BLUPs shrink means in proportion to their unreliability, so families with few trees are pulled more strongly to the mean.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
Model A has log-likelihood −250 with 5 parameters; model B has −246 with 8 parameters. Which has the lower AIC?
D. Model B: 508 against 510 for A
AIC = 2k − 2 ln L: A = 10 + 500 = 510; B = 16 + 492 = 508, so B is lower, though a difference of 2 is modest support. −490 and −476 come from adding 2 ln L instead of subtracting it; 500 and 492 omit the parameter penalty and compare −2 ln L alone.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
Site A has abundances (10, 5, 0) and site B (6, 5, 3) of three species. What is the Bray–Curtis dissimilarity?
C. 0.24
C = 6 + 5 + 0 = 11; BC = 1 − 2 × 11 ÷ (15 + 14) = 1 − 22 ÷ 29 ≈ 0.24. 0.76 is the similarity, 22 ÷ 29.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
An NMDS ordination of plot communities has a stress of 0.25. This indicates:
D. A poor fit; interpret the ordination with caution
Stress above about 0.2 is regarded as poor. Stress is not a proportion of variance explained.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
A researcher runs 10 independent tests and applies a Bonferroni correction for an overall α of 0.05. Each test uses:
C. 0.005
α ÷ m = 0.05 ÷ 10 = 0.005.
Further reading: Gomez K; Montgomery D; Hurlbert S; Williams E
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