Probability, Statistics & Linear Programming

Sampling and Design of Experiments

14 previous-year questions →every option and the correct answer, free Practice this topic →one question at a time, no clock, check as you go

See where the marks are across every paper →

1. SRS, stratified and systematic sampling

Read this as a chapter →

Lecture 53: Selection of Sample — Simple Random Sampling

IIT Kanpur — NPTEL

SRS with and without replacement, and why the finite population correction appears.

Lecture 22: Sampling Techniques-I

NPTEL — IIT Roorkee

Stratified and systematic sampling alongside SRS: when stratification actually reduces variance.

Lec 9: Sampling Strategies

NPTEL — IIT Guwahati

A compact comparison of the standard strategies — good revision before the exam.

Sampling designs free

In plain words

Simple random sampling is the baseline, and a full census has no sampling error. Stratifying helps when the groups differ, and systematic sampling can be far worse if the list has a period.

Simple random sampling

For SRSWOR of size n from N units with population variance $S^{2}$ :

Var(ȳ $) = (S^{2}/n)(1 - n/N)$ — the factor (1 − n/N) is the finite population correction. With replacement there is no fpc, so SRSWOR is more efficient.

ȳ is unbiased for the population mean under both schemes.

Worked: population {2, 4, 6}, samples of size 2 without replacement. The three samples have means 3, 4, 5 each with probability ⅓: $E[M] = 4, Var(M) = 2/3, E[M^{3}] = 72$ .

Stratified sampling

Var(ȳ $_st) = \sum W_{h}^{2} S_{h}^{2}/n_{h}$ with $W_{h} = N_{h}/N$ .

Allocation	$n_{h}$ ∝	Comment
Proportional	$N_{h}$	always at least as good as SRS
Neyman/optimum	$N_{h}S_{h}$	minimises variance for fixed n
Optimum with cost	$N_{h}S_{h}/\sqrt{c}_{h}$	minimises variance for fixed cost

Efficiency order: Neyman ≤ proportional ≤ SRS (in variance). Stratification helps most when strata are internally homogeneous and mutually different.

Systematic sampling

Pick every k-th unit (k = N/n). Simple to run; efficient for a linear trend, but disastrous if the population has period k. Variance cannot be estimated unbiasedly from a single systematic sample.

Ratio and regression estimators

Ratio estimator ȳ_R = (ȳ/x̄)X̄ — biased but consistent, better than ȳ when y is roughly proportional to x (correlation > CV_x/2CV_y). The regression estimator is at least as good asymptotically.

Key takeaways

SRSWOR beats SRSWR via the fpc.
Neyman ≤ proportional ≤ SRS in variance.
Systematic sampling fails when the period matches the sampling interval.

Why stratification helps and systematic sampling can betray youinteractive

This visual is part of Trap Analysis

The variance decomposition behind stratification, and the periodicity that ruins the systematic scheme.

Unlock interactive visuals

Exam focus: Know when SRSWOR beats SRSWR (the finite population correction), and that proportional allocation beats SRS while Neyman allocation beats proportional.

Practice

Part B2025 Decembershow ▾

Consider a finite population of size N = 100. Let

T_{1}

be the sample mean of a study variable based on a sample of size n (1 < n < N) under simple random sampling with replacement scheme. Let

T_{2}

be the sample mean of the same study variable based on a sample of size n under simple random sampling without replacement scheme. If

Var(T_{1}) = 9Var(T_{2})

, then the sample size n equals

A.33
B.69
C.89✓ Correct
D.93

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Consider a dataset of n observations given by A = {

x_{1},x_{2},\dots,x_{n}

}. Create a new dataset, given by B = {

x_{1},x_{2},\dots,x_{n},-x_{1},-x_{2},\dots,-x_{n}

}. Which of the following statements are always true? (Variance is calculated with divisor as the number of observations in the corresponding dataset.)

A.Mean of the observations in dataset B is 0✓ Correct
B.Median of the observations in dataset B is 0✓ Correct
C.Variance of the observations in dataset B ≥ Variance of the observations in dataset A✓ Correct
D.Range of the observations in dataset B ≥ Range of the observations in dataset A✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾

Suppose we want to estimate the population mean Ȳ of a variable for a finite population of size 85, with 34 Statisticians and 51 Biologists. We consider the following sampling scheme: a stratified random sample with 2 strata of Statisticians (Stratum-1) and Biologists (Stratum-2), where 12 Statisticians and 15 Biologists are drawn from Stratum-1 and Stratum-2, respectively, using SRSWOR scheme. Denote ȳ_S, ȳ_B, and ȳ as the mean of the variable among the Statistician sample, Biologist sample, and the combined sample, respectively. Which of the following is an unbiased estimator of Ȳ?

A.ȳ
B.(2ȳ_S + 3ȳ_B)/5✓ Correct
C.(4ȳ_S + 5ȳ_B)/9
D.(ȳ_S/12 + ȳ_B/15)/(1/12 + 1/15)

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Let P be the population proportion of units possessing a certain attribute in a population of N units. Let p be the sample proportion in a simple random sample (without replacement) of n units, (2 ≤ n < N). Then an unbiased estimator of P(1 − P) is

A.((N − n)/(Nn))p(1 − p)
B.((N − n)/((N − 1)n))p(1 − p)
C.(n/(n − 1))p(1 − p)
D.(((N − 1)n)/(N(n − 1)))p(1 − p)✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider a probability proportional to size without replacement sample involving two draws from a population of N (> 6) units with normed size measures

p_{i}'s (p_{i} > 0, i = 1, 2, \dots, N; \sum(i=1..N)p_{i} = 1)

. Then, which of the following statements are true?

A.P(Unit 1 is included in the sample $) = p_{1}$
B.P(Unit 1 and Unit 3 are included in the sample $) = p_{1}p_{3}[1/(1 - p_{1}) + 1/(1 - p_{3})]$ ✓ Correct
C.P(Unit 1, Unit 3 and Unit 5 are included in the sample $) = p_{1}p_{3}p_{5}[1/((1 - p_{1})(1 - p_{3})) + 1/((1 - p_{3})(1 - p_{5})) + 1/((1 - p_{5})(1 - p_{1}))]$
D.Expected number of distinct units in the sample is $2[1 - \sum(i=1..N)(1 - p_{i})^{2}]$

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

The expected number of distinct units in a simple random sample of 3 units drawn with replacement from a population of 100 units is

A. $3 - (99/100)^{3}$
B. $100 - 99^{3}/100^{2}$ ✓ Correct
C. $2 + 99^{2}/100^{3}$
D. $3 - (99/100)^{2}$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Consider a population of 3 units having values 2, 4 and 6. A simple random sample (without replacement) of 2 units is drawn. Let M be the sample mean. Which of the following statements are true?

A.E(M) = 4✓ Correct
B. $E(M^{2}) = 17$
C. $E(M^{3}) = 72$ ✓ Correct
D.Var(M) = 1

Solution

M takes the values 3, 4, 5 with probability 1/3 each: $E(M) = 4, E(M^{2}) = 50/3, E(M^{3}) = 72, Var(M) = 2/3$ .

Save for spaced revision (free account)

Part C2023 Decembershow ▾

Two groups

G_{1}

and

G_{2}

have 10 and 30 patients with mean diastolic blood pressures 80 and 100 mmHg and variances 4 and 2 mmHg

^{2}

. Let

{\bar{X}}, S^{2}, C

and R be the mean, variance (with divisor n), coefficient of variation (%) and range of the combined group. Which of the following statements are true?

A.X̄ = 95✓ Correct
B. $S^{2} = 77$
C.C > 180/19
D.R > 8✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Compare simple random sampling without replacement (size n from N) with stratified and systematic sampling. Which of the following are true?

A.Under SRSWOR the sample mean is unbiased for the population mean.✓ Correct
B.The variance of the sample mean under SRSWOR carries the finite population correction relative to with-replacement sampling.✓ Correct
C.Stratified sampling with proportional allocation has variance no larger than SRS when strata means differ.✓ Correct
D.Systematic sampling is always at least as precise as SRS.

The worked solution and the per-option analysis are in Trap Analysis.

2. CRD, RBD, LSD essentials

Read this as a chapter →

Lecture 38: Introduction to Design of Experiment

NPTEL — IIT Kharagpur

Why designs exist: randomisation, replication, local control — the vocabulary the questions assume.

Lecture 39: Randomized Block Design

NPTEL — IIT Kharagpur

RBD and its ANOVA table; note where the (t−1)(b−1) error degrees of freedom come from.

Lec 26: Randomised Block Design (RBD)

NPTEL — IIT Roorkee

A second pass on RBD with worked computation — useful if the first treatment felt fast.

Design of experiments essentials free

In plain words

Each design takes one more known source of variation out of the error, paying for it in degrees of freedom. Block on something irrelevant and you have spent them for nothing.

The three basic designs

Design	Model	Error d.f.	Controls
CRD	$y_{ij} = \mu + \tau_{i} + \varepsilon_{ij}$	N − k	nothing
RBD	$y_{ij} = \mu + \tau_{i} + \beta_{j} + \varepsilon_{ij}$	(k−1)(r−1)	one nuisance factor
LSD (p×p)	$y_{ijk} = \mu + \tau_{i} + \rho_{j} + \gamma_{k} + \varepsilon$	(p−1)(p−2)	two nuisance factors

In an LSD the number of treatments equals the number of rows and columns (p), and $p^{2}$ units are used. If the error d.f. is 42, then (p−1)(p−2) = 42 gives p = 8, so treatments have 7 d.f.

ANOVA table shape

Total d.f. = N − 1, split into treatment, block(s) and error. The F statistic is MS_treatment/MS_error with (k−1, error) degrees of freedom.

Principles

Randomisation removes bias; replication gives an error estimate; local control (blocking) removes a known source of variability.
Blocking increases precision only if the blocks really differ — otherwise you have merely lost error degrees of freedom.
Relative efficiency of RBD to CRD = (MSE_CRD/MSE_RBD), adjusted for d.f.

Factorial experiments

$A 2^{n}$ factorial estimates main effects and interactions; confounding sacrifices a high-order interaction to reduce block size. Effects are estimated by contrasts of the treatment totals.

Missing plot

For one missing value in an RBD: ŷ = (rB + kT − G)/((r−1)(k−1)), and the error d.f. drops by one.

Key takeaways

Memorise the three error d.f. formulas — they are asked directly.
LSD: p treatments, $p^{2}$ , two nuisance directions.
Blocking trades error d.f. for reduced error variance; it only pays if blocks differ.

CRD, RBD and Latin square: degrees of freedom without memorisinginteractive

This visual is part of Trap Analysis

Derive each error df by subtraction, and know when blocking costs more than it buys.

Unlock interactive visuals

Exam focus: Degrees of freedom are the most-asked item: CRD N − k, RBD (k−1)(r−1), LSD (p−1)(p−2). Blocking removes a nuisance source and costs error degrees of freedom.

Practice

Part C2025 Juneshow ▾

If the incidence matrix of a block design is given by N with rows (1, 1, 1, 0), (1, 1, 0, 1), (1, 0, 1, 1) and (0, 1, 1, 1), then which of the following statements are true?

A.The design is incomplete✓ Correct
B.The design is connected✓ Correct
C.The design is balanced✓ Correct
D.The design is orthogonal

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Consider the following

2^{5}-

factorial design with 8 blocks (Block 1: (1), acd, bce, abde; Block 2: e, acde, bc, abd; and six further blocks). Which of the following statements are true?

A.ABC is confounded with blocks✓ Correct
B.BCD is confounded with blocks✓ Correct
C.CDE is confounded with blocks✓ Correct
D.ABDE is confounded with blocks✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Consider the following design where the columns represent blocks and the letters represent treatments — block 1: A, B; block 2: C, D; block 3: A, C; block 4: B, D; block 5: A, D; block 6: B, C; block 7: A, E; block 8: B, E; block 9: C, E; block 10: D, E. Then, which of the following statements is NOT true?

A.The design is a balanced incomplete block design.
B.The design is connected.
C.The design is binary.
D.The design is symmetric.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Consider the following ANOVA table for a randomized block design — Treatments: sum of squares 48, degrees of freedom 4, mean squares 12, F calculated

\beta

; Blocks: sum of squares 72, degrees of freedom 3, mean squares 24, F calculated 12; Error: sum of squares

\alpha

, degrees of freedom m, mean squares

\gamma

; Total: sum of squares 144, degrees of freedom 19. Which of the following statements are true?

A. $\alpha = 20$
B. $\beta = 6$ ✓ Correct
C.m = 10
D. $\gamma = 2$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the following design where the columns represent blocks and the letters represent treatments — block 1: A, B, E; block 2: C, D, E; block 3: A, C, F; block 4: B, D, F; block 5: A, D, G; block 6: B, C, G; block 7: E, F, G. Then, which of the following statements are true?

A.The design is a balanced incomplete block design.✓ Correct
B.The design is not connected.
C.The design is binary.✓ Correct
D.The design is symmetric.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾

In a Latin square design, the degrees of freedom for the error sum of squares is 42. Then the degrees of freedom for the sum of squares due to treatments is

A.6
B.7✓ Correct
C.8
D.9

Solution

For a p × p Latin square the error d.f. is (p − 1)(p − 2) = 42 ⇒ p = 8, so treatments have p − 1 = 7 d.f.

Save for spaced revision (free account)

Part Cshow ▾

Which of the following statements about standard experimental designs are true?

A.A CRD with t treatments and n total units has error degrees of freedom n − t.✓ Correct
B.An RBD with t treatments and b blocks has error degrees of freedom (t − 1)(b − 1).✓ Correct
C.A Latin square of side t uses $t^{2}$ units and blocks two nuisance directions at once.✓ Correct
D.Blocking improves precision even when the blocks are homogeneous in the response.

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

In a randomised block design with 5 treatments and 4 blocks, the error degrees of freedom is

A.12✓ Correct
B.15
C.19
D.20

Solution

(k−1)(r−1) = 4·3 = 12.

Save for spaced revision (free account)