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ICAR AICE-JRF/SRF Biostatistics, Computer Application & Research Methodology: practice questions

ICAR AICE-JRF/SRF (PhD) · India · NTA for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
31written to this exam in this subject
11syllabus units

What this subject covers

The statistics a veterinarian needs to read a paper, design a trial or analyse herd records: descriptive statistics, tests of significance, regression and experimental design.

In the exams: ICAR AIEEA-PG 14 (statistics unit); ASRB NET

What ICAR AICE-JRF/SRF asks from it

ICAR AICE-JRF/SRF (PhD)

Research methodology, experimental design (RBD, Latin square, factorial, missing plot), sample-size reasoning, multiple comparisons, regression and the interpretation of P values and confidence intervals.

Where the 31 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Biostatistics, Computer Application & Research Methodology. Difficulty: 1 easy, 10 medium, 20 hard.

Sample-size calculation5
Correlation and regression4
Experimental designs: CRD and RBD4
Research methodology4
Analysis of variance3
Computer and spreadsheet basics3
Probability distributions2
Tests of significance: the t-test2
The chi-square test2
Measures of dispersion1
Sampling methods1

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 · The chi-square test · hard

Two diagnostic tests are applied to the same 200 sheep. Which test compares the proportions positive by the two methods?

  1. McNemar's test using the discordant pairs
  2. Pearson's chi-square on a 2 × 4 table
  3. Unpaired two-sample t-test on the proportions
  4. Kruskal-Wallis test on the test results
Answer and working

A. McNemar's test using the discordant pairs

Results on the same animals are paired, so McNemar's test applies: χ² = (b − c)²/(b + c), using only sheep positive on one test and negative on the other. An ordinary chi-square treats the two sets of results as independent samples.

Further reading: Thrusfield M, Christley R. Veterinary Epidemiology, 4th ed.

2 · The chi-square test · hard

Tick counts per animal are grouped into 6 classes and a Poisson distribution, with its mean estimated from the same data, is fitted. How many df does the goodness-of-fit chi-square test have?

  1. 5
  2. 6
  3. 3
  4. 4
Answer and working

D. 4

df = classes − 1 − parameters estimated = 6 − 1 − 1 = 4. 5 forgets the df lost for estimating the Poisson mean; 3 subtracts two parameters, as for a fitted normal.

Further reading: Snedecor GW, Cochran WG. Statistical Methods, 8th ed.

3 · Analysis of variance · medium

A one-way ANOVA of rat liver weights gives treatment SS 54 and error SS 6. What proportion of the total variation is explained by the treatments (eta squared)?

  1. 0.11
  2. 0.90
  3. 0.10
  4. 9.00
Answer and working

B. 0.90

Total SS = 54 + 6 = 60; η² = 54/60 = 0.90. 0.10 is the unexplained share (6/60); 9.00 is the ratio of treatment to error SS and 0.11 its inverse (6/54).

Further reading: Petrie A, Watson P. Statistics for Veterinary and Animal Science, 3rd ed.

4 · Correlation and regression · medium

Farms with more cattle report both more milk sold and more cases of mastitis, giving a strong correlation between milk sold and mastitis cases. What is the best interpretation?

  1. Selling more milk causes mastitis through over-milking
  2. Mastitis causes farmers to sell more milk
  3. The correlation is likely spurious, both depending on herd size
  4. The correlation proves that the two are causally linked
Answer and working

C. The correlation is likely spurious, both depending on herd size

Both variables rise with a third variable, herd size, so their correlation does not show that one causes the other. Analysing per-cow rates or adjusting for herd size would remove the confounding.

Further reading: Dohoo I, Martin W, Stryhn H. Veterinary Epidemiologic Research, 2nd ed.

5 · Correlation and regression · hard

In a logistic regression of calf mortality on colostrum feeding, the coefficient for 'no colostrum within 6 hours' is 0.693. What does this mean?

  1. Risk of death rises by 69.3%
  2. Odds of death rise by 0.693
  3. Probability of death is 0.693
  4. Odds of death are doubled
Answer and working

D. Odds of death are doubled

In logistic regression, exp(coefficient) is the odds ratio: e^0.693 = 2.0, so the odds of death are twice as high. The coefficient is on the log-odds scale, not a probability or a percentage change in risk.

Further reading: Dohoo I, Martin W, Stryhn H. Veterinary Epidemiologic Research, 2nd ed.

6 · Experimental designs: CRD and RBD · hard

A 2³ factorial experiment is laid out in a randomised block design with 3 blocks. What are the error degrees of freedom?

  1. 14
  2. 16
  3. 21
  4. 7
Answer and working

A. 14

8 treatment combinations: treatment df 7, block df 2, error df 7 × 2 = 14, total 23. 16 is the CRD error df 8 × 2; 7 is the treatment df; 21 subtracts only the block df from the total (23 − 2).

Further reading: Gomez KA, Gomez AA. Statistical Procedures for Agricultural Research, 2nd ed.

7 · Sample-size calculation · hard

A simple-random-sample size of 246 cattle is needed. The survey will instead sample 10 cattle per herd, and the intra-herd correlation is 0.2. Using design effect DE = 1 + (m − 1)ρ, how many cattle are needed?

  1. 689
  2. 492
  3. 295
  4. 2460
Answer and working

A. 689

DE = 1 + 9 × 0.2 = 2.8; n = 246 × 2.8 = 688.8, rounded up to 689. 492 assumes DE = 2; 295 uses DE = 1.2 (ρ without the m − 1 factor; 295.2); 2460 multiplies by m.

Further reading: Dohoo I, Martin W, Stryhn H. Veterinary Epidemiologic Research, 2nd ed.

8 · Sample-size calculation · hard

A vaccine trial expects 30% morbidity in controls and 15% in vaccinated calves. Using n = (zα/2 + zβ)²[p1q1 + p2q2]/(p1 − p2)² per group, α = 0.05 two-sided and 80% power (z values 1.96 and 0.842), how many calves per group are needed?

  1. 236
  2. 58
  3. 118
  4. 59
Answer and working

C. 118

(2.802)² = 7.849; p1q1 + p2q2 = 0.21 + 0.1275 = 0.3375; n = 7.849 × 0.3375/0.0225 = 117.7, so 118 per group. 58 ignores power; 236 is the total for both groups and 59 wrongly halves the per-group figure. Pooled-variance or continuity-corrected formulas give somewhat larger numbers.

Further reading: Thrusfield M, Christley R. Veterinary Epidemiology, 4th ed.

9 · Research methodology · medium

In a clinical trial of an analgesic in dogs, neither the owners nor the clinicians scoring pain know which dog received the drug or the placebo. What is the purpose of this?

  1. To ensure the groups are equal at the start
  2. To prevent bias in assessing the outcome
  3. To reduce the number of dogs that are needed
  4. To remove the need for an ethics approval
Answer and working

B. To prevent bias in assessing the outcome

Blinding protects against measurement (observer) and performance bias, especially for subjective outcomes such as pain scores. Comparability at baseline comes from randomisation, not blinding.

Further reading: Petrie A, Watson P. Statistics for Veterinary and Animal Science, 3rd ed.

10 · Research methodology · medium

A trial in one university herd of high-yielding cows shows a clear benefit of a ration. Smallholders ask whether it applies to their low-yielding indigenous cows. This question concerns:

  1. Internal validity of the trial
  2. Statistical power of the trial
  3. External validity of the trial
  4. Reliability of the measurements
Answer and working

C. External validity of the trial

External validity (generalisability) asks whether results apply beyond the study population. Internal validity asks whether the result is true for the animals studied.

Further reading: Petrie A, Watson P. Statistics for Veterinary and Animal Science, 3rd ed.

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Biostatistics, Computer Application & Research MethodologyThe full study manual, free ICAR AICE-JRF/SRFFormat, sections and how it has changed

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