I85PREP
⌘ K
GATE ST study materials
GATE ST study resources

GATE ST practice questions

One question at a time. Build your GATE ST understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Statistics (ST) test paper

For a simple random sample of size n from a population with finite variance, which quantity is an unbiased estimator of the population variance σ²?

  1. (1/(n−1))·Σ(xᵢ − x̄)² Correct answer.
  2. (1/(n+1))·Σ(xᵢ − x̄)²
  3. (1/(n−1))·Σ(xᵢ − μ)²
  4. (1/n)·Σ(xᵢ − x̄)²

Explanation

Dividing the sum of squared deviations about the sample mean by n−1 corrects for the fact that x̄ is itself estimated from the data, making the expected value exactly σ². Bessel's correction is the standard unbiased choice for the sample variance.

Question 2Statistics (ST) test paper

A random variable X has moment generating function M(t) = exp(3t + 2t²). What are the mean and variance of X?

  1. Mean 6, variance 2
  2. Mean 3, variance 2
  3. Mean 2, variance 3
  4. Mean 3, variance 4 Correct answer.

Explanation

This is the mgf of a normal distribution with mean μ = 3 and variance σ² = 2·2 = 4, since the normal mgf is exp(μt + σ²t²/2). Matching coefficients gives μ = 3 and σ²/2 = 2, hence σ² = 4.

Question 3Statistics (ST) test paper

In a hypothesis test, the p-value is defined as which of the following?

  1. The largest significance level at which the alternative is rejected
  2. The probability that the null hypothesis is true given the observed data
  3. The probability of obtaining a test statistic at least as extreme as the observed one, computed under the null hypothesis Correct answer.
  4. The probability of a Type II error at the observed test statistic

Explanation

The p-value is a tail probability of the test statistic evaluated assuming H₀ holds; it measures how incompatible the data are with the null. It is not the posterior probability of H₀ and does not by itself give error probabilities for a decision.

Question 4Statistics (ST) test paper

In simple linear regression with an intercept, the ordinary least squares residuals satisfy which property?

  1. Their sum equals the residual standard error
  2. They are all positive when the slope is positive
  3. They equal the errors εᵢ exactly
  4. Their sum is zero and they are uncorrelated with the fitted values Correct answer.

Explanation

The normal equations force the residuals to sum to zero and to be orthogonal to both the fitted values and the regressor. These are algebraic consequences of least squares, not probabilistic assumptions.

Question 5Statistics (ST) test paper

A fair six-sided die is rolled twice. Given that the sum of the two rolls is 8, what is the probability that the first roll was 3?

  1. 1/4
  2. 1/5 Correct answer.
  3. 1/3
  4. 1/6

Explanation

The sums equal to 8 arise from the five ordered pairs (2,6), (3,5), (4,4), (5,3), (6,2). Exactly one of these has first roll 3, so the conditional probability is 1/5.

Question 6Statistics (ST) test paper

Which statement about the median and the mean of a distribution is correct?

  1. The median always equals the mean for any continuous distribution
  2. The median is more sensitive to extreme outliers than the mean
  3. The mean is undefined whenever the distribution is skewed
  4. The mean minimizes the sum of squared deviations, while the median minimizes the sum of absolute deviations Correct answer.

Explanation

The mean is the least-squares centre and the median is the least-absolute-deviations centre, which explains why the median resists outliers. Equality of mean and median holds for symmetric distributions, not in general.

Question 7Statistics (ST) test paper

If X and Y are independent random variables, which equality always holds?

  1. Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y) with Cov(X, Y) > 0
  2. Var(X + Y) = Var(X) + Var(Y) Correct answer.
  3. Var(X − Y) = Var(X) − Var(Y)
  4. E(XY) = E(X) + E(Y)

Explanation

Independence implies zero covariance, so the cross term 2Cov(X, Y) in the variance of a sum vanishes. Hence variances add for independent variables.

Question 8Statistics (ST) test paper

A statistic T is called sufficient for a parameter θ if which condition holds?

  1. T has minimum variance among all unbiased estimators
  2. T is a consistent estimator of θ
  3. T is unbiased for θ
  4. The conditional distribution of the sample given T does not depend on θ Correct answer.

Explanation

Sufficiency means the statistic captures all information about θ in the sample, so the conditional law of the data given T is free of θ. Unbiasedness and efficiency are separate properties that a sufficient statistic need not possess.

Question 9Statistics (ST) test paper

Let X₁, …, Xₙ be iid Bernoulli(p). The maximum likelihood estimator of p is which quantity?

  1. The sample median of the observations
  2. The sample mean of the observations Correct answer.
  3. The largest observed value
  4. The reciprocal of the sample variance

Explanation

The Bernoulli likelihood is p^(Σxᵢ)(1−p)^(n−Σxᵢ), and maximising it yields p̂ equal to the proportion of successes, which is the sample mean. Differentiating the log-likelihood and setting it to zero gives this closed form.

Question 10Statistics (ST) test paper

A random sample of size 16 is drawn from a normal population with unknown variance. To test a hypothesis about the mean, which reference distribution applies to the t statistic?

  1. F distribution with 1 and 16 degrees of freedom
  2. t distribution with 15 degrees of freedom Correct answer.
  3. Chi-square distribution with 16 degrees of freedom
  4. Standard normal distribution

Explanation

When σ is estimated by the sample standard deviation, the standardized mean follows a t distribution with n−1 degrees of freedom. Here n = 16, so there are 15 degrees of freedom.

Question 11Statistics (ST) test paper

In a one-way analysis of variance with k groups and total sample size N, the F statistic compares the mean square between groups with the mean square within groups. What are the correct degrees of freedom?

  1. k and N − k − 1
  2. N − k and k − 1
  3. k − 1 and N − 1
  4. k − 1 and N − k Correct answer.

Explanation

The between-group sum of squares has k − 1 degrees of freedom and the within-group sum of squares has N − k, so the F statistic uses these two values. The total sum of squares has N − 1 degrees of freedom, which splits into the two components.

Question 12Statistics (ST) test paper

If X has a Poisson distribution with mean λ, then for large λ the standardized variable (X − λ)/√λ is approximately distributed as which distribution?

  1. Exponential with mean 1
  2. Poisson with mean 0
  3. Standard normal Correct answer.
  4. Binomial with parameters n and λ

Explanation

The Poisson mean and variance both equal λ, and the central limit theorem makes the standardized count approximately standard normal for large λ. This is the basis for normal approximations to Poisson tail probabilities.

Question 13Statistics (ST) test paper

A test has significance level 0.05 and power 0.80 against a particular alternative. If the alternative is true, what is the probability of failing to reject the null?

  1. 0.20 Correct answer.
  2. 0.05
  3. 0.95
  4. 0.80

Explanation

Power is the probability of rejecting H₀ when the alternative holds, so the complementary probability of a Type II error is 1 − 0.80 = 0.20. The significance level 0.05 governs Type I error under the null, not under the alternative.

Question 14Statistics (ST) test paper

For a bivariate normal vector with correlation coefficient ρ, the conditional distribution of Y given X = x has mean μ_Y + ρ(σ_Y/σ_X)(x − μ_X). What is its variance?

  1. ρ²σ_Y²
  2. σ_Y²(1 − ρ²) Correct answer.
  3. σ_Y²
  4. σ_Y² + σ_X²

Explanation

Conditioning on X removes the portion of Y's variability explained by the linear relationship, leaving variance σ_Y²(1 − ρ²). This residual variance is constant in x for the bivariate normal.

Question 15Statistics (ST) test paper

A statistician constructs a 95% confidence interval for a population mean and obtains (12.3, 18.7). Which interpretation is correct?

  1. 95% of the population values lie between 12.3 and 18.7
  2. There is a 95% probability that the population mean lies between 12.3 and 18.7
  3. The sample mean lies in the interval with probability 0.95
  4. In repeated sampling, about 95% of intervals built this way would contain the true mean Correct answer.

Explanation

Frequentist coverage refers to the long-run performance of the procedure: 95% of intervals generated by the same rule capture μ. Once computed, the specific interval either contains μ or does not, so no probability attaches to it.

Question 16Statistics (ST) test paper

Let X₁, …, Xₙ be iid with mean μ and variance σ². By Chebyshev's inequality, what does the sample mean X̄ satisfy for any ε > 0?

  1. P(|X̄ − μ| ≥ ε) ≤ σ/(nε)
  2. P(|X̄ − μ| ≥ ε) ≤ σ²/(nε²) Correct answer.
  3. P(|X̄ − μ| ≥ ε) ≥ σ²/(nε²)
  4. P(|X̄ − μ| ≥ ε) = σ²/(nε²) exactly

Explanation

Chebyshev's inequality applied to X̄, which has variance σ²/n, bounds the tail probability by Var(X̄)/ε² = σ²/(nε²). This bound drives the weak law of large numbers as n grows.

Question 17Statistics (ST) test paper

In multiple linear regression Y = Xβ + ε with ε ~ N(0, σ²I), the least squares estimator β̂ has covariance matrix given by which expression?

  1. σ²I
  2. σ²(XᵗX)⁻¹ Correct answer.
  3. (XᵗX)⁻¹
  4. σ²XᵗX

Explanation

Since β̂ = (XᵗX)⁻¹XᵗY, its covariance is (XᵗX)⁻¹Xᵗ(σ²I)X(XᵗX)⁻¹, which simplifies to σ²(XᵗX)⁻¹. The diagonal entries of this matrix give the variances of the individual coefficient estimates.

Question 18Statistics (ST) test paper

A Markov chain on a finite state space is irreducible, aperiodic and positive recurrent. Which statement about its limiting behaviour is correct?

  1. The limiting distribution depends on the starting state
  2. The chain converges only if started in the stationary distribution
  3. The chain converges to the stationary distribution from every starting state Correct answer.
  4. The chain has no stationary distribution because it is aperiodic

Explanation

For an irreducible, aperiodic, positive recurrent chain, the limiting probabilities exist and equal the unique stationary distribution regardless of the initial state. This is the ergodic theorem for Markov chains.

Question 19Statistics (ST) test paper

A random variable X has density f(x) = 2x for 0 < x < 1 and zero elsewhere. What is E(1/X)?

  1. 1
  2. 2 Correct answer.
  3. 3/2
  4. 4/3

Explanation

Compute E(1/X) = ∫₀¹ (1/x)(2x) dx = ∫₀¹ 2 dx = 2. The integrand simplifies because the x factors cancel.

Question 20Statistics (ST) test paper

Two independent samples of sizes 10 and 12 are drawn from normal populations with equal but unknown variance. To test equality of means, the pooled t statistic has how many degrees of freedom?

  1. 10
  2. 22
  3. 20 Correct answer.
  4. 21

Explanation

The pooled variance combines n₁ − 1 = 9 and n₂ − 1 = 11 degrees of freedom, giving 20. The t statistic then follows a t distribution with 20 degrees of freedom under the null.

Question 21Statistics (ST) test paper

Let X and Y be independent standard exponential random variables. What is P(X < Y)?

  1. 2/3
  2. 3/4
  3. 1/4
  4. 1/2 Correct answer.

Explanation

By symmetry of two iid continuous variables, P(X < Y) = P(Y < X), and ties have probability zero, so each probability is 1/2. This holds for any iid continuous pair.

Question 22Statistics (ST) test paper

The Cramér–Rao lower bound for the variance of an unbiased estimator is the reciprocal of which quantity?

  1. The sample size
  2. The square of the log-likelihood
  3. The Fisher information in the sample Correct answer.
  4. The variance of the sufficient statistic

Explanation

The bound states Var(T) ≥ 1/I(θ), where I(θ) is the Fisher information contained in the sample. An estimator attaining this bound is efficient.

Question 23Statistics (ST) test paper

A quality engineer inspects items arriving independently with defect probability 0.02 and wants the exact distribution of the number of inspections until the first defective item. Which distribution applies?

  1. Negative binomial
  2. Poisson
  3. Geometric Correct answer.
  4. Binomial

Explanation

The number of independent Bernoulli trials up to and including the first success follows a geometric distribution with parameter p = 0.02. Its support is the positive integers and its mean is 1/p.

Question 24Statistics (ST) test paper

A likelihood ratio test compares a restricted model with a more general nested model. Under the null and standard regularity conditions, −2 log Λ is asymptotically distributed as which distribution?

  1. F distribution with 1 and n − 1 degrees of freedom
  2. t distribution with n − 1 degrees of freedom
  3. Chi-square with degrees of freedom equal to the number of restrictions Correct answer.
  4. Standard normal

Explanation

Wilks' theorem states that −2 log Λ converges in distribution to a chi-square with degrees of freedom equal to the difference in the number of free parameters. This makes the test widely applicable without specifying the alternative.

Question 25Statistics (ST) test paper

Let X₁, …, Xₙ be iid Uniform(0, θ) and let M = max(Xᵢ). Which quantity is a pivotal quantity for θ?

  1. M − θ
  2. M/θ Correct answer.
  3. θ/M²
  4. M + θ

Explanation

The ratio M/θ has distribution P(M/θ ≤ t) = tⁿ on (0,1), which does not involve θ, so it is a pivot. Pivots of this form allow exact construction of confidence intervals for θ.

Question 26Statistics (ST) test paper

In a simple random sample drawn without replacement from a finite population, which quantity is an unbiased estimator of the population variance σ²?

  1. Σ(xᵢ − x̄)² / (n − 1) Correct answer.
  2. Σ(xᵢ − x̄)² / n
  3. Σ(xᵢ − x̄)² / (n + 1)
  4. Σ(xᵢ − x̄)² / n²

Explanation

Dividing the sum of squared deviations by n − 1 corrects the downward bias caused by measuring deviations from the sample mean rather than the true mean. This is Bessel's correction, and it holds regardless of whether the sample is taken with or without replacement. Dividing by n gives the maximum likelihood estimator under normality, which is biased.

Question 27Statistics (ST) test paper

A random variable X has moment generating function M(t) = exp(3t + 2t²) for t in a neighbourhood of zero. What are the mean and variance of X?

  1. Mean 3, variance 2
  2. Mean 6, variance 2
  3. Mean 3, variance 4 Correct answer.
  4. Mean 2, variance 3

Explanation

The mgf has the form of a normal distribution with mean μ and variance σ², where the exponent is μt + σ²t²/2. Matching coefficients gives μ = 3 and σ²/2 = 2, so σ² = 4. Differentiating twice also confirms M'(0) = 3 and M''(0) − (M'(0))² = 4.

Question 28Statistics (ST) test paper

For a random sample of size n from a normal population with unknown mean and unknown variance, which pivot leads to the exact t confidence interval for the mean?

  1. (x̄ − μ)/s, which follows a t distribution with n degrees of freedom
  2. n(x̄ − μ)²/s², which follows an F distribution with 1 and n − 1 degrees of freedom
  3. (x̄ − μ)/(σ/√n), which follows a standard normal distribution
  4. (x̄ − μ)/(s/√n), which follows a t distribution with n − 1 degrees of freedom Correct answer.

Explanation

Replacing the unknown σ by the sample standard deviation s and dividing by s/√n yields a statistic that is t distributed with n − 1 degrees of freedom. The degrees of freedom come from the sample variance, which uses one estimated mean. This pivot is free of unknown parameters and can be inverted to give the interval x̄ ± t_{n−1,1−α/2} s/√n.

Question 29Statistics (ST) test paper

Let X₁, …, Xₙ be iid with a Gamma(α, β) distribution, where the density is proportional to x^{α−1} e^{−x/β}. Which statistic is jointly sufficient for (α, β)?

  1. The pair (Σ log Xᵢ, Σ Xᵢ) Correct answer.
  2. The sample maximum
  3. The pair (min Xᵢ, max Xᵢ)
  4. The sample mean alone

Explanation

Writing the joint density in exponential-family form shows that it factors into a term depending on the data only through Σ log Xᵢ and Σ Xᵢ, multiplied by a term free of the parameters. By the Fisher–Neyman factorisation theorem this pair is sufficient. The log term carries the information about the shape parameter α.

Question 30Statistics (ST) test paper

A random variable X takes values 0, 1, 2, 3 with probabilities 0.1, 0.2, 0.3, 0.4 respectively. What is E(X²)?

  1. 3.0
  2. 2.0
  3. 4.0
  4. 5.0 Correct answer.

Explanation

Compute E(X²) = Σ x² P(X = x) = 0(0.1) + 1(0.2) + 4(0.3) + 9(0.4) = 0 + 0.2 + 1.2 + 3.6 = 5.0. Note that E(X²) is not the square of E(X), which is 2 here. The variance would be 5.0 − 4 = 1.0.

Question 31Statistics (ST) test paper

In a simple linear regression with n observations, the fitted line passes through (x̄, ȳ). What is the sum of the residuals Σ(yᵢ − ŷᵢ) when the model includes an intercept?

  1. Always negative
  2. Zero Correct answer.
  3. Equal to the total sum of squares
  4. Equal to the residual sum of squares

Explanation

The least squares normal equations for a model with an intercept include the condition that the residuals sum to zero. Equivalently, the fitted line contains the point of means, so the average fitted value equals the average response. This property fails if the intercept is forced through the origin.

Question 32Statistics (ST) test paper

A test statistic T has a discrete null distribution. A statistician wants the test to have size at most 0.05 and therefore uses a randomised test. What does randomisation accomplish here?

  1. It increases the power of the test beyond what any non-randomised test can achieve
  2. It removes the need for the null distribution of the test statistic
  3. It allows the exact size to equal the nominal level even when no critical value gives exactly that probability Correct answer.
  4. It guarantees the test is unbiased for all alternatives

Explanation

With a discrete statistic the rejection probability jumps as the critical value moves, so a fixed rejection region often has size strictly below the nominal level. Randomising on the boundary value lets the rejection probability be tuned continuously, achieving exactly the desired size. Power need not exceed that of the best non-randomised test of the same size.

Question 33Statistics (ST) test paper

If X and Y are independent random variables with finite variances, what is Var(X − Y)?

  1. Var(X) + Var(Y) + 2Cov(X, Y)
  2. Var(X) − Var(Y)
  3. Var(X) + Var(Y) − 2Cov(X, Y)
  4. Var(X) + Var(Y) Correct answer.

Explanation

The general formula is Var(X − Y) = Var(X) + Var(Y) − 2Cov(X, Y). Independence implies Cov(X, Y) = 0, so the covariance term vanishes and the variances add. Variances never subtract, because the difference of two independent quantities fluctuates more than either one alone.

Question 34Statistics (ST) test paper

A Markov chain has transition matrix P = [[0.6, 0.4], [0.3, 0.7]] on states {1, 2}. What is the stationary distribution π = (π₁, π₂)?

  1. (0.5, 0.5)
  2. (3/7, 4/7) Correct answer.
  3. (0.4, 0.6)
  4. (0.6, 0.4)

Explanation

Solving πP = π with π₁ + π₂ = 1 gives 0.6π₁ + 0.3π₂ = π₁, so 0.3π₂ = 0.4π₁ and π₁/π₂ = 3/4. Normalising yields π = (3/7, 4/7). The chain is irreducible and aperiodic, so this stationary distribution is also its limiting distribution.

Question 35Statistics (ST) test paper

Let X₁, …, Xₙ be iid Uniform(0, θ). The maximum likelihood estimator is θ̂ = max(Xᵢ). What is the bias of θ̂?

  1. −θ/(n + 1) Correct answer.
  2. −θ/n
  3. Zero, because θ̂ is unbiased
  4. θ/(n + 1)

Explanation

The maximum M has density n x^{n−1}/θ^n on (0, θ), so E(M) = nθ/(n + 1). The bias is E(M) − θ = −θ/(n + 1), meaning the MLE systematically underestimates θ. Multiplying M by (n + 1)/n removes this bias.

Question 36Statistics (ST) test paper

In a one-sample sign test for the median of a continuous distribution, the null hypothesis is that the median equals m₀. Under the null, the number of observations exceeding m₀ follows which distribution?

  1. A normal distribution with mean n/2 and variance n/2
  2. A Poisson distribution with mean n/2
  3. A hypergeometric distribution
  4. A binomial distribution with parameters n and 1/2 Correct answer.

Explanation

Under the null each observation independently lies above m₀ with probability 1/2, so the count of values above m₀ is Binomial(n, 1/2) for continuous data with no ties. Exact critical values come from this binomial distribution, and the normal approximation is used only for large n. The test is distribution-free because it uses only the signs of deviations.

Question 37Statistics (ST) test paper

A random variable X has density f(x) = λe^{−λx} for x > 0. What is P(X > s + t | X > s) for s, t > 0?

  1. P(X > t) Correct answer.
  2. P(X > s + t)
  3. P(X > t)/P(X > s)
  4. P(X > s)

Explanation

The exponential distribution is memoryless: given survival past s, the remaining lifetime has the same distribution as a fresh exponential variable. Algebraically, P(X > s + t)/P(X > s) = e^{−λ(s+t)}/e^{−λs} = e^{−λt} = P(X > t). This property characterises the exponential law among continuous distributions.

Question 38Statistics (ST) test paper

A statistician computes a 95% confidence interval for a proportion from a sample of size 400 and obtains a margin of error of 0.04. Approximately what sample size would halve the margin of error, keeping the same confidence level?

  1. 800
  2. 1600 Correct answer.
  3. 200
  4. 1200

Explanation

The margin of error is proportional to 1/√n, so halving it requires quadrupling the sample size. Four times 400 is 1600. The estimated proportion may shift slightly with a new sample, but the scaling law governs the approximate answer.

Question 39Statistics (ST) test paper

Let X₁, …, Xₙ be iid N(μ, σ²). Which statement about the sample variance S² and the sample mean X̄ is correct?

  1. They are independent, and (n − 1)S²/σ² follows a chi-square distribution with n degrees of freedom
  2. They are independent, and (n − 1)S²/σ² follows a chi-square distribution with n − 1 degrees of freedom Correct answer.
  3. They are dependent because both are computed from the same data
  4. They are independent only when μ = 0

Explanation

For normal samples, the sample mean and the deviations from it are independent, a result known as Fisher's lemma. Consequently X̄ and S² are independent, and the scaled sum of squared deviations follows a chi-square with n − 1 degrees of freedom because one degree is lost to estimating μ. This independence underlies the exact t and F procedures.

Question 40Statistics (ST) test paper

A random sample of size n is used to estimate the mean of a population with known variance σ². A statistician proposes the estimator T = (X₁ + X₂)/2, using only the first two observations. How does Var(T) compare with the variance of the sample mean X̄?

  1. Var(T) equals Var(X̄) because both are unbiased
  2. Var(T) is smaller because it uses fewer observations
  3. Var(T) is n/2 times Var(X̄) Correct answer.
  4. Var(T) is 2/n times Var(X̄)

Explanation

Var(T) = σ²/2 while Var(X̄) = σ²/n, so the ratio Var(T)/Var(X̄) = n/2. Both estimators are unbiased, which shows that unbiasedness alone does not make an estimator efficient. The sample mean uses all n observations and therefore has much smaller variance when n exceeds 2.

Question 41Statistics (ST) test paper

In a two-way layout with one observation per cell and an additive model with no interaction, the residual degrees of freedom equal which expression, where a and b are the numbers of levels of the two factors?

  1. (a − 1)(b − 1) Correct answer.
  2. ab − 1
  3. a + b − 1
  4. ab − a − b

Explanation

The total degrees of freedom ab − 1 split into a − 1 for the first factor, b − 1 for the second factor, and the remainder for error. The remainder is (a − 1)(b − 1), which is the error degrees of freedom when there is exactly one observation per cell. With replication, interaction and pure error terms can be separated.

Question 42Statistics (ST) test paper

Let X have a binomial distribution with parameters n and p. For large n, which transformation of the sample proportion p̂ = X/n has a variance that is approximately stabilised, independent of p?

  1. The arcsine square-root transformation, arcsin(√p̂) Correct answer.
  2. The logit transformation, log(p̂/(1 − p̂))
  3. The square-root transformation, √p̂
  4. The reciprocal transformation, 1/p̂

Explanation

The delta method shows that arcsin(√p̂) has approximate variance 1/(4n), which no longer depends on p. This variance-stabilising property makes the arcsine transformation useful for constructing confidence intervals for proportions near 0 or 1. The logit also removes the dependence on p from the mean structure but its variance still varies with p.

Question 43Statistics (ST) test paper

A random variable X has the probability generating function G(z) = (0.3 + 0.7z)³. What is P(X = 0)?

  1. 0.343
  2. 0.21
  3. 0.027 Correct answer.
  4. 0.3

Explanation

The pgf has the form (q + pz)ⁿ with n = 3, p = 0.7 and q = 0.3, so X is Binomial(3, 0.7). Then P(X = 0) = G(0) = 0.3³ = 0.027. Evaluating the pgf at zero always returns the probability of a zero count.

Question 44Statistics (ST) test paper

Which statement correctly describes the relationship between a confidence interval and the corresponding two-sided hypothesis test at the same level?

  1. A 95% confidence interval contains exactly the values of μ not rejected by a two-sided test at the 5% level Correct answer.
  2. A 95% confidence interval contains exactly the values of μ rejected at the 5% level
  3. The confidence level must be 50% for the correspondence to hold
  4. A confidence interval can be used only for one-sided tests

Explanation

Inverting a two-sided test gives the set of parameter values that are not rejected, which is precisely the confidence interval at level 1 − α. A value outside the interval is rejected by the test, and a value inside is not. This duality requires the test and interval to use the same level and the same two-sided construction.

Question 45Statistics (ST) test paper

Let X and Y be jointly continuous with joint density f(x, y). Which expression gives the conditional density of Y given X = x?

  1. f(x, y) · f_X(x)
  2. f(x, y) / f_X(x) Correct answer.
  3. f(x, y) / f_Y(y)
  4. f_X(x) f_Y(y)

Explanation

The conditional density is obtained by dividing the joint density by the marginal density of the conditioning variable, evaluated at the conditioning value x. This follows from P(Y ∈ dy | X = x) = f(x, y)dy / f_X(x). The result integrates to one over y for each fixed x where f_X(x) > 0.

Question 46Statistics (ST) test paper

A random variable X is said to be stochastically larger than Y if P(X > t) ≥ P(Y > t) for all t. If X is stochastically larger than Y, which conclusion about their expectations is valid when both are integrable?

  1. E(X) = E(Y) always
  2. E(X) ≥ E(Y) Correct answer.
  3. E(X) ≤ E(Y)
  4. No ordering of expectations can be inferred

Explanation

Stochastic ordering implies an ordering of expectations for integrable variables because E(X) − E(Y) can be written as the integral of the difference in survival functions. Since each survival difference is non-negative, the integral is non-negative. The converse fails, so equal or ordered means do not imply stochastic ordering.

Question 47Statistics (ST) test paper

In a simple random sample of size n from a population, the sample median is used to estimate the population median. Compared with the sample mean for a normal population, the sample median has which property?

  1. No asymptotic normal distribution
  2. Smaller asymptotic variance, so it is more efficient
  3. The same asymptotic variance, so the two are equally efficient
  4. Larger asymptotic variance, so it is less efficient Correct answer.

Explanation

For a normal population the sample median is asymptotically normal with variance 1/(4n f(m)²), which equals πσ²/(2n), compared with σ²/n for the mean. The ratio π/2 is greater than one, so the median is less efficient under normality. The median is more robust to outliers, which can offset this loss outside the normal model.

Question 48Statistics (ST) test paper

Let X₁, …, Xₙ be iid with density f(x; θ) = θ x^{θ−1} for 0 < x < 1, where θ > 0. What is the method-of-moments estimator of θ based on the first moment?

  1. 2x̄ − 1
  2. x̄/(1 + x̄)
  3. x̄/(1 − x̄) Correct answer.
  4. (1 − x̄)/x̄

Explanation

The mean of this Beta-type density is E(X) = θ/(θ + 1). Setting this equal to x̄ and solving gives θ̂ = x̄/(1 − x̄). The estimator is positive whenever the sample mean lies between 0 and 1, as it must for data in the unit interval.

Question 49Statistics (ST) test paper

A random variable X has a distribution with P(X = k) = 1/N for k = 1, …, N. What is the variance of X?

  1. (N − 1)²/12
  2. (N² − 1)/12 Correct answer.
  3. N²/12
  4. (N + 1)/2

Explanation

For a discrete uniform distribution on 1, …, N, the mean is (N + 1)/2 and the variance is (N² − 1)/12. This follows from E(X²) = (N + 1)(2N + 1)/6 combined with the squared mean. The result grows quadratically with N, as the spread of the support does.

Question 50Statistics (ST) test paper

In a sequential probability ratio test with log-likelihood ratio increments Zᵢ having mean μ under the alternative, the expected sample size at the alternative is approximately (A + B)/μ for boundary constants A and B. Which feature of the SPRT does this formula illustrate?

  1. The SPRT always uses a fixed sample size determined in advance
  2. The SPRT typically requires fewer observations than a fixed-sample test with the same error probabilities Correct answer.
  3. The SPRT has no error probabilities of either type
  4. The SPRT requires the sample size to grow with the boundary constants

Explanation

Wald's approximation shows that the average number of observations needed under the alternative is roughly (A + B)/μ, which is usually smaller than the fixed-sample size achieving the same α and β. The saving arises because the test stops as soon as the evidence is decisive. The boundaries control the error probabilities, not the sample size directly.