Identify the correct match between the terms in Column-I below with the definitions given in Column-II.Column-IColumn-II(i) AOQ(\(\alpha\)) Number of items per lot that will be inspected on average(ii) AQL(\(\beta\)) A lower bound on the incoming quality for which a lot is to be rejected(iii) ATI(\(\gamma\)) Proportion of defective items remaining in the lot after inspection(iv) RQL(\(\delta\)) An upper bound on the incoming quality for which a lot to be accepted

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.

Sampling Theory Questions and Answers

Q: Under systematic sampling :

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 \(k\).

Q: Post stratification is used when :

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.

Q: Double sampling is useful :

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.

Q: Suppose \(S_{Y}^{2} = \dfrac{1}{n-1}\sum_{r=1}^{n} (Y_{r} - \bar{Y})^{2}\)\(s_{y}^{2} = \dfrac{1}{N-1}\sum_{i=1}^{N} (y_{i} - \bar{y})^{2}\)\(\bar{Y}\) is sample mean under SRSWR\(\bar{y}\) is population mean\(\bar{Y}_{sy}\) is sample mean under systematic samplingWhich of the following statements is not true ?

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 \(E(\bar{Y}_{sy}) = \bar{y}\).

Q: Consider a population of 50 units, {1, 2, 3, ....., 50} and suppose that 50 possible samples are listed as {1}, {1, 2}, {1, 2, 3}, ....., {1, 2, 3, ....., 50}. One of these samples is choosen at random. Let \(\pi_{i}\) be the probability that unit i of the population is in the selected sample. Then which of the following is necessarily true ?

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.)

Q: Consider the following hypothetical population of 6 units. What is the estimated value of population total of y-variable based on SRSWOR sample of 2 units selected as unit 4 and unit 6 from the population using ratio method of estimation :Unit No.xy10121333114618582961046

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.

Q: In stratified random sampling with a linear cost function\[ C = C_{0} + \sum_{h=1}^{L} C_{h}n_{h}, \]the variance of estimated mean \((\bar{Y}_{st})\) is minimum for a specified cost C and the cost is minimum for a specified variance of \(\bar{Y}_{st}\) when :

The optimum allocation is nh &prop; WhSh/√Ch.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.

Q: If population size N is very large, then the approximate value of minimum sample size for estimating population mean with desired accuracy under SRSWOR design is :where \(\alpha\) and d are \(P[|\overline{Y} - \overline{y}| \lt d] \ge 1 - \alpha\)

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.

21.

Identify the correct match between the terms in Column-I below with the definitions given in Column-II.

Column-I	Column-II
(i) AOQ	(\(\alpha\)) Number of items per lot that will be inspected on average
(ii) AQL	(\(\beta\)) A lower bound on the incoming quality for which a lot is to be rejected
(iii) ATI	(\(\gamma\)) Proportion of defective items remaining in the lot after inspection
(iv) RQL	(\(\delta\)) An upper bound on the incoming quality for which a lot to be accepted

A.i-\(\gamma\), ii-\(\delta\), iii-\(\alpha\), iv-\(\beta\)
B.i-\(\gamma\), ii-\(\beta\), iii-\(\alpha\), iv-\(\delta\)
C.i-\(\beta\), ii-\(\delta\), iii-\(\alpha\), iv-\(\gamma\)
D.i-\(\alpha\), ii-\(\delta\), iii-\(\gamma\), iv-\(\beta\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

Discuss in Forum

22.

Under systematic sampling :

A.First unit in the sample is selected by probability proportionate to size (pps) sampling and remaining \((n - 1)\) units are selected by simple random sampling
B.Variance of sample mean is always less than that of sample mean from SRSWOR
C.Each population unit has unequal chance of being selected in the sample
D.Sample mean is unbiased estimator for population mean

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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 \(k\).

Discuss in Forum

23.

Under Midzuno's sampling design the inclusion probability of \(i\)th element \(\pi_{i}\) is at least :

A.\(\frac{n - 1}{N - 1}\)
B.\(\frac{n}{N}\)
C.\(\frac{n(n - 1)}{N - 1}\)
D.\(\frac{n(n - 1)}{N}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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.

Discuss in Forum

24.

Post stratification is used when :

A.Stratum sizes are equal
B.Number of strata of the population is not known
C.Population size and stratum sizes are not known
D.Stratum to which a unit belongs is not known before sampling

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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.

Discuss in Forum

25.

Double sampling is useful :

A.When complete information on auxiliary variable is not available
B.When population is heterogeneous
C.To improve the performance of ratio estimator for population mean
D.To improve the performance of regression estimator for population mean

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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.

Discuss in Forum

26.

Suppose \(S_{Y}^{2} = \dfrac{1}{n-1}\sum_{r=1}^{n} (Y_{r} - \bar{Y})^{2}\)

\(s_{y}^{2} = \dfrac{1}{N-1}\sum_{i=1}^{N} (y_{i} - \bar{y})^{2}\)

\(\bar{Y}\) is sample mean under SRSWR
\(\bar{y}\) is population mean
\(\bar{Y}_{sy}\) is sample mean under systematic sampling

Which of the following statements is not true ?

A.Probability of selecting ith element of the population on the rth draw is \(\dfrac{1}{N}\) under SRSWOR design for \(i = 1, 2, 3, \ldots, N\) and \(r = 1, 2, 3, \ldots, n\)
B.\(ES_{Y}^{2} = s_{y}^{2}\) under SRSWOR
C.\(E(\bar{Y}_{sy}) = \bar{y}\) under systematic sampling
D.\(V(\bar{Y}) = \dfrac{N-1}{n} s_{y}^{2}\) under SRSWR

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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 \(E(\bar{Y}_{sy}) = \bar{y}\).

Discuss in Forum

27.

Consider a population of 50 units, {1, 2, 3, ....., 50} and suppose that 50 possible samples are listed as {1}, {1, 2}, {1, 2, 3}, ....., {1, 2, 3, ....., 50}. One of these samples is choosen at random. Let \(\pi_{i}\) be the probability that unit i of the population is in the selected sample. Then which of the following is necessarily true ?

A.The expected sample size is 25.5
B.\(\sum_{i=1}^{50} \pi_{i} = 1\)
C.\(\sum_{i=1}^{50} \pi_{i} = 25\)
D.The expected sample size is 25

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.)

Discuss in Forum

28.

Consider the following hypothetical population of 6 units. What is the estimated value of population total of y-variable based on SRSWOR sample of 2 units selected as unit 4 and unit 6 from the population using ratio method of estimation :

Unit No.	x	y
1	0	1
2	1	3
3	3	11
4	6	18
5	8	29
6	10	46

A.27
B.112
C.198
D.432

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

29.

In stratified random sampling with a linear cost function

\[ C = C_{0} + \sum_{h=1}^{L} C_{h}n_{h}, \]

the variance of estimated mean \((\bar{Y}_{st})\) is minimum for a specified cost C and the cost is minimum for a specified variance of \(\bar{Y}_{st}\) when :

A.\(n_{h} \propto \left\{\dfrac{w_{h}s_{h}}{\sqrt{C_{h}}}\right\}\)
B.\(n_{h} \propto \left\{\dfrac{\sum_{h=1}^{L} w_{h}s_{h}}{C_{h}}\right\}\)
C.\(n_{h} \propto \left\{\dfrac{N_{h}s_{h}}{s_{h}}\right\}\)
D.\(n_{h} \propto \left\{\dfrac{n\, s_{h}}{\sqrt{C_{h}}}\right\}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

30.

If population size N is very large, then the approximate value of minimum sample size for estimating population mean with desired accuracy under SRSWOR design is :

where \(\alpha\) and d are \(P[|\overline{Y} - \overline{y}| \lt d] \ge 1 - \alpha\)

A.\(\left(\frac{s_{y} Z_{\alpha/2}}{d}\right)^{1/2}\)
B.\(\sqrt{N}\)
C.\(\frac{(s_{y} Z_{\alpha/2})^{2}}{d}\)
D.\(\left(\frac{s_{y} Z_{\alpha/2}}{d}\right)^{2}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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