Sampling Theory Questions and Answers

Q: Let us consider the following methods of estimation, all based on SRSWOR of the same sample size. I Sample mean II Ratio estimator III Regression estimator. Let the corresponding variances be denoted by \(V_{1}^{2}\), \(V_{2}^{2}\) and \(V_{3}^{2}\). Assuming that the sample size is sufficiently large, we have :

The answer is none of the above.To first order in large samples, with \(\rho\) the correlation and \(C_{x}, C_{y}\) the coefficients of variation,\[ \begin{aligned} V_{1}^{2} &\propto C_{y}^{2} \\ V_{2}^{2} &\propto C_{y}^{2} + C_{x}^{2} - 2\rho C_{x}C_{y} \\ V_{3}^{2} &\propto C_{y}^{2}(1 - \rho^{2}) \end{aligned} \]The regression estimator is always at least as good as both others, so \(V_{3}^{2}\) is smallest and option A is wrong. But the order of the other two depends on the data: the ratio estimator beats the sample mean only when \(\rho \gt C_{x}/(2C_{y})\). So option B holds for some populations and option C for others, and neither ordering holds in general.

Q: Consider a population of \(N = nk\) units with \(k \gt 1\). A sample of size \(n\) is obtained by the following method :(i) A unit is drawn from the set of units labelled \(\{k, 2k, \ldots, N\}\) with probability \(1/n\) and is included in the sample.(ii) A SRSWOR of size \(n - 1\) is then obtained from the remaining \(N - 1\) units in the population and these units are included in the sample.Then,

The answer is inclusion probability is not the same for all the population units.Let \(S = \{k, 2k, \ldots, N\}\), which has \(n\) units. A unit in S can enter at the first draw (probability \(1/n\)) or later, while a unit outside S enters only at the second stage:\[ \begin{aligned} \pi_{S} &= \frac{1}{n} + \left(1 - \frac{1}{n}\right)\frac{n - 1}{N - 1} \\ \pi_{\text{other}} &= \frac{n - 1}{N - 1} \end{aligned} \]Since \(N \gt n\), \(\pi_{S} \gt \pi_{\text{other}}\). The design is not systematic sampling, and because inclusion probabilities differ, the plain sample mean is biased. But all inclusion probabilities are positive, so the Horvitz-Thompson estimator is unbiased for the total, which rules out option B.

Q: Let \(V_{SY}^{2}\) be the variance of the usual estimator in a systematic sampling design and \(V_{SRS}^{2}\) the variance of the sample mean in a SRSWOR design. It is known that the population has a linear trend, i.e., \(y_{i}\) is approximately equal to \(ic\), for a positive \(c\), \(i = 1, 2, \ldots, N\). Then :

The answer is \(V_{SY}^{2} \lt V_{SRS}^{2}\).When the population has a linear trend, a systematic sample takes one unit from each successive stretch of \(k\) units, so it automatically spreads over the whole range, from small to large values. Its mean stays close to the population mean. A simple random sample can by chance cluster at one end of the trend.The standard result (for example in Cochran's treatment of populations with linear trend) is that systematic sampling is more precise than SRS in this case, though less precise than stratified sampling with one unit per stratum. So the variances are not equal, and they can be compared.

Q: A systematic sample of size 2 is drawn from a finite population with \(N = 4\) and \(k = 2\). The samples are drawn independently of each other. The probability that the two units (2, 4) are included in sample equals :

The probability is 1/2.With \(N = 4\) and sampling interval \(k = 2\), a random start \(r \in \{1, 2\}\) is chosen with probability 1/2 each, and the sample is \(\{r, r + 2\}\). The only possible samples are {1, 3} and {2, 4}, each with probability 1/2.So the units 2 and 4 appear together exactly when the start is 2, with probability 1/2. It is not 0, since {2, 4} is a possible sample, and not 1/4, which would treat the six pairs of a simple random sample as if they were possible here.

Q: Let us consider the two sampling strategies, both based on a sample of size \(n\) : I. SRSWOR II post-stratification. To estimate the population total :

The answer is post-stratification is better than SRSWOR if the stratum sizes are large and errors in \(W_{h}\) can be ignored.In post-stratification the sample is drawn by SRSWOR and then classified into strata with known weights \(W_{h}\). When the expected sample size in each stratum is large, its variance is approximately that of proportional stratified sampling, plus a small term of order \(1/n^{2}\) from the random stratum sample sizes. Stratification with proportional allocation is never worse than SRS, so under these conditions post-stratification is at least as good as SRSWOR.Neither method is better always: with small strata the random sample sizes can make post-stratification worse. A large \(n\) alone is not the stated condition; it is the stratum sizes and the accuracy of the weights that matter.

Q: In a survey of finite population, the following method is applied to obtain a sample of size \(n\).(i) In the 1st draw, the ith unit is selected with probability \(P_{i} \gt 0\), \(i = 1, \ldots, N\), \(\sum_{i=1}^{N} Pi = 1\) and \(P_{i} \neq \frac{1}{N}\) for some i.(ii) The remaining \((n - 1)\) units are obtained by SRSWOR method from remaining N - 1 units of the population.Let \(\overline{Y}\) denote the sample mean, then :

The key's answer is that there exists an unbiased estimator of the population mean and it is different from the sample mean.This is the Midzuno-Sen scheme. Because the first unit is drawn with unequal probabilities \(P_{i}\), the sample mean \(\bar{y}\) gives units different weights from their population shares and is biased (A fails). But the scheme has known inclusion probabilities, so the Horvitz-Thompson estimator, and the ratio-type estimator \(\bar{y}/\big(\sum_{i \in s} P_{i} \cdot \frac{N}{n}\big)\) used with this scheme, are unbiased, which also rules out D.The procedure draws from the whole population at each stage; it does not divide the population into strata, so C is wrong.

Q: Let the variances of the regression estimator and the sample mean, both computed from SRSWOR of the same sample size, be denoted by \(V_{\mathrm{Reg}}^{2}\) and \(V_{\mathrm{SRS}}^{2}\) respectively. Then :

The correct statement is \(V_{Reg}^{2} \le V_{SRS}^{2}\) for large \(n\).To the first order of approximation, the variance of the regression estimator is \(V_{SRS}^{2}(1 - \rho^{2})\), where \(\rho\) is the correlation between the study and auxiliary variables. Since \(1 - \rho^{2} \le 1\), it is at least as precise as the sample mean in large samples.This is a large-sample result: in small samples the regression estimator is biased and its variance can exceed that of \(\bar{y}\), so D (any \(n\)) and C (both unbiased) fail. The variances are equal only when \(\rho = 0\), so B is not true in general.

Q: Let N = 4. Two units are selected for inclusion in the sample based on the following design. The \(i\)-th unit is selected with probability \(p_{i}\) where \(p_{1} = 1/6\), \(p_{2} = 1/6\), \(p_{3} = 2/3\). The second unit is selected randomly from among the remaining three units in the population. Then, the first unit is included in the sample with probability :

The probability is 4/9.Since \(p_{1} + p_{2} + p_{3} = 1\), the fourth unit has \(p_{4} = 0\). Unit 1 enters the sample either on the first draw, or on the second draw after another unit is drawn first:\[ \begin{aligned} P &= p_{1} + (1 - p_{1}) \times \frac{1}{3} \\ &= \frac{1}{6} + \frac{5}{6} \times \frac{1}{3} = \frac{3}{18} + \frac{5}{18} = \frac{4}{9} \end{aligned} \]2/9 forgets the first draw, and 6/9 and 7/9 overcount the second.

Q: The regression estimator of the sample mean based on a SRSWOR :

The regression estimator is better than the sample mean if \(n\) is sufficiently large.In large samples its variance is approximately \(V(\bar{y})(1 - \rho^{2})\), never larger than that of \(\bar{y}\), and smaller whenever the study and auxiliary variables are correlated. In small samples the regression estimator is biased and its gain is not guaranteed.So it is not 'always' better (A), it does not have the same efficiency except when \(\rho = 0\) (B), and it is not worse (C) in large samples.

Q: Let \(p\) be the observed proportion of units with the attribute A in a SRSWOR of size \(n\) from a finite population size N. Let P be the population proportion and Q = 1 - P. Then, Var(\(p\)) equals :

\(\mathrm{Var}(p) = \) \(\dfrac{PQ}{n}\dfrac{N - n}{N - 1}\).A sample proportion is the mean of 0-1 values whose population variance (with divisor N) is PQ. Under SRSWOR the variance of a sample mean is \(\frac{\sigma^{2}}{n}\frac{N - n}{N - 1}\), so\[ \mathrm{Var}(p) = \frac{PQ}{n} \cdot \frac{N - n}{N - 1} \]The factor \((N - n)/(N - 1)\) is the finite population correction. \(PQ/n\) is the with-replacement variance, and the other two options use the wrong factor or denominator.

11.

Let us consider the following methods of estimation, all based on SRSWOR of the same sample size. I Sample mean II Ratio estimator III Regression estimator. Let the corresponding variances be denoted by \(V_{1}^{2}\), \(V_{2}^{2}\) and \(V_{3}^{2}\). Assuming that the sample size is sufficiently large, we have :

A.\(V_{1}^{2} \le V_{3}^{2} \le V_{2}^{2}\)
B.\(V_{3}^{2} \le V_{2}^{2} \le V_{1}^{2}\)
C.\(V_{3}^{2} \le V_{1}^{2} \le V_{2}^{2}\)
D.None of the above

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The answer is none of the above.

To first order in large samples, with \(\rho\) the correlation and \(C_{x}, C_{y}\) the coefficients of variation,

\[ \begin{aligned} V_{1}^{2} &\propto C_{y}^{2} \\ V_{2}^{2} &\propto C_{y}^{2} + C_{x}^{2} - 2\rho C_{x}C_{y} \\ V_{3}^{2} &\propto C_{y}^{2}(1 - \rho^{2}) \end{aligned} \]

The regression estimator is always at least as good as both others, so \(V_{3}^{2}\) is smallest and option A is wrong. But the order of the other two depends on the data: the ratio estimator beats the sample mean only when \(\rho \gt C_{x}/(2C_{y})\). So option B holds for some populations and option C for others, and neither ordering holds in general.

Discuss in Forum

12.

Consider a population of \(N = nk\) units with \(k \gt 1\). A sample of size \(n\) is obtained by the following method :

(i) A unit is drawn from the set of units labelled \(\{k, 2k, \ldots, N\}\) with probability \(1/n\) and is included in the sample.
(ii) A SRSWOR of size \(n - 1\) is then obtained from the remaining \(N - 1\) units in the population and these units are included in the sample.

Then,

A.The sampling method corresponds to a systematic sample
B.There is no unbiased estimator of the population total
C.Inclusion probability is not the same for all the population units
D.The sample mean is an unbiased estimator of the population mean

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The answer is inclusion probability is not the same for all the population units.

Let \(S = \{k, 2k, \ldots, N\}\), which has \(n\) units. A unit in S can enter at the first draw (probability \(1/n\)) or later, while a unit outside S enters only at the second stage:

\[ \begin{aligned} \pi_{S} &= \frac{1}{n} + \left(1 - \frac{1}{n}\right)\frac{n - 1}{N - 1} \\ \pi_{\text{other}} &= \frac{n - 1}{N - 1} \end{aligned} \]

Since \(N \gt n\), \(\pi_{S} \gt \pi_{\text{other}}\). The design is not systematic sampling, and because inclusion probabilities differ, the plain sample mean is biased. But all inclusion probabilities are positive, so the Horvitz-Thompson estimator is unbiased for the total, which rules out option B.

Discuss in Forum

13.

Let \(V_{SY}^{2}\) be the variance of the usual estimator in a systematic sampling design and \(V_{SRS}^{2}\) the variance of the sample mean in a SRSWOR design. It is known that the population has a linear trend, i.e., \(y_{i}\) is approximately equal to \(ic\), for a positive \(c\), \(i = 1, 2, \ldots, N\). Then :

A.\(V_{SY}^{2} = V_{SRS}^{2}\)
B.\(V_{SY}^{2} \lt V_{SRS}^{2}\)
C.\(V_{SY}^{2} \gt V_{SRS}^{2}\)
D.It is not possible to compare the two estimators based on the given information

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The answer is \(V_{SY}^{2} \lt V_{SRS}^{2}\).

When the population has a linear trend, a systematic sample takes one unit from each successive stretch of \(k\) units, so it automatically spreads over the whole range, from small to large values. Its mean stays close to the population mean. A simple random sample can by chance cluster at one end of the trend.

The standard result (for example in Cochran's treatment of populations with linear trend) is that systematic sampling is more precise than SRS in this case, though less precise than stratified sampling with one unit per stratum. So the variances are not equal, and they can be compared.

Discuss in Forum

14.

A systematic sample of size 2 is drawn from a finite population with \(N = 4\) and \(k = 2\). The samples are drawn independently of each other. The probability that the two units (2, 4) are included in sample equals :

A.0
B.\(\dfrac{1}{2}\)
C.1
D.\(\dfrac{1}{4}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The probability is 1/2.

With \(N = 4\) and sampling interval \(k = 2\), a random start \(r \in \{1, 2\}\) is chosen with probability 1/2 each, and the sample is \(\{r, r + 2\}\). The only possible samples are {1, 3} and {2, 4}, each with probability 1/2.

So the units 2 and 4 appear together exactly when the start is 2, with probability 1/2. It is not 0, since {2, 4} is a possible sample, and not 1/4, which would treat the six pairs of a simple random sample as if they were possible here.

Discuss in Forum

15.

Let us consider the two sampling strategies, both based on a sample of size \(n\) : I. SRSWOR II post-stratification. To estimate the population total :

A.If the stratum sizes are sufficiently large for each stratum and if the error effects in \(W_{h}\) can be ignored, post-stratification is better than SRSWOR
B.SRSWOR is always better than post-stratification
C.If \(n\) is sufficiently large, post-stratification is better than SRSWOR
D.Post-stratification is always better than SRSWOR

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The answer is post-stratification is better than SRSWOR if the stratum sizes are large and errors in \(W_{h}\) can be ignored.

In post-stratification the sample is drawn by SRSWOR and then classified into strata with known weights \(W_{h}\). When the expected sample size in each stratum is large, its variance is approximately that of proportional stratified sampling, plus a small term of order \(1/n^{2}\) from the random stratum sample sizes. Stratification with proportional allocation is never worse than SRS, so under these conditions post-stratification is at least as good as SRSWOR.

Neither method is better always: with small strata the random sample sizes can make post-stratification worse. A large \(n\) alone is not the stated condition; it is the stratum sizes and the accuracy of the weights that matter.

Discuss in Forum

16.

In a survey of finite population, the following method is applied to obtain a sample of size \(n\).

(i) In the 1st draw, the ith unit is selected with probability \(P_{i} \gt 0\), \(i = 1, \ldots, N\), \(\sum_{i=1}^{N} Pi = 1\) and \(P_{i} \neq \frac{1}{N}\) for some i.
(ii) The remaining \((n - 1)\) units are obtained by SRSWOR method from remaining N - 1 units of the population.

Let \(\overline{Y}\) denote the sample mean, then :

A.\(\overline{Y}\) is an unbiased estimator of population mean
B.There exists an unbiased estimator of the population mean and it is different from the sample mean
C.This is a stratified sampling procedure
D.There does not exist an unbiased estimator of the population mean

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The key's answer is that there exists an unbiased estimator of the population mean and it is different from the sample mean.

This is the Midzuno-Sen scheme. Because the first unit is drawn with unequal probabilities \(P_{i}\), the sample mean \(\bar{y}\) gives units different weights from their population shares and is biased (A fails). But the scheme has known inclusion probabilities, so the Horvitz-Thompson estimator, and the ratio-type estimator \(\bar{y}/\big(\sum_{i \in s} P_{i} \cdot \frac{N}{n}\big)\) used with this scheme, are unbiased, which also rules out D.

The procedure draws from the whole population at each stage; it does not divide the population into strata, so C is wrong.

Discuss in Forum

17.

Let the variances of the regression estimator and the sample mean, both computed from SRSWOR of the same sample size, be denoted by \(V_{\mathrm{Reg}}^{2}\) and \(V_{\mathrm{SRS}}^{2}\) respectively. Then :

A.\(V_{\mathrm{Reg}}^{2} \le V_{\mathrm{SRS}}^{2}\) for a large \(n\).
B.\(V_{\mathrm{Reg}}^{2} = V_{\mathrm{SRS}}^{2}\)
C.Both the estimators of the population mean are unbiased
D.\(V_{\mathrm{Reg}}^{2} \le V_{\mathrm{SRS}}^{2}\) for any \(n\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The correct statement is \(V_{Reg}^{2} \le V_{SRS}^{2}\) for large \(n\).

To the first order of approximation, the variance of the regression estimator is \(V_{SRS}^{2}(1 - \rho^{2})\), where \(\rho\) is the correlation between the study and auxiliary variables. Since \(1 - \rho^{2} \le 1\), it is at least as precise as the sample mean in large samples.

This is a large-sample result: in small samples the regression estimator is biased and its variance can exceed that of \(\bar{y}\), so D (any \(n\)) and C (both unbiased) fail. The variances are equal only when \(\rho = 0\), so B is not true in general.

Discuss in Forum

18.

Let N = 4. Two units are selected for inclusion in the sample based on the following design. The \(i\)-th unit is selected with probability \(p_{i}\) where \(p_{1} = 1/6\), \(p_{2} = 1/6\), \(p_{3} = 2/3\). The second unit is selected randomly from among the remaining three units in the population. Then, the first unit is included in the sample with probability :

A.2/9
B.4/9
C.6/9
D.7/9

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The probability is 4/9.

Since \(p_{1} + p_{2} + p_{3} = 1\), the fourth unit has \(p_{4} = 0\). Unit 1 enters the sample either on the first draw, or on the second draw after another unit is drawn first:

\[ \begin{aligned} P &= p_{1} + (1 - p_{1}) \times \frac{1}{3} \\ &= \frac{1}{6} + \frac{5}{6} \times \frac{1}{3} = \frac{3}{18} + \frac{5}{18} = \frac{4}{9} \end{aligned} \]

2/9 forgets the first draw, and 6/9 and 7/9 overcount the second.

Discuss in Forum

19.

The regression estimator of the sample mean based on a SRSWOR :

A.is always better than the sample mean
B.and the sample mean have the same efficiency
C.is worse than the sample mean
D.is better than the sample mean if \(n\), the sample size, is sufficiently large

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The regression estimator is better than the sample mean if \(n\) is sufficiently large.

In large samples its variance is approximately \(V(\bar{y})(1 - \rho^{2})\), never larger than that of \(\bar{y}\), and smaller whenever the study and auxiliary variables are correlated. In small samples the regression estimator is biased and its gain is not guaranteed.

So it is not 'always' better (A), it does not have the same efficiency except when \(\rho = 0\) (B), and it is not worse (C) in large samples.

Discuss in Forum

20.

Let \(p\) be the observed proportion of units with the attribute A in a SRSWOR of size \(n\) from a finite population size N. Let P be the population proportion and Q = 1 - P. Then, Var(\(p\)) equals :

A.\(\dfrac{PQ}{n}\dfrac{\mathrm{N} - n}{\mathrm{N} - 1}\)
B.\(\dfrac{PQ}{n}\)
C.\(\dfrac{PQ}{n}\dfrac{1}{\mathrm{N} - 1}\)
D.\(\dfrac{PQ}{\mathrm{N}}\dfrac{\mathrm{N} - n}{\mathrm{N} - 1}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

\(\mathrm{Var}(p) = \) \(\dfrac{PQ}{n}\dfrac{N - n}{N - 1}\).

A sample proportion is the mean of 0-1 values whose population variance (with divisor N) is PQ. Under SRSWOR the variance of a sample mean is \(\frac{\sigma^{2}}{n}\frac{N - n}{N - 1}\), so

\[ \mathrm{Var}(p) = \frac{PQ}{n} \cdot \frac{N - n}{N - 1} \]

The factor \((N - n)/(N - 1)\) is the finite population correction. \(PQ/n\) is the with-replacement variance, and the other two options use the wrong factor or denominator.

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