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Part CCSIR NET December 2023

SRS, stratified and systematic sampling

Consider a population of 3 units having values 2, 4 and 6. A simple random sample (without replacement) of 2 units is drawn. Let M be the sample mean. Which of the following statements are true?

  1. A.E(M) = 4✓ Correct
  2. B.E(M2)=17E(M^{2}) = 17
  3. C.E(M3)=72E(M^{3}) = 72✓ Correct
  4. D.Var(M) = 1

The idea, in plain words

The population has only three values and the sample only two, so there are just three possible samples. List them, compute the middle value for each, and every average asked for is a sum of three equally likely numbers.

Solution

M takes the values 3, 4, 5 with probability 1/3 each: E(M)=4,E(M2)=50/3,E(M3)=72,Var(M)=2/3E(M) = 4, E(M^{2}) = 50/3, E(M^{3}) = 72, Var(M) = 2/3.

Why each option is right or wrong

Checked against 4 symbolic computations
  • A.Correct

    True.

    There are exactly three samples of size 2: {2,4}, {2,6} and {4,6}, with means 3, 4 and 5, each with probability 1/3.

    So E(M) = (3 + 4 + 5)/3 = 4 (verified)

    which equals the population mean — the sample mean is unbiased under simple random sampling without replacement.

    The trigger: with a population this small, enumerate all samples rather than reaching for formulas.

  • B.WrongExecution slip

    False.

    E(M2)=(9+16+25)/3=50/3≈16.67(E(M^{2}) = (9 + 16 + 25)/3 = 50/3 \approx 16.67 (verified), not 17.

    The near-miss is what makes this tempting — 50/3 rounds to 16.67, and 17 is what you get by rounding up or by mis-adding the squares to 51.

  • C.Correct

    True.

    E(M3)=(27+64+125)/3=216/3=72(E(M^{3}) = (27 + 64 + 125)/3 = 216/3 = 72 (verified).

    The cubes happen to sum to a multiple of 3

    which is why this one comes out as a whole number while the second moment does not.

  • D.WrongExecution slip

    False.

    Var(M)=E(M2)−(EM)2=50/3−16=2/3(Var(M) = E(M^{2}) - (E M)^{2} = 50/3 - 16 = 2/3 (verified), not 1.

    Note that Var(M) = 2/3 is much smaller than the population variance 8/3

    reflecting both the averaging and the finite population correction.

2 of these 2 wrong options correspond to a named reasoning error — the same errors recur across subjects.

Related counterexample: Systematic sampling is always at least as efficient as SRS

More on this topic

The chapter behind this: Sampling designs — free to read

From Sampling and Design of Experiments › SRS, stratified and systematic sampling

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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