LAISHA
Open Laisha

UGC-NET Business Analytics & Digital Agribusiness: practice questions

UGC-NET (June 2025) · India · National Testing Agency

Original practice questions written to the published syllabus · not past papers
10worked here, free
52written to this exam in this subject
4syllabus units

What this subject covers

Using 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)

What UGC-NET asks from it

UGC-NET Management (business statistics, research methods and marketing parts)

Conceptual questions on multivariate techniques, regression, digital marketing terms and metrics.

Where the 52 questions fall

How our questions for UGC-NET spread across the units of Business Analytics & Digital Agribusiness. Difficulty: 13 easy, 27 medium, 12 hard.

Supervised learning: regression21
Digital marketing and brand building12
Principal component analysis10
Linear discriminant analysis9

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 · Supervised learning: regression · medium

A demand model has errors of 4, −2, 6 and −8 tonnes on four test weeks. What is the root mean squared error (RMSE)?

  1. 5.48
  2. 5.00
  3. 0.00
  4. 30.00
Answer and working

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.

2 · Supervised learning: regression · medium

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?

  1. Members have a 0.693 probability of adopting, regardless of their other traits
  2. Members are 69.3 percentage points more likely to adopt than non-members
  3. Members have twice the odds of adopting compared with non-members
  4. Membership explains 69.3% of the variation in adoption across the sample farms
Answer and working

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.

3 · Supervised learning: regression · medium

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)?

  1. 0.80
  2. 0.67
  3. 0.85
  4. 0.93
Answer and working

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.

4 · Supervised learning: regression · medium

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?

  1. 0.73
  2. 0.85
  3. 0.53
  4. 0.67
Answer and working

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.

5 · Supervised learning: regression · medium

To test a price-forecasting model for onions on six years of weekly data, which evaluation design is correct?

  1. Fit on the first five years and test on the sixth, moving forward in time
  2. Fit on the sixth year and test on the first five years to obtain a larger test set
  3. Fit and test on all six years so that every seasonal pattern appears in both sets
  4. Randomly select 20% of weeks from all years as a test set and fit on the rest
Answer and working

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.

6 · Linear discriminant analysis · medium

When is logistic regression usually preferred over linear discriminant analysis for a two-group problem?

  1. When the groups share equal covariance matrices and the predictors are normal
  2. When predictors include dummy variables and are clearly not multivariate normal
  3. When there are more than two groups and a low-dimensional plot of them is wanted
  4. When the sample is small and the most statistically efficient estimates are needed
Answer and working

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.

7 · Linear discriminant analysis · hard

In discriminant output, why do analysts often interpret the structure matrix (loadings) rather than the standardised coefficients when predictors are correlated?

  1. The structure matrix is the only output that remains valid when the groups have unequal sizes
  2. Correlated predictors make coefficients unstable or flip sign; loadings are plain correlations
  3. Standardised coefficients are expressed in the original units and cannot be compared at all
  4. Loadings always sum to one, so they show each predictor's exact share of discrimination
Answer and working

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.

8 · Digital marketing and brand building · medium

A food brand spent 50,000 on ads that generated 180,000 of attributed revenue at a 20% gross margin. Which statement is correct?

  1. ROAS is 3.6, yet the ads lost money because gross profit of 36,000 is below the spend
  2. ROAS is 3.6, so the campaign earned a net profit of 130,000 after paying for all the ads
  3. ROAS is 0.28, because ad spend is divided by the revenue the ads generated
  4. ROAS is 0.72, because return on ad spend is always calculated from gross profit
Answer and working

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.

9 · Digital marketing and brand building · hard

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?

  1. 4,000
  2. 2,100
  3. 3,000
  4. 840
Answer and working

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

10 · Digital marketing and brand building · hard

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?

  1. 81,600
  2. 82
  3. 816
  4. 8,200
Answer and working

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.

Ten here, 52 in the app. The Laisha app has every practice question we have written to UGC-NET in this subject, marked as you go, with timed papers, a jump-to-unanswered list and a certificate when you pass at 70 per cent.

Free for ten days, then one subscription of US$10 a month. We would rather say so here than have you find it later.

Practise the full set in Laisha →

Read the subject

Business Analytics & Digital AgribusinessThe full study manual, free UGC-NETFormat, sections and how it has changed

More UGC-NET practice

Production & Operations ManagementEconometricsPublic FinanceStrategic ManagementInstitutional EconomicsAgribusiness Law & EthicsDevelopment EconomicsMarketing ManagementMacroeconomics & PolicyRural Development, Local Governance & LivelihoodsQuality Management & Food StandardsEconomic Development and Agricultural Economies of the WorldBusiness Communication

Business Analytics & Digital Agribusiness for other exams

ICAR AICE-JRF/SRF · India

Ask Laisha about this

Opens Laisha's chat with your question ready to send. Sign in or start free.