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Sampling Theory, Estimation & Statistical Inference (Statistics) Solved Questions & Notes (2026) - Apex Rankers

Statistics & Data Science > Statistics > Sampling Theory, Estimation & Statistical Inference

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Sampling Theory, Estimation & Statistical Inference

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Q. 1 Statistics
Difficulty: easy (1 Mark)
In Simple Random Sampling Without Replacement (SRSWOR) of size $n$ from a finite population of size $N$, what is the variance of the sample mean $\bar{y}$?
A
$\text{Var}(\bar{y}) = \frac{S^2}{n}$
B
$\text{Var}(\bar{y}) = \left(\frac{N - n}{N}\right) \frac{S^2}{n}$
✓ Correct
C
$\text{Var}(\bar{y}) = \left(\frac{N - 1}{N}\right) \frac{S^2}{n}$
D
$\text{Var}(\bar{y}) = \frac{S^2}{Nn}$
💡 Step-by-Step Explanation & Concept Rationale
In SRSWOR, the variance of the sample mean includes the Finite Population Correction (FPC) factor: $\text{Var}(\bar{y}) = \left(1 - \frac{n}{N}\right) \frac{S^2}{n} = \left(\frac{N - n}{N}\right) \frac{S^2}{n}$, where $S^2 = \frac{1}{N-1}\sum (Y_i - \bar{Y})^2$.
Q. 2 Statistics
Difficulty: easy (1 Mark)
What is the Finite Population Correction (FPC) factor defined as in survey sampling?
A
$1 - \frac{n}{N}$ (or $\frac{N - n}{N}$)
✓ Correct
B
$\frac{n}{N}$
C
$\frac{N}{n}$
D
$1 + \frac{n}{N}$
💡 Step-by-Step Explanation & Concept Rationale
The FPC factor is $1 - f = 1 - \frac{n}{N} = \frac{N-n}{N}$, where $f = n/N$ is the sampling fraction. When $f < 0.05$ (less than 5% of the population), FPC is often omitted as it approaches 1.
Q. 3 Statistics
Difficulty: medium (1 Mark)
Under Neyman's Optimum Allocation in Stratified Random Sampling, how is the sample size $n_h$ in stratum $h$ allocated when sampling costs per unit are equal?
A
$n_h = n \frac{N_h}{N}$
B
$n_h = n \frac{N_h S_h}{\sum N_h S_h}$
✓ Correct
C
$n_h = n \frac{S_h}{\sum S_h}$
D
$n_h = \frac{n}{L}$
💡 Step-by-Step Explanation & Concept Rationale
Neyman allocation determines that the stratum sample size $n_h$ is directly proportional to both the stratum size $N_h$ and the stratum standard deviation $S_h$: $n_h = n \frac{N_h S_h}{\sum_{i=1}^L N_i S_i}$, minimizing the variance of the stratified estimator for a fixed total sample size.
Q. 4 Statistics
Difficulty: medium (1 Mark)
When is Stratified Random Sampling most effective at reducing the variance of the overall population mean estimator compared to Simple Random Sampling?
A
When strata are heterogeneous within and homogeneous between
B
When strata are homogeneous within (low internal variance $S_h^2$) and heterogeneous between (large differences among stratum means)
✓ Correct
C
When all strata have identical sample sizes
D
When the population size $N$ is less than 50
💡 Step-by-Step Explanation & Concept Rationale
Stratification maximizes precision when elements within each stratum are as homogeneous as possible (minimizing within-stratum variance), and means between different strata are widely separated.
Q. 5 Statistics
Difficulty: easy (1 Mark)
What is the fundamental condition for an estimator $T_n$ to be 'Consistent' for parameter $\theta$?
A
$E[T_n] = \theta$ for all sample sizes $n$
B
$T_n$ converges in probability to $\theta$ as $n \to \infty$ (i.e., $\text{plim}_{n \to \infty} T_n = \theta$)
✓ Correct
C
$\text{Var}(T_n) = 0$ for all $n$
D
$T_n$ follows a normal distribution for any $n$
💡 Step-by-Step Explanation & Concept Rationale
Consistency is an asymptotic property stating that as sample size $n$ grows infinitely large, the probability of the estimator deviating from the true parameter by any $\epsilon > 0$ approaches zero: $\lim_{n \to \infty} P(|T_n - \theta| > \epsilon) = 0$.
Q. 6 Statistics
Difficulty: medium (1 Mark)
If an estimator $T_n$ is asymptotically unbiased ($\\lim_{n \to \infty} E[T_n] = \theta$) and its variance vanishes as $n \to \infty$ ($\\lim_{n \to \infty} \text{Var}(T_n) = 0$), does this guarantee consistency?
A
No, consistency requires higher moments to exist
B
Yes, by Chebyshev's inequality, a vanishing mean squared error (MSE) is a sufficient condition for consistency
✓ Correct
C
Only if the population is normally distributed
D
No, it only guarantees unbiasedness
💡 Step-by-Step Explanation & Concept Rationale
Since $\text{MSE}(T_n) = \text{Var}(T_n) + [\text{Bias}(T_n)]^2$, if both variance and bias vanish as $n \to \infty$, $\text{MSE} \to 0$, which by Chebyshev's inequality ensures convergence in probability (consistency).
Q. 7 Statistics
Difficulty: medium (1 Mark)
What does the Neyman-Pearson Factorization Theorem state regarding a statistic $T = T(\mathbf{X})$ and parameter $\theta$?
A
$T$ is unbiased if and only if $E[T] = \theta$
B
$T$ is a sufficient statistic for $\theta$ if and only if the joint likelihood can be factored as $L(\mathbf{x}; \theta) = g(T(\mathbf{x}), \theta) \cdot h(\mathbf{x})$, where $h(\mathbf{x})$ does not depend on $\theta$
✓ Correct
C
$T$ is efficient if its variance achieves the Cramér-Rao lower bound
D
$T$ is consistent if sample size exceeds 30
💡 Step-by-Step Explanation & Concept Rationale
The Factorization Criterion provides the definitive mathematical test for sufficiency: the likelihood factors into a function $g(T,\theta)$ depending on the data only through $T$, and a factor $h(\mathbf{x})$ completely free of the parameter $\theta$.
Q. 8 Statistics
Difficulty: easy (1 Mark)
What is the sufficient statistic for parameter $\lambda$ in a random sample $X_1, X_2, \dots, X_n$ from a Poisson distribution $\text{Poisson}(\lambda)$?
A
Sample maximum $X_{(n)}$
B
Sample sum $\sum_{i=1}^n X_i$ (or sample mean $\bar{X}$)
✓ Correct
C
Sample median
D
Sample variance $S^2$
💡 Step-by-Step Explanation & Concept Rationale
The joint likelihood is $L = \frac{e^{-n\lambda} \lambda^{\sum x_i}}{\prod x_i!} = \left[e^{-n\lambda} \lambda^{\sum x_i}\right] \left[\frac{1}{\prod x_i!}\right]$. By the Factorization Theorem, $T = \sum_{i=1}^n X_i$ is jointly sufficient for $\lambda$.
Q. 9 Statistics
Difficulty: medium (1 Mark)
What is the minimal sufficient statistic for $(\mu, \sigma^2)$ in a random sample from a Normal distribution $N(\mu, \sigma^2)$?
A
$(\bar{X}, S^2)$ or equivalently $\left(\sum_{i=1}^n X_i, \sum_{i=1}^n X_i^2\right)$
✓ Correct
B
Only the sample mean $\bar{X}$
C
Only the sample variance $S^2$
D
The sample range and median
💡 Step-by-Step Explanation & Concept Rationale
Factoring the joint normal likelihood reveals that the pair of statistics $\left(\sum X_i, \sum X_i^2\right)$ (or equivalently sample mean $\bar{X}$ and sample variance $S^2$) captures all information about both unknown parameters $(\mu, \sigma^2)$.
Q. 10 Statistics
Difficulty: hard (1 Mark)
What does the Rao-Blackwell Theorem state regarding an unbiased estimator $T$ and a sufficient statistic $S$?
A
The estimator $T$ should be discarded in favor of the sample median
B
The conditional expectation $\delta = E[T \mid S]$ is an unbiased estimator of $\theta$ whose variance is strictly less than or equal to the variance of $T$ ($\text{Var}(\delta) \le \text{Var}(T)$)
✓ Correct
C
A sufficient statistic can never be unbiased
D
Conditioning on a sufficient statistic doubles the mean squared error
💡 Step-by-Step Explanation & Concept Rationale
Rao-Blackwellization improves any crude unbiased estimator $T$ by conditioning it on a sufficient statistic $S$: the resulting estimator $\delta = E[T \mid S]$ is independent of $\theta$, unbiased, and has variance $\le \text{Var}(T)$.
Q. 11 Statistics
Difficulty: hard (1 Mark)
What does the Lehmann-Scheffé Theorem establish when an estimator is a function of a 'Complete and Sufficient' statistic?
A
The estimator is asymptotically normal
B
If $S$ is a complete sufficient statistic and $g(S)$ is an unbiased estimator of $\theta$, then $g(S)$ is the unique Uniformly Minimum Variance Unbiased Estimator (UMVUE) of $\theta$
✓ Correct
C
The estimator achieves the Cramér-Rao lower bound in all finite samples
D
The estimator has maximum likelihood properties
💡 Step-by-Step Explanation & Concept Rationale
Lehmann-Scheffé proves that conditioning any unbiased estimator on a complete sufficient statistic yields the unique, optimal UMVUE (no other unbiased estimator can achieve a smaller variance anywhere in parameter space).
Q. 12 Statistics
Difficulty: medium (1 Mark)
What is Fisher Information $I(\theta)$ for a single observation from a probability distribution $f(x; \theta)$ defined as under standard regularity conditions?
A
$I(\theta) = E\left[\left(\frac{\partial \ln f(X; \theta)}{\partial \theta}\right)^2\right] = -E\left[\frac{\partial^2 \ln f(X; \theta)}{\partial \theta^2}\right]$
✓ Correct
B
$I(\theta) = \text{Var}(X) \cdot \theta$
C
$I(\theta) = \frac{1}{E[X^2]}$
D
$I(\theta) = \int f(x; \theta) dx$
💡 Step-by-Step Explanation & Concept Rationale
Fisher Information measures the amount of information an observable random variable carries about an unknown parameter, defined as the variance of the score function (or the negative expected second derivative of the log-likelihood).
Q. 13 Statistics
Difficulty: medium (1 Mark)
What is the Cramér-Rao Lower Bound (CRLB) for the variance of any unbiased estimator $T$ of a scalar parameter $\theta$ from a sample of $n$ i.i.d. observations?
A
$\text{Var}(T) \ge \frac{1}{n I(\theta)}$
✓ Correct
B
$\text{Var}(T) \le \frac{1}{n I(\theta)}$
C
$\text{Var}(T) \ge \frac{I(\theta)}{n}$
D
$\text{Var}(T) = \frac{\sigma^2}{n}$
💡 Step-by-Step Explanation & Concept Rationale
The Cramér-Rao inequality establishes an absolute lower theoretical bound for the variance of any unbiased estimator: $\text{Var}(T) \ge \frac{1}{I_n(\theta)} = \frac{1}{n I(\theta)}$. An unbiased estimator attaining this bound is called 'Efficient'.
Q. 14 Statistics
Difficulty: medium (1 Mark)
What is the Fisher Information $I(\lambda)$ for a single observation from a Poisson distribution $P(X=x) = \frac{e^{-\lambda}\lambda^x}{x!}$?
A
$\lambda$
B
$\frac{1}{\lambda}$
✓ Correct
C
$\lambda^2$
D
$\frac{1}{\lambda^2}$
💡 Step-by-Step Explanation & Concept Rationale
$\ln f(x; \lambda) = -\lambda + x \ln\lambda - \ln(x!)$. The second derivative is $\frac{\partial^2 \ln f}{\partial \lambda^2} = -\frac{x}{\lambda^2}$. Taking the negative expectation: $-E\left[-\frac{X}{\lambda^2}\right] = \frac{E[X]}{\lambda^2} = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}$.
Q. 15 Statistics
Difficulty: medium (1 Mark)
For a random sample of size $n$ from $N(\mu, \sigma^2)$ with known $\sigma^2$, does the sample mean $\bar{X}$ attain the Cramér-Rao Lower Bound for estimating $\mu$?
A
No, its variance is higher than the CRLB
B
Yes, $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$, which is exactly equal to the CRLB $\frac{1}{n I(\mu)}$
✓ Correct
C
Only when $n \ge 100$
D
The CRLB cannot be computed for normal distributions
💡 Step-by-Step Explanation & Concept Rationale
For $N(\mu, \sigma^2)$, $I(\mu) = \frac{1}{\sigma^2}$. The CRLB is $\frac{1}{n(1/\sigma^2)} = \frac{\sigma^2}{n}$. Since $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$, $\bar{X}$ attains the bound with 100% efficiency.
Q. 16 Statistics
Difficulty: easy (1 Mark)
What is the Invariance Property of Maximum Likelihood Estimators (MLE)?
A
The MLE never changes when new data is added
B
If $\hat{\theta}$ is the MLE of parameter $\theta$, then for any function $g(\theta)$, the MLE of $g(\theta)$ is $g(\hat{\theta})$
✓ Correct
C
The variance of the MLE is invariant to sample size
D
The MLE is invariant under change of coordinate origin only
💡 Step-by-Step Explanation & Concept Rationale
The Zehna Invariance Property states that if $\hat{\theta}$ is the MLE of $\theta$, then $g(\hat{\theta})$ is automatically the MLE of $g(\theta)$ (e.g., if $\hat{\sigma}$ is the MLE of $\sigma$, then $\hat{\sigma}^2$ is the MLE of $\sigma^2$).
Q. 17 Statistics
Difficulty: medium (1 Mark)
What is the Maximum Likelihood Estimator (MLE) of the population variance $\sigma^2$ for a random sample from $N(\mu, \sigma^2)$ when both $\mu$ and $\sigma^2$ are unknown?
A
$\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2$
✓ Correct
B
$S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2$
C
$\hat{\sigma}^2 = \frac{1}{n+1} \sum_{i=1}^n (X_i - \bar{X})^2$
D
$\hat{\sigma}^2 = \sqrt{\frac{1}{n} \sum X_i^2}$
💡 Step-by-Step Explanation & Concept Rationale
Maximizing the normal log-likelihood yields the divisor $n$: $\hat{\sigma}_{MLE}^2 = \frac{1}{n} \sum (X_i - \bar{X})^2$. Note that this MLE is biased in finite samples ($E[\hat{\sigma}^2] = \frac{n-1}{n}\sigma^2$), but asymptotically unbiased and consistent.
Q. 18 Statistics
Difficulty: easy (1 Mark)
What is the expectation of the sample variance $S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2$ for any population with finite variance $\sigma^2$?
A
$E[S^2] = \frac{n-1}{n} \sigma^2$
B
$E[S^2] = \sigma^2$ (it is strictly unbiased)
✓ Correct
C
$E[S^2] = \sigma^2 + \mu^2$
D
$E[S^2] = \frac{\sigma^2}{n}$
💡 Step-by-Step Explanation & Concept Rationale
Bessel's correction using $n-1$ in the denominator corrects the downward bias caused by estimating the population mean with the sample mean, ensuring $E[S^2] = \sigma^2$ for all sample sizes.
Q. 19 Statistics
Difficulty: hard (1 Mark)
Under standard regularity conditions, what asymptotic distribution does the Maximum Likelihood Estimator $\hat{\theta}_n$ follow as $n \to \infty$?
A
Uniform distribution
B
Asymptotically normal: $\sqrt{n}(\hat{\theta}_n - \theta) \xrightarrow{d} N\left(0, \frac{1}{I(\theta)}\right)$
✓ Correct
C
Chi-square distribution with 1 degree of freedom
D
Student's t-distribution with $n-1$ degrees of freedom
💡 Step-by-Step Explanation & Concept Rationale
Asymptotic normality and asymptotic efficiency are hallmark properties of MLEs: as $n \to \infty$, $\hat{\theta}$ converges in distribution to a normal variable with mean $\theta$ and variance attaining the CRLB $\frac{1}{n I(\theta)}$.
Q. 20 Statistics
Difficulty: hard (1 Mark)
In a sample of $n$ observations from a Uniform distribution $U(0, \theta)$, what is the Maximum Likelihood Estimator of $\theta$?
A
$2\bar{X}$
B
Sample maximum $X_{(n)} = \max(X_1, X_2, \dots, X_n)$
✓ Correct
C
Sample minimum $X_{(1)}$
D
$\frac{X_{(1)} + X_{(n)}}{2}$
💡 Step-by-Step Explanation & Concept Rationale
The likelihood $L(\theta) = \frac{1}{\theta^n}$ for $\theta \ge X_{(n)}$ is a monotonically decreasing function of $\theta$. It is maximized at the smallest allowable value of $\theta$, which is the sample maximum $X_{(n)}$. (Note: Regularity conditions do not hold here as support depends on $\theta$).
Q. 21 Statistics
Difficulty: easy (1 Mark)
For $X_1, X_2, \dots, X_n \sim U(0, \theta)$, what is the Method of Moments estimator $\tilde{\theta}_{MOM}$?
A
$\tilde{\theta}_{MOM} = 2\bar{X}$
✓ Correct
B
$\tilde{\theta}_{MOM} = X_{(n)}$
C
$\tilde{\theta}_{MOM} = \bar{X} + S$
D
$\tilde{\theta}_{MOM} = \frac{\bar{X}}{2}$
💡 Step-by-Step Explanation & Concept Rationale
The population mean for $U(0, \theta)$ is $E[X] = \frac{\theta}{2}$. Equating sample mean to theoretical mean: $\bar{X} = \frac{\tilde{\theta}}{2} \implies \tilde{\theta}_{MOM} = 2\bar{X}$.
Q. 22 Statistics
Difficulty: hard (1 Mark)
For $X \sim U(0, \theta)$, how does the Method of Moments estimator $\tilde{\theta} = 2\bar{X}$ compare to the adjusted MLE $\hat{\theta}^* = \frac{n+1}{n} X_{(n)}$?
A
The MOM estimator has smaller variance than the MLE
B
The adjusted MLE is the UMVUE and has variance $O(1/n^2)$, whereas MOM has much larger variance $O(1/n)$
✓ Correct
C
Both estimators are identical in all samples
D
The MOM estimator is sufficient, while the MLE is not
💡 Step-by-Step Explanation & Concept Rationale
Because the order statistic $X_{(n)}$ converges at rate $n$ (variance $\propto 1/n^2$), the adjusted MLE $\frac{n+1}{n}X_{(n)}$ is drastically superior to the MOM estimator $2\bar{X}$ (variance $\propto 1/n$).
Q. 23 Statistics
Difficulty: easy (1 Mark)
What is the Method of Moments (MOM) estimation principle formulated by Karl Pearson?
A
Maximizing the log-likelihood function using calculus
B
Equating the first $k$ sample moments $m_r = \frac{1}{n}\sum X_i^r$ to the corresponding theoretical population moments $\mu_r'(\theta_1, \dots, \theta_k)$ and solving the system of equations
✓ Correct
C
Minimizing the sum of squared errors between observations
D
Conditioning on a minimal sufficient statistic
💡 Step-by-Step Explanation & Concept Rationale
Method of Moments estimates parameters by matching theoretical moments $E[X^k]$ with empirical sample moments $\frac{1}{n}\sum X_i^k$ and solving the resulting equations for the parameters.
Q. 24 Statistics
Difficulty: medium (1 Mark)
What is a 'Pivotal Quantity' in interval estimation?
A
A statistic whose value is always equal to 0
B
A function of the sample data and the parameter of interest whose probability distribution is completely known and does NOT depend on any unknown parameters
✓ Correct
C
The ratio of the sample mean to the sample median
D
A parameter that determines the shape of the prior distribution
💡 Step-by-Step Explanation & Concept Rationale
A pivotal quantity $Q(\mathbf{X}, \theta)$ has a known distribution free of $\theta$ (e.g., $Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim N(0, 1)$ or $T = \frac{\bar{X} - \mu}{S/\sqrt{n}} \sim t_{(n-1)}$), allowing exact algebraic construction of confidence intervals.
Q. 25 Statistics
Difficulty: easy (1 Mark)
What is the exact $(1 - \alpha)$ confidence interval for the population mean $\mu$ of a normal distribution with UNKNOWN variance $\sigma^2$, based on a sample of size $n$?
A
$\bar{X} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$
B
$\bar{X} \pm t_{\alpha/2, n-1} \frac{S}{\sqrt{n}}$
✓ Correct
C
$\bar{X} \pm \chi^2_{\alpha/2, n-1} \frac{S}{n}$
D
$\bar{X} \pm F_{\alpha/2} S^2$
💡 Step-by-Step Explanation & Concept Rationale
When population variance $\sigma^2$ is unknown and estimated by sample standard deviation $S$, the pivotal quantity follows Student's t-distribution with $n-1$ degrees of freedom, yielding $\bar{X} \pm t_{\alpha/2, n-1} \frac{S}{\sqrt{n}}$.
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