Sampling Theory Questions and Answers

Answer: A

The correct match is i-γ, ii-δ, iii-α, iv-β.

• AOQ, average outgoing quality: the proportion of defectives remaining in lots after inspection (γ).
• AQL, acceptable quality level: the poorest incoming quality still considered acceptable, an upper bound for acceptance (δ).
• ATI, average total inspection: the number of items per lot inspected on average (α).
• RQL, rejectable quality level (LTPD): the quality at or beyond which lots should be rejected, a lower bound for rejection (β).

The other codes swap AQL and RQL or move ATI and AOQ to the wrong definitions.

Answer: D

Under systematic sampling, the sample mean is an unbiased estimator of the population mean (in the usual case where the population size is a multiple of the sample size).

With \(N = nk\), linear systematic sampling chooses a random start \(r\) from 1 to \(k\) and takes units \(r, r + k, r + 2k, \ldots\). Each of the \(k\) possible samples has probability \(1/k\), and every unit lies in exactly one of them, so every unit has inclusion probability \(1/k = n/N\). Averaging the \(k\) possible sample means gives the population mean exactly.

The other statements are false. The first unit is chosen at random with equal probabilities, not by pps, and the rest follow deterministically, not by SRS. Each unit has an equal chance, \(n/N\), of selection. And the variance of the systematic mean can be larger than that of SRSWOR, for example when the population has a periodic pattern with period \(k\).

Answer: A

The inclusion probability is at least \(\dfrac{n - 1}{N - 1}\).

In Midzuno's scheme the first unit is drawn with probability proportional to size, \(p_{i}\), and the remaining \(n - 1\) units by SRSWOR from the other \(N - 1\). Unit \(i\) is included either as the first draw or as one of the later ones:

\[ \begin{aligned} \pi_{i} &= p_{i} + (1 - p_{i})\frac{n - 1}{N - 1} \\ &= \frac{n - 1}{N - 1} + \frac{N - n}{N - 1}\,p_{i} \end{aligned} \]

Since \(p_{i} \ge 0\), \(\pi_{i} \ge (n - 1)/(N - 1)\), with equality only for a unit of size zero.

\(n/N\) is the inclusion probability under SRSWOR, which is not a lower bound here (a small unit can fall below it). The forms \(n(n - 1)/(N - 1)\) and \(n(n - 1)/N\) are not probabilities in general, since they can exceed 1.

Answer: D

Post-stratification is used when the stratum to which a unit belongs is not known before sampling.

In ordinary stratified sampling the frame lists each unit's stratum, so the sample can be drawn separately within strata. When stratum membership is only discovered on observing the unit (age group, for example, found out at interview), we draw a simple random sample and classify the sampled units afterwards. The estimate then weights the stratum sample means by the known stratum proportions \(W_{h} = N_{h}/N\).

So post-stratification does need the stratum sizes (or proportions) and the strata themselves to be known, which rules out the options saying these are unknown. Equal stratum sizes are not a requirement of the method.

Answer: A

Double sampling is useful when complete information on the auxiliary variable is not available.

Ratio and regression estimators need the population mean (or total) of an auxiliary variable \(x\). When it is unknown, double (two-phase) sampling first takes a large, cheap sample to measure \(x\) alone and estimate its mean, then a smaller subsample on which the expensive study variable \(y\) is also measured. The first-phase estimate of \(\bar{X}\) stands in for the unknown population value.

So the purpose is to supply missing auxiliary information, not to improve a ratio or regression estimator that already has it (with the true \(\bar{X}\) known, double sampling only adds variance). Heterogeneity of the population is the motivation for stratification, not for double sampling.

Answer: D

The statement that is not true is \(V(\bar{Y}) = \frac{N-1}{n}s_{y}^{2}\) under SRSWR.

Under SRSWR the draws are independent, each with population variance \(\sigma^{2} = \frac{N - 1}{N}s_{y}^{2}\). So

\[ \begin{aligned} V(\bar{Y}) &= \frac{\sigma^{2}}{n} \\ &= \frac{N - 1}{Nn}s_{y}^{2} \end{aligned} \]

The statement leaves out the N in the denominator, so it is false.

The others are standard results. Under SRSWOR each unit has probability 1/N of being selected at any particular draw, and the sample mean square is unbiased for \(s_{y}^{2}\). In systematic sampling with N = nk, each of the k possible samples is equally likely and every unit appears in exactly one, so \(E(\bar{Y}_{sy}) = \bar{y}\).

Answer: A

The necessarily true statement is that the expected sample size is 25.5.

The 50 samples have sizes 1, 2, ..., 50, each chosen with probability 1/50, so

\[ \begin{aligned} E(\text{size}) &= \frac{1 + 2 + \ldots + 50}{50} \\ &= \frac{1275}{50} \\ &= 25.5 \end{aligned} \]

The inclusion probabilities always add up to the expected sample size, \(\sum \pi_{i} = E(n) = 25.5\), so neither 1 nor 25 is right for the sum. (Directly, unit i is in the samples of size i or more, so \(\pi_{i} = (51 - i)/50\), and these add to 25.5.)

Answer: B

The ratio estimate of the total is 112.

The sample is units 4 and 6, with x values 6 and 10 and y values 18 and 46. The ratio estimate is

\[ \hat{R} = \frac{18 + 46}{6 + 10} = \frac{64}{16} = 4 \]

The population total of x is \(0 + 1 + 3 + 6 + 8 + 10 = 28\), so

\[ \hat{Y} = \hat{R}X = 4 \times 28 = 112 \]

The other figures are not ratio estimates from this sample; the ratio method, which scales the known x total by the sample ratio of y to x, gives 112.

Answer: A

The optimum allocation is n_h ∝ W_hS_h/√C_h.

Minimising \(V(\bar{Y}_{st}) = \sum W_{h}^{2}S_{h}^{2}/n_{h}\) (ignoring the finite population correction terms, which do not depend on \(n_{h}\)) subject to \(\sum C_{h}n_{h}\) fixed, or the cost subject to a fixed variance, leads by the Cauchy-Schwarz inequality or a Lagrange multiplier to

\[ n_{h} \propto \frac{W_{h}S_{h}}{\sqrt{C_{h}}} \]

So a stratum gets a larger sample when it is larger, more variable or cheaper to sample. Option B puts a sum over strata in the numerator and \(C_{h}\) without a square root; option C cancels to \(N_{h}\), which is proportional allocation; and option D drops the stratum weight.

Answer: D

The minimum sample size is approximately \(\left(\frac{s_{y} Z_{\alpha/2}}{d}\right)^{2}\).

Under SRSWOR, \(\mathrm{Var}(\overline{y}) = \left(\frac{1}{n} - \frac{1}{N}\right)s_{y}^{2}\). The accuracy condition \(P[|\overline{Y} - \overline{y}| \lt d] \ge 1 - \alpha\) requires \(Z_{\alpha/2}\sqrt{\mathrm{Var}(\overline{y})} \le d\):

\[ \begin{aligned} \frac{1}{n} - \frac{1}{N} &= \frac{d^{2}}{Z_{\alpha/2}^{2} s_{y}^{2}} \\ n_{0} &= \left(\frac{s_{y} Z_{\alpha/2}}{d}\right)^{2} \text{ when N is large} \end{aligned} \]

For finite N the correction is \(n = \frac{n_{0}}{1 + n_{0}/N}\).

The square root (option A) and dividing by d rather than \(d^{2}\) (option C) get the dimensions wrong, and \(\sqrt{N}\) has nothing to do with the accuracy required.