ISS 2016 Statistics Paper-2 Solution: Question 50 (Standard Error and Precision)

This post continues our ongoing question-by-question walkthrough of the ISS (Indian Statistical Service) Statistics Paper-2 previous year papers. Today we're working through the 2016 paper, and we land on a question that trips up a surprising number of otherwise well-prepared students: the relationship between standard error and the precision of an estimator.

Quick Summary

• Topic:Sampling distribution of the mean, standard error, and estimator precision

• Question Reference:ISS 2016, Statistics Paper-2, Question 50

• Correct Answer:Option (d) — As the standard error of mean decreases, the precision of the sample mean as an estimator of the population mean increases

The Question

Which one of the following statements is correct?

• (a) The mean of the sampling distribution of mean is greater than the population mean.

• (b) The sampling distribution of mean approaches normality as the sample size decreases.

• (c) As the standard error of mean increases, the value of any sample mean will be closer to the value of the population mean.

• (d) As the standard error of mean decreases, the precision of the sample mean as an estimator of the population mean increases.

Step 1: Set Up the Sampling Distribution of the Mean

Suppose we draw a random sample X1, X2, ..., Xn of size n from a population with mean μ and variance σ². The sample mean is defined as:

X̄ = (1/n) Σ Xi

If we imagine repeating this sampling process over and over — drawing many different samples of size n and computing X̄ each time — we get a whole distribution of possible sample mean values. This is called the sampling distribution of the mean, and it is the central idea this question is testing.

Step 2: Find the Mean of the Sampling Distribution

We compute the expected value of X̄ using linearity of expectation. Since each Xi has expectation μ:

E(X̄) = E[(1/n) Σ Xi] = (1/n) Σ E(Xi) = (1/n)(nμ) = μ

So the mean of the sampling distribution of the sample mean is exactly equal to the population mean μ — not greater than it. This is simply the statement that the sample mean is an unbiased estimator of μ. So option (a) is false: there is no general reason for the sampling mean to exceed μ; it equals μ on average, by construction.

Step 3: Find the Variance and Standard Error

Assuming the Xi are independent (as in simple random sampling), we compute the variance of X̄:

Var(X̄) = Var[(1/n) Σ Xi] = (1/n²) Σ Var(Xi) = (1/n²)(nσ²) = σ²/n

Taking the square root gives the standard error (SE)of the mean:

SE(X̄) = σ/√n

Notice immediately that SE is a decreasing function of the sample size n: as n grows, √n grows, and σ/√n shrinks. This single formula is the key to resolving every option in this question.

Step 4: Check Option (b) — Normality and Sample Size

Option (b) claims the sampling distribution of the mean approaches normality as sample size decreases. This is backwards. The Central Limit Theorem tells us the opposite: as n increases, the sampling distribution of X̄ approaches a Normal distribution, regardless of the shape of the parent population (under mild regularity conditions). A small sample size gives a sampling distribution that may be far from normal, especially if the population itself is skewed. So option (b) is false — it has the direction of the relationship reversed.

Step 5: Check Option (c) — Standard Error and Closeness to μ

Option (c) says that as SE increases, any sample mean gets closer to the population mean. Since SE = σ/√n measures the spread (standard deviation) of the sampling distribution of X̄ around μ, a larger SE means the sample means are more spread out around μ — individual sample means are, on average, farther from μ, not closer. So option (c) states the exact opposite of what standard error actually means, and is false.

Pro Tip: The trickiest part of this question is not the formula — it's resisting the pull of option (c), which sounds intuitive if you read "standard error" loosely as just "some error term that goes with being wrong." Always anchor standard error to its precise meaning: it is the standard deviation of the sampling distribution of an estimator. A bigger standard deviation always means more spread, never less. The way this distinction is usually drilled into students is by forcing them to rederive SE = σ/√n from scratch every time it appears in a question, rather than memorizing the formula in isolation — that habit of rebuilding it from Var(X̄) is what makes options like (c) obviously wrong on sight instead of a coin flip.

Step 6: Confirm Option (d) and the Final Answer

Precision of an estimator refers to how tightly the estimator's sampling distribution is clustered around the true parameter — in other words, precision is inversely related to the variance (or standard error) of the estimator. Since SE(X̄) = σ/√n decreases as n increases, a smaller SE directly means the sample means are more tightly clustered around μ, which is exactly what we mean by higher precision.

So option (d) — "As the standard error of mean decreases, the precision of the sample mean as an estimator of the population mean increases" — is a direct, correct restatement of the inverse relationship between standard error and precision.

The correct answer is (d), and this matches the official ISS 2016 Statistics Paper-2 answer key (Series A), which lists option D for Question 50.

Why This Question Matters

This question is a favourite in ISS, IIT JAM, and GATE Statistics papers because it tests conceptual understanding rather than computation. Many candidates can recite "SE = σ/√n" but stumble when asked to reason about what increasing or decreasing SE actually implies about an estimator's behaviour. Questions like this are designed to catch students who have memorized formulas without internalizing what standard error, bias, and precision actually measure — a gap that shows up again and again across hypothesis testing, confidence intervals, and estimation theory questions in Paper-2.

Frequently Asked Questions

What exactly is the difference between standard deviation and standard error?

Standard deviation (σ) measures the spread of individual observations in the population. Standard error measures the spread of a statistic's sampling distribution, such as SE(X̄) = σ/√n for the sample mean. Standard error is always about a statistic computed from a sample, not about individual data points.

Why does standard error decrease as sample size increases?

Because SE(X̄) = σ/√n, and √n grows as n grows while σ (the population standard deviation) stays fixed. Larger samples average out random fluctuations more effectively, so the sample mean varies less from sample to sample, shrinking the standard error.

Is the sample mean always an unbiased estimator of the population mean?

Yes, for simple random sampling with replacement (or from an infinite population), E(X̄) = μ always holds regardless of sample size, as shown by the linearity-of-expectation derivation above. This unbiasedness does not depend on the sample size being large.

How is precision different from accuracy in this context?

Precision refers to how tightly an estimator's possible values cluster together (low variance/SE), while accuracy (closely tied to bias) refers to whether the estimator is centered on the true parameter value. An estimator can be precise but inaccurate (biased with low variance) or accurate but imprecise (unbiased with high variance); the ideal estimator is both.

Does the Central Limit Theorem require the population to be normal?

No. The Central Limit Theorem states that the sampling distribution of X̄ approaches normality as n increases, regardless of the population's original distribution shape, provided the population has finite variance. This is precisely why option (b) in this question, which ties normality to decreasing sample size, is incorrect.

Why is this type of "which statement is correct" question harder than a direct numerical question?

Because each option sounds plausible unless you check it against the exact formula and definition. There's no shortcut besides deriving E(X̄) and Var(X̄) from first principles each time, which is why this derivation-first approach is worth practicing until it becomes automatic rather than something to re-derive under exam pressure.

If anything in this derivation felt unclear, or you'd like another previous year question solved in the comments, drop a comment below — and if this helped, consider sharing it with a fellow ISS aspirant who might be stuck on the same concept.