Sampling Theory Questions and Answers

Q: For an SRSWOR (N, n), the probability that a specified unit is included in the sample is :

The inclusion probability is \(\dfrac{n}{N}\).All \({}^{N}C_{n}\) samples are equally likely. A specified unit is in a sample when the other \(n - 1\) units are chosen from the remaining \(N - 1\), so\[ \pi = \frac{{}^{N-1}C_{n-1}}{{}^{N}C_{n}} = \frac{n}{N} \]\(1/{}^{N}C_{n}\) is the probability of one particular sample, not of one unit. \(1/N\) is the probability that the unit is chosen at one particular draw. \(1/N(N - 1)\) is the probability of one particular ordered pair at the first two draws.

Q: If the regression estimator of \(\bar{Y}\) is \(\bar{y} + b(\bar{X} - \bar{x})\) where \(b = \dfrac{S_{xy}}{S_{x}^{2}}\), then an exact expression for bias of regression estimator is :

The exact bias is \(-Cov(b, \bar{x})\).\(\bar{y}\) is unbiased for \(\bar{Y}\), so the bias of the regression estimator is \(E[b(\bar{X} - \bar{x})]\). Using \(E(b\bar{x}) = Cov(b, \bar{x}) + E(b)E(\bar{x})\) and \(E(\bar{x}) = \bar{X}\):\[ \begin{aligned} E[b(\bar{X} - \bar{x})] &= \bar{X}E(b) - E(b\bar{x}) \\ &= \bar{X}E(b) - Cov(b, \bar{x}) - \bar{X}E(b) \\ &= -Cov(b, \bar{x}) \end{aligned} \]The bias comes from the randomness of \(b\) being correlated with \(\bar{x}\); with a fixed \(b\) the estimator would be unbiased. Options involving \(\bar{y}\) are wrong because \(\bar{y}\) enters linearly and contributes no bias, and option C has the sign reversed.

Q: The bais in ratio estimator decreases with :

The bias decreases with increasing the sample size \(n\).The ratio estimator \(\hat{R}\bar{X} = (\bar{y}/\bar{x})\bar{X}\) is biased because the expectation of a ratio is not the ratio of expectations. To first order its bias is\[ \begin{aligned} B &\approx \frac{1 - f}{n}\bar{Y} \\ &\quad (C_{x}^{2} - \rho C_{x}C_{y}) \end{aligned} \]which is of order \(1/n\), so it shrinks as \(n\) grows, and the ratio estimator is consistent. Hartley and Ross showed that the bias divided by the standard error of \(\hat{R}\) is at most the coefficient of variation of \(\bar{x}\), which also tends to 0 as \(n\) increases.Decreasing \(n\) increases the bias, so 'both' is wrong, and a larger population with \(n\) fixed only moves the factor \(1 - f\) towards 1.

Q: If \(\pi_{i}\) and \(\pi_{ij}\) are respectively the first order and second order inclusion probabilities of a sampling design in PPSWOR, then which of the following relation is true ?

The true relation is \(\sum_{j}\pi_{ij} = (n - 1)\pi_{i}\), the sum over \(j \ne i\).For any fixed-size design of size \(n\), count the pairs containing unit \(i\). Whenever \(i\) is in the sample, there are exactly \(n - 1\) other units with it, so\[ \begin{aligned} \sum_{j \ne i}\pi_{ij} &= E\left[I_{i}\sum_{j \ne i}I_{j}\right] \\ &= (n - 1)\pi_{i} \end{aligned} \]where \(I_{i}\) indicates that unit \(i\) is selected. This identity is used to derive the Yates-Grundy form of the Horvitz-Thompson variance. Summing it over \(i\) gives \(n(n - 1)\), which is the total over all ordered pairs, not the sum for one \(i\) (D); A and C contradict the identity.

Q: PPSWR sampling reduced to SRS if the probability of proportion to size i.e. \(p_{i}\) is :

PPSWR reduces to SRS (with replacement) when \(p_{i} = 1/N\).In PPS sampling with replacement, unit \(i\) is chosen on each draw with probability \(p_{i}\) proportional to its size measure. If every unit has the same size, all \(p_{i}\) are equal, and since they sum to 1 over the N units, \(p_{i} = 1/N\). Every draw is then equally likely to pick any unit, which is simple random sampling with replacement, and the Hansen-Hurwitz estimator becomes \(N\bar{y}\).\(1/n\) and \(n/N\) do not sum to 1 over N units (\(n/N\) is the inclusion probability under SRSWOR), and \(p_{i} = 1\) would always select the same unit.

Q: Consider a finite population of \(N = 2n\) units. A sample of size 2 is drawn as follows :(i) The population is randomly divided into 2 groups each of size \(n\).(ii) From each of the two groups as obtained above, one unit is drawn with probability \(1/n\).These two units form the sample. Then, the correct statement is :

The sample is a SRSWOR of size 2.Fix two units \(u\) and \(v\). They can both be chosen only if the random split puts them in different groups, which happens with probability \(n/(2n - 1)\) (of the other \(2n - 1\) places, \(n\) are in the other group). Then each is drawn from its group with probability \(1/n\):\[ \begin{aligned} P(\{u, v\}) &= \frac{n}{2n - 1} \cdot \frac{1}{n^{2}} \\ &= \frac{1}{n(2n - 1)} = \frac{1}{\binom{2n}{2}} \end{aligned} \]Every pair has the same probability, which is the definition of simple random sampling without replacement. It is not a stratified sample, because the groups are formed at random rather than fixed in advance, and the sample mean is unbiased, as it is under SRSWOR.

Q: Consider the following sampling designs : a stratified sample scheme with SRSWOR within each stratum and a SRSWOR. Both have the same sample size. We are interested in estimating the population mean. Let \(V_{\text{prop}}^{2}\) and \(V_{\text{SRS}}^{2}\) be the corresponding variances of the usual estimators of the population mean. Then,

The keyed relation is \(V_{\text{prop}}^{2} \le V_{\text{SRS}}^{2}\).Ignoring finite-population terms of order \(1/N_{h}\), the variance of the mean under SRSWOR splits into within-stratum and between-stratum parts:\[ \begin{aligned} &V_{\text{SRS}} - V_{\text{prop}} \\ &\approx \frac{1 - f}{n}\sum_{h} W_{h}(\bar{Y}_{h} - \bar{Y})^{2} \ge 0 \end{aligned} \]Proportional allocation removes the between-strata part, so it is never worse, and the two are equal when all stratum means coincide. That is why '\(\le\)' rather than strict '\(\lt\)' is the safe statement, and '\(=\)' or '\(\gt\)' is wrong in general. (With very small strata the exact formula can reverse the inequality slightly; the textbook result assumes the strata are not tiny.)

Q: A probability proportional to size (PPS) without replacement sample of size 2 is to be drawn from a population of N = 3 units. Let \(z_{1} = 1/4\), \(z_{2} = 1/4\), \(z_{3} = 1/2\). The \(i\)-th unit is selected with probability \(z_{i}\), \(i = 1, 2, 3\). If the first unit selected is \(i\), then the \(j\)-th unit is selected for inclusion in the sample with probability \(z_{j}/(1 - z_{i})\), \(j \ne i\). Then, the probability that the third unit is the population is included in the sample is given by :

The probability is 5/6.Unit 3 is in the sample if it is drawn first, or if unit 1 or unit 2 is drawn first and unit 3 second:\[ \begin{aligned} \pi_{3} &= z_{3} + z_{1}\frac{z_{3}}{1 - z_{1}} + z_{2}\frac{z_{3}}{1 - z_{2}} \\ &= \tfrac{1}{2} + 2 \cdot \tfrac{1}{4}\cdot\tfrac{1/2}{3/4} \\ &= \tfrac{1}{2} + \tfrac{1}{3} = \tfrac{5}{6} \end{aligned} \]The paper's 'the third unit is the population' means 'in the population'. 1/2 is only the first-draw chance; 2/3 is the inclusion probability with equal probabilities; and 1 would hold only if unit 3 could not be missed. As a check, \(\pi_{1} = \pi_{2} = 7/12\), and \(7/12 + 7/12 + 5/6 = 2\), the sample size.

Q: Let \(r = \bar{y}/\bar{x}\) denote the ratio of the sample means of the variable \(y\) and of the auxiliary variable \(x\). The sample is a SRSWOR sample of size \(n\). Let \(f = n/N\) and \(\bar{X}\) be the population mean of the auxiliary variable \(x\). Let \(R = \bar{Y}/\bar{X}\). By considering \(\mathrm{Cov}(r, \bar{x})\), the bias of \(r\), as an estimator of R, is given by :

The bias of \(r\) is \(-\dfrac{1}{\bar{X}}\mathrm{Cov}(r, \bar{x})\).Since \(r\bar{x} = \bar{y}\) and \(E(\bar{y}) = \bar{Y}\) under SRSWOR,\[ \begin{aligned} \mathrm{Cov}(r, \bar{x}) &= E(r\bar{x}) - E(r)E(\bar{x}) \\ &= \bar{Y} - E(r)\bar{X} \end{aligned} \]Dividing by \(\bar{X}\) gives \(R - E(r) = \mathrm{Cov}(r, \bar{x})/\bar{X}\), so the bias \(E(r) - R = -\mathrm{Cov}(r, \bar{x})/\bar{X}\). This is the exact expression; by the Cauchy-Schwarz inequality it also shows \(|\text{bias}| \le \sigma_{r}\sigma_{\bar{x}}/\bar{X}\), small in large samples.The other options are pieces of the approximate bias formula \(\tfrac{1 - f}{n}R(C_{x}^{2} - \rho C_{x}C_{y})\), with the factor in brackets missing, so they are not the bias.

If \(\rho_{wsy}\) denotes the intra-class correlation coefficient between pairs of units that are in the same systematic sample and if \(\rho_{wsy} = 0\), then :

A.Systematic sampling is as efficient as SRSWOR
B.Systematic sampling is more efficient than SRSWOR
C.Systematic sampling is as efficient as SRSWR but less efficient than SRSWOR
D.Systematic sampling is more efficient than SRSWR

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

Systematic sampling is as efficient as SRSWR but less efficient than SRSWOR.

For a population of size \(N = nk\), the variance of the systematic sample mean in terms of the intra-class correlation is

\[ \begin{aligned} V(\bar{y}_{sy}) &= \frac{\sigma^{2}}{n}\,[1 + (n - 1)\rho_{wsy}] \\ \sigma^{2} &= \frac{N - 1}{N}S^{2} \end{aligned} \]

With \(\rho_{wsy} = 0\) this is \(\sigma^{2}/n\), exactly the variance of the mean under SRSWR. Under SRSWOR the variance is \(\dfrac{N - n}{N} \cdot \dfrac{S^{2}}{n}\), which is smaller than \(\dfrac{N - 1}{N} \cdot \dfrac{S^{2}}{n}\) whenever \(n \gt 1\), because of the finite population correction.

So systematic sampling matches SRSWR and falls just short of SRSWOR. It beats SRSWOR only when \(\rho_{wsy}\) is negative enough, which rules out the options claiming it is as efficient as or more efficient than SRSWOR, or more efficient than SRSWR.

Discuss in Forum

For an SRSWOR (N, n), the probability that a specified unit is included in the sample is :

A.\(\dfrac{1}{N}\)
B.\(\dfrac{n}{N}\)
C.\(\dfrac{1}{({}^{N}C_{n})}\)
D.\(\dfrac{1}{N(N - 1)}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The inclusion probability is \(\dfrac{n}{N}\).

All \({}^{N}C_{n}\) samples are equally likely. A specified unit is in a sample when the other \(n - 1\) units are chosen from the remaining \(N - 1\), so

\[ \pi = \frac{{}^{N-1}C_{n-1}}{{}^{N}C_{n}} = \frac{n}{N} \]

\(1/{}^{N}C_{n}\) is the probability of one particular sample, not of one unit. \(1/N\) is the probability that the unit is chosen at one particular draw. \(1/N(N - 1)\) is the probability of one particular ordered pair at the first two draws.

Discuss in Forum

If the regression estimator of \(\bar{Y}\) is \(\bar{y} + b(\bar{X} - \bar{x})\) where \(b = \dfrac{S_{xy}}{S_{x}^{2}}\), then an exact expression for bias of regression estimator is :

A.\(-Cov(\bar{y}, b)\)
B.\(Cov(\bar{y}, b)\)
C.\(Cov(b, \bar{x})\)
D.\(-Cov(b, \bar{x})\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The exact bias is \(-Cov(b, \bar{x})\).

\(\bar{y}\) is unbiased for \(\bar{Y}\), so the bias of the regression estimator is \(E[b(\bar{X} - \bar{x})]\). Using \(E(b\bar{x}) = Cov(b, \bar{x}) + E(b)E(\bar{x})\) and \(E(\bar{x}) = \bar{X}\):

\[ \begin{aligned} E[b(\bar{X} - \bar{x})] &= \bar{X}E(b) - E(b\bar{x}) \\ &= \bar{X}E(b) - Cov(b, \bar{x}) - \bar{X}E(b) \\ &= -Cov(b, \bar{x}) \end{aligned} \]

The bias comes from the randomness of \(b\) being correlated with \(\bar{x}\); with a fixed \(b\) the estimator would be unbiased. Options involving \(\bar{y}\) are wrong because \(\bar{y}\) enters linearly and contributes no bias, and option C has the sign reversed.

Discuss in Forum

The bais in ratio estimator decreases with :

A.increasing the sample size \(n\)
B.decreasing the sample size \(n\)
C.both (A) and (B)
D.increase in the population size

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The bias decreases with increasing the sample size \(n\).

The ratio estimator \(\hat{R}\bar{X} = (\bar{y}/\bar{x})\bar{X}\) is biased because the expectation of a ratio is not the ratio of expectations. To first order its bias is

\[ \begin{aligned} B &\approx \frac{1 - f}{n}\bar{Y} \\ &\quad (C_{x}^{2} - \rho C_{x}C_{y}) \end{aligned} \]

which is of order \(1/n\), so it shrinks as \(n\) grows, and the ratio estimator is consistent. Hartley and Ross showed that the bias divided by the standard error of \(\hat{R}\) is at most the coefficient of variation of \(\bar{x}\), which also tends to 0 as \(n\) increases.

Decreasing \(n\) increases the bias, so 'both' is wrong, and a larger population with \(n\) fixed only moves the factor \(1 - f\) towards 1.

Discuss in Forum

If \(\pi_{i}\) and \(\pi_{ij}\) are respectively the first order and second order inclusion probabilities of a sampling design in PPSWOR, then which of the following relation is true ?

A.\(\sum_{j}\pi_{ij} = \pi_{i}\)
B.\(\sum_{j}\pi_{ij} = (n - 1)\pi_{i}\)
C.\(\sum_{j}\pi_{ij} \ne \pi_{i}\)
D.\(\sum_{j}\pi_{ij} = n(n - 1)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The true relation is \(\sum_{j}\pi_{ij} = (n - 1)\pi_{i}\), the sum over \(j \ne i\).

For any fixed-size design of size \(n\), count the pairs containing unit \(i\). Whenever \(i\) is in the sample, there are exactly \(n - 1\) other units with it, so

\[ \begin{aligned} \sum_{j \ne i}\pi_{ij} &= E\left[I_{i}\sum_{j \ne i}I_{j}\right] \\ &= (n - 1)\pi_{i} \end{aligned} \]

where \(I_{i}\) indicates that unit \(i\) is selected. This identity is used to derive the Yates-Grundy form of the Horvitz-Thompson variance. Summing it over \(i\) gives \(n(n - 1)\), which is the total over all ordered pairs, not the sum for one \(i\) (D); A and C contradict the identity.

Discuss in Forum

PPSWR sampling reduced to SRS if the probability of proportion to size i.e. \(p_{i}\) is :

A.\(1/n\)
B.1
C.\(1/N\)
D.\(n/N\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

PPSWR reduces to SRS (with replacement) when \(p_{i} = 1/N\).

In PPS sampling with replacement, unit \(i\) is chosen on each draw with probability \(p_{i}\) proportional to its size measure. If every unit has the same size, all \(p_{i}\) are equal, and since they sum to 1 over the N units, \(p_{i} = 1/N\). Every draw is then equally likely to pick any unit, which is simple random sampling with replacement, and the Hansen-Hurwitz estimator becomes \(N\bar{y}\).

\(1/n\) and \(n/N\) do not sum to 1 over N units (\(n/N\) is the inclusion probability under SRSWOR), and \(p_{i} = 1\) would always select the same unit.

Discuss in Forum

Consider a finite population of \(N = 2n\) units. A sample of size 2 is drawn as follows :

(i) The population is randomly divided into 2 groups each of size \(n\).
(ii) From each of the two groups as obtained above, one unit is drawn with probability \(1/n\).

These two units form the sample. Then, the correct statement is :

A.The sample is a stratified sample with two strata.
B.The sample mean is not an unbiased estimator of the population mean.
C.The sample is a SRSWOR of size 2.
D.The sample is not randomly obtained.

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The sample is a SRSWOR of size 2.

Fix two units \(u\) and \(v\). They can both be chosen only if the random split puts them in different groups, which happens with probability \(n/(2n - 1)\) (of the other \(2n - 1\) places, \(n\) are in the other group). Then each is drawn from its group with probability \(1/n\):

\[ \begin{aligned} P(\{u, v\}) &= \frac{n}{2n - 1} \cdot \frac{1}{n^{2}} \\ &= \frac{1}{n(2n - 1)} = \frac{1}{\binom{2n}{2}} \end{aligned} \]

Every pair has the same probability, which is the definition of simple random sampling without replacement. It is not a stratified sample, because the groups are formed at random rather than fixed in advance, and the sample mean is unbiased, as it is under SRSWOR.

Discuss in Forum

Consider the following sampling designs : a stratified sample scheme with SRSWOR within each stratum and a SRSWOR. Both have the same sample size. We are interested in estimating the population mean. Let \(V_{\text{prop}}^{2}\) and \(V_{\text{SRS}}^{2}\) be the corresponding variances of the usual estimators of the population mean. Then,

A.\(V_{\text{prop}}^{2} \lt V_{\text{SRS}}^{2}\)
B.\(V_{\text{prop}}^{2} = V_{\text{SRS}}^{2}\)
C.\(V_{\text{prop}}^{2} \gt V_{\text{SRS}}^{2}\)
D.\(V_{\text{prop}}^{2} \le V_{\text{SRS}}^{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The keyed relation is \(V_{\text{prop}}^{2} \le V_{\text{SRS}}^{2}\).

Ignoring finite-population terms of order \(1/N_{h}\), the variance of the mean under SRSWOR splits into within-stratum and between-stratum parts:

\[ \begin{aligned} &V_{\text{SRS}} - V_{\text{prop}} \\ &\approx \frac{1 - f}{n}\sum_{h} W_{h}(\bar{Y}_{h} - \bar{Y})^{2} \ge 0 \end{aligned} \]

Proportional allocation removes the between-strata part, so it is never worse, and the two are equal when all stratum means coincide. That is why '\(\le\)' rather than strict '\(\lt\)' is the safe statement, and '\(=\)' or '\(\gt\)' is wrong in general. (With very small strata the exact formula can reverse the inequality slightly; the textbook result assumes the strata are not tiny.)

Discuss in Forum

A probability proportional to size (PPS) without replacement sample of size 2 is to be drawn from a population of N = 3 units. Let \(z_{1} = 1/4\), \(z_{2} = 1/4\), \(z_{3} = 1/2\). The \(i\)-th unit is selected with probability \(z_{i}\), \(i = 1, 2, 3\). If the first unit selected is \(i\), then the \(j\)-th unit is selected for inclusion in the sample with probability \(z_{j}/(1 - z_{i})\), \(j \ne i\). Then, the probability that the third unit is the population is included in the sample is given by :

A.1
B.1/2
C.5/6
D.2/3

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The probability is 5/6.

Unit 3 is in the sample if it is drawn first, or if unit 1 or unit 2 is drawn first and unit 3 second:

\[ \begin{aligned} \pi_{3} &= z_{3} + z_{1}\frac{z_{3}}{1 - z_{1}} + z_{2}\frac{z_{3}}{1 - z_{2}} \\ &= \tfrac{1}{2} + 2 \cdot \tfrac{1}{4}\cdot\tfrac{1/2}{3/4} \\ &= \tfrac{1}{2} + \tfrac{1}{3} = \tfrac{5}{6} \end{aligned} \]

The paper's 'the third unit is the population' means 'in the population'. 1/2 is only the first-draw chance; 2/3 is the inclusion probability with equal probabilities; and 1 would hold only if unit 3 could not be missed. As a check, \(\pi_{1} = \pi_{2} = 7/12\), and \(7/12 + 7/12 + 5/6 = 2\), the sample size.

Discuss in Forum

10.

Let \(r = \bar{y}/\bar{x}\) denote the ratio of the sample means of the variable \(y\) and of the auxiliary variable \(x\). The sample is a SRSWOR sample of size \(n\). Let \(f = n/N\) and \(\bar{X}\) be the population mean of the auxiliary variable \(x\). Let \(R = \bar{Y}/\bar{X}\). By considering \(\mathrm{Cov}(r, \bar{x})\), the bias of \(r\), as an estimator of R, is given by :

A.\(\dfrac{1 - f}{n}\)
B.\(\dfrac{1 - f}{nR}\)
C.\(\dfrac{1 - f}{n}R\)
D.\(-\dfrac{1}{\bar{X}}\mathrm{Cov}(r, \bar{x})\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The bias of \(r\) is \(-\dfrac{1}{\bar{X}}\mathrm{Cov}(r, \bar{x})\).

Since \(r\bar{x} = \bar{y}\) and \(E(\bar{y}) = \bar{Y}\) under SRSWOR,

\[ \begin{aligned} \mathrm{Cov}(r, \bar{x}) &= E(r\bar{x}) - E(r)E(\bar{x}) \\ &= \bar{Y} - E(r)\bar{X} \end{aligned} \]

Dividing by \(\bar{X}\) gives \(R - E(r) = \mathrm{Cov}(r, \bar{x})/\bar{X}\), so the bias \(E(r) - R = -\mathrm{Cov}(r, \bar{x})/\bar{X}\). This is the exact expression; by the Cauchy-Schwarz inequality it also shows \(|\text{bias}| \le \sigma_{r}\sigma_{\bar{x}}/\bar{X}\), small in large samples.

The other options are pieces of the approximate bias formula \(\tfrac{1 - f}{n}R(C_{x}^{2} - \rho C_{x}C_{y})\), with the factor in brackets missing, so they are not the bias.

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