ASRB NET Bioassay, Survival & Epidemiological Statistics: practice questions

ASRB NET - Agricultural Statistics (and Computer Application / Bioinformatics) · India · Agricultural Scientists Recruitment Board (ICAR)

Original practice questions written to the published syllabus · not past papers

10worked here, free

80written to this exam in this subject

11syllabus units

What this subject covers

Dose and response, time to event and disease in populations: direct and indirect assays, probit and logit analysis of LD50 and LC50, parallel-line and slope-ratio assays, Kaplan-Meier and Cox models, and the statistical side of plant and animal epidemiology. Introductory veterinary epidemiology lives in the Veterinary track; this is the statistics behind it.

In the exams: Light in AICE-JRF Ag Stat; ICAR MSc/PhD elective; plant-protection and veterinary research (applied); UPSC ISS (light)

Direct and indirect bioassays
Quantal response: probit and logit analysis, LD50 and LC50
Parallel-line and slope-ratio assays; Fieller's theorem
Survival data: censoring and truncation
Kaplan-Meier estimator and the log-rank test
Cox proportional hazards model
Parametric survival models: exponential and Weibull
Measures of disease frequency and association: risk ratio and odds ratio
Study designs: cohort, case-control and cross-sectional
Diagnostic-test evaluation: sensitivity, specificity and ROC curves
Disease progress curves and the area under the disease progress curve

What ASRB NET asks from it

ASRB NET / ARS and research-institute recruitment (plant protection, animal science)

Applied choice and interpretation: resistance ratios, Abbott's correction, Kaplan-Meier and log-rank comparisons, hazard ratios, diagnostic-test predictive values and AUDPC.

Where the 80 questions fall

How our questions for ASRB NET spread across the units of Bioassay, Survival & Epidemiological Statistics. Difficulty: 16 easy, 48 medium, 16 hard.

Parallel-line and slope-ratio assays; Fieller's theorem	9
Diagnostic-test evaluation: sensitivity, specificity and ROC curves	9
Disease progress curves and the area under the disease progress curve	9
Direct and indirect bioassays	8
Quantal response: probit and logit analysis, LD50 and LC50	8
Measures of disease frequency and association: risk ratio and odds ratio	8
Study designs: cohort, case-control and cross-sectional	8
Kaplan-Meier estimator and the log-rank test	6
Survival data: censoring and truncation	5
Cox proportional hazards model	5
Parametric survival models: exponential and Weibull	5

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Quantal response: probit and logit analysis, LD50 and LC50 · hard

If log tolerances are normal with mean μ and SD σ, what is the slope of the probit line against log dose?

1/σ
1/μ
μ/σ
σ

Answer and working

A. 1/σ

Φ⁻¹(P) = (x − μ)/σ, so the slope is 1/σ: a steep line means homogeneous tolerances. The LD50 is at x = μ.

Further reading: Finney, Statistical Method in Biological Assay; Klein and Moeschberger, Survival Analysis; Dohoo, Martin and Stryhn, Veterinary Epidemiologic Research; Madden, Hughes and van den Bosch, The Study of Plant Disease Epidemics

2 · Quantal response: probit and logit analysis, LD50 and LC50 · medium

A field aphid population has LC50 = 120 mg/L and a susceptible strain 8 mg/L. What is the resistance ratio?

15
128
112
0.067

Answer and working

A. 15

Resistance ratio = LC50(field)/LC50(susceptible) = 120/8 = 15. 112 is a difference, not a ratio.

3 · Quantal response: probit and logit analysis, LD50 and LC50 · medium

Why are doses giving 0% or 100% mortality of little use in estimating a probit line?

They cause the LD50 to equal zero
Little information on the line
They make the probit undefined in software
They are always excluded by regulation

Answer and working

B. Little information on the line

Binomial information np(1 − p) and the probit weights vanish as p approaches 0 or 1, so such groups barely constrain the fit; design should target 10-90% responses.

4 · Parametric survival models: exponential and Weibull · medium

An exponential survival model has hazard 0.02 per day. What is the median survival time?

25.0 days
0.0139 days
50.0 days
34.7 days

Answer and working

D. 34.7 days

Median = (log 2)/λ = 0.693/0.02 = 34.7 days; 50 days is the mean.

5 · Measures of disease frequency and association: risk ratio and odds ratio · easy

Thirty of 200 animals tested on one day are positive. What is this measure?

Point prevalence, 0.15
Incidence rate, 0.15 per day
Incidence risk, 0.15
Attack rate, 0.30

Answer and working

A. Point prevalence, 0.15

A proportion diseased at one time is a prevalence; incidence needs new cases over a period.

6 · Measures of disease frequency and association: risk ratio and odds ratio · medium

In plant pathology, what does disease 'incidence' usually measure?

The rate of new infections per day
The proportion of plant units showing disease
The proportion of tissue area affected
The yield loss per infected plant

Answer and working

B. The proportion of plant units showing disease

Plant-pathology incidence counts diseased units (plants, leaves) — a prevalence in the epidemiological sense; severity measures tissue affected.

7 · Study designs: cohort, case-control and cross-sectional · easy

Which design selects subjects by disease status and compares their past exposures?

Randomised field trial
Cross-sectional
Case-control
Cohort

Answer and working

C. Case-control

Cases and controls are sampled by outcome; cohort studies sample by exposure and follow forward.

8 · Diagnostic-test evaluation: sensitivity, specificity and ROC curves · medium

Positives from a screening test are confirmed by a second test before culling. Which property increases?

Prevalence
Sensitivity
Apparent prevalence
Specificity

Answer and working

D. Specificity

Serial interpretation (positive only if both positive) raises specificity and lowers sensitivity.

9 · Disease progress curves and the area under the disease progress curve · medium

An AUDPC of 315 %-days covers 14 days. What is the standardised AUDPC?

22.5%
45%
4.4%
315%

Answer and working

A. 22.5%

Standardised AUDPC = AUDPC/duration = 315/14 = 22.5%, the average severity over the period.

10 · Disease progress curves and the area under the disease progress curve · medium

Which model suits a monocyclic soil-borne disease whose inoculum does not increase in the season?

Gompertz
Logistic
Monomolecular
Exponential

Answer and working

C. Monomolecular

Monomolecular growth dy/dt = r(1 − y) reflects infection from a fixed inoculum; logistic growth suits polycyclic spread.

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