GATE - Statistics (ST) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersThe general linear model in matrix form: estimability, the Gauss-Markov theorem, least squares with full and less-than-full rank, the analysis of variance as a linear model, model checking and selection, and regression diagnostics. It is the theory under every design in Phase 3.
In the exams: AICE-JRF Ag Stat Units 2 and 4 (heavy); GATE ST 'Regression Analysis' (heavy); UPSC ISS Statistics II (heavy); GATE DA regression (present)
GATE Statistics (ST)
Simple and multiple linear regression, R squared and adjusted R squared, distributions of quadratic forms, the Fisher-Cochran and Gauss-Markov theorems, tests and confidence intervals for regression coefficients.
GATE Data Science and AI (DA)
Simple and multiple linear regression and ridge regression as supervised learning methods.
How our questions for GATE spread across the units of Linear Models & Regression Analysis. Difficulty: 7 easy, 17 medium, 4 hard.
| The general linear model and least squares in matrix form | 9 | |
| Testing linear hypotheses; R-squared and adjusted R-squared | 6 | |
| Distribution of quadratic forms and the Fisher-Cochran theorem | 5 | |
| Estimable functions and the Gauss-Markov theorem | 2 | |
| Examination of residuals and model adequacy | 2 | |
| Influence and leverage: Cook's distance and hat values | 2 | |
| Multicollinearity and ridge regression | 1 | |
| Weighted and generalised least squares | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
For n = 12 observations, Sxx = 40, Sxy = 60, x-bar = 5 and y-bar = 20. What is the least squares intercept?
A. 12.5
b = Sxy/Sxx = 1.5; a = y-bar - b x-bar = 20 - 1.5 x 5 = 12.5. Using Sxx/Sxy as the slope gives 16.7; adding b x-bar gives 27.5.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
When X has full column rank, what is the least squares estimator of beta in y = X beta + e?
B. (X'X)^(-1) X'y
The normal equations X'X beta-hat = X'y give beta-hat = (X'X)^(-1) X'y. The other forms have wrong order or dimension.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
What is the variance matrix of the least squares estimator under V(e) = sigma^2 I?
C. sigma^2 (X'X)^(-1)
V((X'X)^(-1)X'y) = (X'X)^(-1)X' (sigma^2 I) X(X'X)^(-1) = sigma^2 (X'X)^(-1). sigma^2 H is the variance of the fitted values.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
In simple linear regression, the variance of the fitted mean response at x0 is smallest when:
C. x0 equals the mean of the x values
V = sigma^2 [1/n + (x0 - x-bar)^2/Sxx], minimised at x0 = x-bar; precision falls as x0 moves away, especially outside the data.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
In the one-way model y_ij = mu + tau_i + e_ij (written without constraints), which function is estimable?
C. tau_1 - tau_2
Contrasts among treatment effects (coefficients summing to zero) and treatment means mu + tau_i are estimable; mu, a single tau_i or a non-contrast sum like tau_1 + tau_2 are not.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
If y is N(0, sigma^2 I) and A is symmetric, y'Ay/sigma^2 has a chi-square distribution if and only if:
D. A is idempotent
The chi-square property holds iff A^2 = A; the degrees of freedom are rank(A) = trace(A).
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
Two quadratic forms y'Ay and y'By in y ~ N(mu, sigma^2 I) are independent if and only if:
A. AB = 0
Independence of quadratic forms in normal vectors requires AB = 0 (Craig's theorem). Commuting matrices are not enough.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
A multiple regression with k = 4 predictors and n = 25 observations has R^2 = 0.8. What is the adjusted R^2?
C. 0.760
Adjusted R^2 = 1 - (1 - R^2)(n - 1)/(n - k - 1) = 1 - 0.20 x 24/20 = 0.760. Inverting the ratio gives 0.833, larger than R^2, which is impossible.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
What happens to R^2 when an extra predictor is added to a regression model?
A. It never decreases
RSS cannot increase when a column is added, so R^2 cannot fall; adjusted R^2 can fall.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
An observation has raw residual 3, the residual standard deviation is s = 2, and its leverage is 0.19. What is its internally studentised residual?
B. 1.667
r = e/(s sqrt(1 - h)) = 3/(2 x 0.9) = 1.667. e/s = 1.5 is the standardised residual ignoring leverage.
Further reading: Searle, Linear Models; Draper and Smith, Applied Regression Analysis
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