Sampling Theory Questions and Answers

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.