Sampling Theory Questions and Answers

Q: Consider a population of size N = 6 as {1, 2, 3, 4, 5, 6} with their auxiliary variable values as :\(x_{1} = 100\), \(x_{2} = 200\), \(x_{3} = 300\), \(x_{4} = 400\), \(x_{5} = 500\) and \(x_{6} = 500\).We want to draw a random sample of n = 3 using PPSWOR design. Let \(U_{r}\) denote the unit selected at the rth draw for r = 1, 2, 3.Then :\[ P[U_{3} = 2 \mid U_{1} = 5, U_{2} = 3] = \]

The probability is \(\frac{1}{6}\).In PPSWOR each draw selects among the remaining units with probability proportional to their sizes. The total size is \(100 + 200 + 300 + 400 + 500 + 500 = 2000\). After units 5 (size 500) and 3 (size 300) are drawn, the remaining total is\[ 2000 - 500 - 300 = 1200 \]so\[ \begin{aligned} P[U_{3} = 2 \mid U_{1} = 5, U_{2} = 3] &= \frac{200}{1200} \\ &= \frac{1}{6} \end{aligned} \]\(\frac{1}{10}\) is the unconditional first-draw probability \(\frac{200}{2000}\), which ignores the units already removed. \(\frac{2}{11}\) and \(\frac{2}{13}\) remove the wrong sizes from the total.

Q: For a sampling design \((\mathfrak{U}, \mathfrak{S}, P)\), the Horvitz-Thompson estimator for population total is given by :

The Horvitz-Thompson estimator of the population total is \(\sum_{i=1}^{N} \frac{y_{i} T_{i}}{\pi_{i}}\).Here \(T_{i}\) is the indicator that unit i is in the sample and \(\pi_{i} = P(T_{i} = 1)\) its inclusion probability. Each sampled value is weighted by \(\frac{1}{\pi_{i}}\), and since \(E(T_{i}) = \pi_{i}\),\[ \begin{aligned} E\left(\sum_{i=1}^{N} \frac{y_{i} T_{i}}{\pi_{i}}\right) &= \sum_{i=1}^{N} y_{i} \\ &= Y \end{aligned} \]so the estimator is unbiased for the total.Dividing by N (option A) estimates the mean instead. Options C and D drop the indicator \(T_{i}\), so they sum over every unit of the population, sampled or not, which is not a statistic.

31.

Consider a population of size N = 6 as {1, 2, 3, 4, 5, 6} with their auxiliary variable values as :

\(x_{1} = 100\), \(x_{2} = 200\), \(x_{3} = 300\), \(x_{4} = 400\), \(x_{5} = 500\) and \(x_{6} = 500\).

We want to draw a random sample of n = 3 using PPSWOR design. Let \(U_{r}\) denote the unit selected at the rth draw for r = 1, 2, 3.

Then :

\[ P[U_{3} = 2 \mid U_{1} = 5, U_{2} = 3] = \]

A.\(\frac{1}{6}\)
B.\(\frac{1}{10}\)
C.\(\frac{2}{11}\)
D.\(\frac{2}{13}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The probability is \(\frac{1}{6}\).

In PPSWOR each draw selects among the remaining units with probability proportional to their sizes. The total size is \(100 + 200 + 300 + 400 + 500 + 500 = 2000\). After units 5 (size 500) and 3 (size 300) are drawn, the remaining total is

\[ 2000 - 500 - 300 = 1200 \]

\[ \begin{aligned} P[U_{3} = 2 \mid U_{1} = 5, U_{2} = 3] &= \frac{200}{1200} \\ &= \frac{1}{6} \end{aligned} \]

\(\frac{1}{10}\) is the unconditional first-draw probability \(\frac{200}{2000}\), which ignores the units already removed. \(\frac{2}{11}\) and \(\frac{2}{13}\) remove the wrong sizes from the total.

Discuss in Forum

32.

For a sampling design \((\mathfrak{U}, \mathfrak{S}, P)\), the Horvitz-Thompson estimator for population total is given by :

A.\(\frac{1}{N} \sum_{i=1}^{N} \frac{y_{i} T_{i}}{\pi_{i}}\)
B.\(\sum_{i=1}^{N} \frac{y_{i} T_{i}}{\pi_{i}}\)
C.\(\frac{N}{n} \sum_{i=1}^{N} \frac{y_{i}}{\pi_{i}}\)
D.\(\sum_{i=1}^{N} \frac{y_{i}}{N\pi_{i}}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The Horvitz-Thompson estimator of the population total is \(\sum_{i=1}^{N} \frac{y_{i} T_{i}}{\pi_{i}}\).

Here \(T_{i}\) is the indicator that unit i is in the sample and \(\pi_{i} = P(T_{i} = 1)\) its inclusion probability. Each sampled value is weighted by \(\frac{1}{\pi_{i}}\), and since \(E(T_{i}) = \pi_{i}\),

\[ \begin{aligned} E\left(\sum_{i=1}^{N} \frac{y_{i} T_{i}}{\pi_{i}}\right) &= \sum_{i=1}^{N} y_{i} \\ &= Y \end{aligned} \]

so the estimator is unbiased for the total.

Dividing by N (option A) estimates the mean instead. Options C and D drop the indicator \(T_{i}\), so they sum over every unit of the population, sampled or not, which is not a statistic.

Discuss in Forum

33.

Which of the following is NOT true for stratified random sampling ?

A larger sample would be required from a stratum if :

A.stratum size is large
B.stratum variability is large
C.sampling cost per unit is large in the stratum
D.sampling cost per unit is small in the stratum

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The statement that is NOT true is that a larger sample is needed when the sampling cost per unit is large in the stratum.

Under optimum allocation the sample size in stratum h is

\[ n_{h} \propto \frac{N_{h} S_{h}}{\sqrt{c_{h}}} \]

It grows with the stratum size \(N_{h}\) and the stratum variability \(S_{h}\), and falls as the cost per unit \(c_{h}\) rises. So a costly stratum gets a smaller sample, not a larger one.

Options A, B and D (large size, large variability, low cost) all correctly call for a larger sample.

Sampling Theory Questions and Answers

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