UGC-NET (June 2025) · India · National Testing Agency
Original practice questions written to the published syllabus · not past papersUsing data, GIS, ERP and digital channels in agribusiness: the data-science workflow, basic machine learning and ICT for development.
In the exams: agribusiness management PhD entrance and fellowship examinations (light)
UGC-NET Management (business statistics, research methods and marketing parts)
Conceptual questions on multivariate techniques, regression, digital marketing terms and metrics.
How our questions for UGC-NET spread across the units of Business Analytics & Digital Agribusiness. Difficulty: 13 easy, 27 medium, 12 hard.
| Supervised learning: regression | 21 | |
| Digital marketing and brand building | 12 | |
| Principal component analysis | 10 | |
| Linear discriminant analysis | 9 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A demand model has errors of 4, −2, 6 and −8 tonnes on four test weeks. What is the root mean squared error (RMSE)?
A. 5.48
RMSE = √[(16 + 4 + 36 + 64)/4] = √(120/4) = √30 = 5.48 tonnes. 5.00 is the MAE (average of absolute errors 20/4), which RMSE always equals or exceeds. 0.00 is the mean error after positive and negative errors cancel. 30.00 is the MSE before taking the root.
In a logistic regression of technology adoption, the coefficient on 'member of a farmer producer organisation' (1 = yes) is 0.693. How is it interpreted?
C. Members have twice the odds of adopting compared with non-members
In logistic regression e^b is the odds ratio: e^0.693 ≈ 2.0, so membership doubles the odds. A coefficient is neither a change in probability in percentage points nor a probability itself, and it is not a share of variance explained.
A crop-disease classifier is tested on 200 fields: 40 true positives, 10 false positives, 20 false negatives and 130 true negatives. What is its recall (sensitivity)?
B. 0.67
Recall = TP/(TP + FN) = 40/(40 + 20) = 0.67. 0.80 is precision, TP/(TP + FP) = 40/50. 0.85 is accuracy, (40 + 130)/200. 0.93 is specificity, 130/140.
A crop-disease classifier is tested on 200 fields: 40 true positives, 10 false positives, 20 false negatives and 130 true negatives. What is its F1 score?
A. 0.73
Precision = 40/50 = 0.80; recall = 40/60 = 0.667. F1 = 2 × 0.80 × 0.667/(0.80 + 0.667) = 1.067/1.467 = 0.73. 0.53 is the product of precision and recall (0.80 × 0.667), not their harmonic mean. 0.85 is accuracy and 0.67 is recall alone.
To test a price-forecasting model for onions on six years of weekly data, which evaluation design is correct?
A. Fit on the first five years and test on the sixth, moving forward in time
Forecasts must be evaluated forward in time: fit on earlier periods and test on later ones. Random weeks let the model use future information next to each test week, testing on the past reverses time, and testing on the fitting data measures fit, not forecasting ability.
When is logistic regression usually preferred over linear discriminant analysis for a two-group problem?
B. When predictors include dummy variables and are clearly not multivariate normal
Logistic regression does not require normal predictors or equal covariances, so it suits mixed and categorical predictors. When LDA's assumptions hold, LDA is more efficient, and it extends naturally to several groups with a low-dimensional view; those conditions favour LDA, not logistic regression.
In discriminant output, why do analysts often interpret the structure matrix (loadings) rather than the standardised coefficients when predictors are correlated?
B. Correlated predictors make coefficients unstable or flip sign; loadings are plain correlations
When predictors are correlated, standardised coefficients share credit unpredictably and can flip sign; loadings, the correlations of each predictor with the function, are steadier for naming the function. Standardised coefficients are unit-free (that is why they are standardised), group size does not decide which output is valid, and loadings do not sum to one.
A food brand spent 50,000 on ads that generated 180,000 of attributed revenue at a 20% gross margin. Which statement is correct?
A. ROAS is 3.6, yet the ads lost money because gross profit of 36,000 is below the spend
ROAS = revenue/ad spend = 180,000/50,000 = 3.6. Gross profit = 20% × 180,000 = 36,000, which is less than the 50,000 spend, so the campaign lost 14,000 despite ROAS above 1. 130,000 treats revenue as profit; 0.72 uses profit, which is not how ROAS is defined; 0.28 inverts the ratio.
An animal-feed brand earns an annual margin of 1,200 per retained dairy customer. Annual retention is 70% and the discount rate 10%. Using CLV = m × r/(1 + d − r), what is customer lifetime value?
B. 2,100
CLV = 1,200 × 0.70/(1 + 0.10 − 0.70) = 840/0.40 = 2,100. 3,000 drops r from the numerator (1,200/0.40), 840 is only the first-year expected margin, and 4,000 uses 1,200/(1 − 0.70) and ignores discounting.
Further reading: Gupta and Lehmann, Customers as assets
A team wants to detect an increase in conversion from 5% to 6% with a two-sided 5% test and 80% power. Using n = (1.96 + 0.84)² × [p1(1 − p1) + p2(1 − p2)]/(p2 − p1)², about how many visitors are needed per version?
D. 8,200
(1.96 + 0.84)² = 7.84. p1(1 − p1) + p2(1 − p2) = 0.0475 + 0.0564 = 0.1039. (p2 − p1)² = 0.0001. n = 7.84 × 0.1039/0.0001 ≈ 8,146, about 8,200 per version (8,155 with z = 0.8416). 816 and 82 slip the squared difference (using 0.001 or 0.01), and 81,600 slips a factor of ten the other way. Small lifts on low baselines need thousands of users per arm.
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