ISS 2016 Statistics Paper-1 Solution: Question 31 (Chi-Square Distribution of Sample Variance)

This post continues our question-by-question walkthrough of the ISS (Indian Statistical Service) Statistics Paper-1 previous year papers. Today we pick up right where the last post left off and tackle Question 31 from the 2016 paper, which tests one of the most fundamental results in sampling theory: the distribution of the sample variance drawn from a normal population.

Quick Summary

• Topic: Sampling distribution of the sample variance (Chi-square distribution)

• Question Reference: ISS 2016, Statistics Paper-1, Question 31

• Correct Answer: Option (c) — χ² with (n − 1) degrees of freedom

The Question, As Asked

If X̄ and S² are the mean and variance of a random sample of size n from a normal population with mean μ and standard deviation σ, then the random variable

Y = (n − 1)S² / σ²

follows

• (a) χ² with (n − 2) d.f.

• (b) t with (n − 1) d.f.

• (c) χ² with (n − 1) d.f.

• (d) t with (n − 2) d.f.

Step 1: Write Down What We Are Given

We are told X1, X2, ..., Xn is a random sample of size n drawn from a Normal population with mean μ and standard deviation σ. In symbols, each Xi is independently and identically distributed as N(μ, σ²).

From this sample we compute two quantities:

• The sample mean: X̄ = (1/n) ∑ Xi

• The sample variance: S² = 1/(n − 1) ∑ (Xi − X̄)²

This is the standard (unbiased) definition of sample variance used throughout the ISS syllabus, where the divisor is (n − 1) rather than n. Keep this definition firmly in mind — it is the entire key to answering this question correctly.

Step 2: Rewrite Y In Terms Of The Raw Sum Of Squares

We are asked about the distribution of Y = (n − 1)S² / σ². Let’s substitute the formula for S² directly into this expression.

Since S² = 1/(n − 1) ∑ (Xi − X̄)², multiplying both sides by (n − 1) gives us:

(n − 1)S² = ∑ (Xi − X̄)²

So our random variable Y simplifies beautifully to:

Y = ∑ (Xi − X̄)² / σ²

This is just the sum of squared deviations of each observation from the sample mean, scaled by the population variance σ². Notice that the messy (n − 1) factor has completely cancelled out — it was only ever there because of how S² itself is defined.

Step 3: Recall Why A Sum Of Squared Standard Normals Is Chi-Square

Before finishing, let’s remind ourselves of the basic building block this whole result rests on. If Z1, Z2, ..., Zk are k independent standard normal variables (each N(0,1)), then by definition:

Z1² + Z2² + ... + Zk² follows a χ² distribution with k degrees of freedom.

If we naively standardised each observation using the true population mean μ instead of the sample mean X̄, we would get exactly n independent standard normal variables: Zi = (Xi − μ)/σ. Squaring and summing all n of them, ∑ Zi² = ∑ (Xi − μ)²/σ², would follow a χ² distribution with n degrees of freedom — not n − 1. This is the trap many students fall into on first reading this question: they assume the answer must involve n degrees of freedom because there are n observations.

Step 4: Why We Lose Exactly One Degree Of Freedom

But our expression from Step 2 uses (Xi − X̄), the deviation from the sample mean, not (Xi − μ), the deviation from the true population mean. This distinction matters enormously.

The n deviations (X1 − X̄), (X2 − X̄), ..., (Xn − X̄) are not free to vary independently. They must always satisfy one linear constraint:

∑ (Xi − X̄) = 0

This is simply because X̄ is, by construction, the average of the Xi’s, so the deviations from it must sum to zero. Once you know any (n − 1) of these deviations, the last one is completely determined — it has no freedom left. So out of the n squared terms in ∑ (Xi − X̄)², only (n − 1) of them carry independent information.

Pro Tip: Whenever you see a chi-square question involving deviations from a

This is a direct consequence of a deeper result called Cochran’s theorem, which formally proves that ∑ (Xi − X̄)²/σ² follows exactly a χ² distribution with (n − 1) degrees of freedom when the Xi are i.i.d. N(μ, σ²), and further, that this quantity is statistically independent of X̄ itself. For the purposes of this MCQ, it is enough to recognise the pattern and the one-degree-of-freedom loss.

Step 5: State The Final Answer

Putting it all together:

Y = (n − 1)S² / σ² = ∑ (Xi − X̄)² / σ², which follows a χ² distribution with (n − 1) degrees of freedom.

This matches option (c) χ² with (n − 1) d.f., which is also exactly what the official ISS 2016 answer key confirms for Question 31. Our independent derivation and the official key are in full agreement — no discrepancy here.

Why This Question Matters

This result is not just an isolated MCQ fact — it is the theoretical backbone behind confidence intervals and hypothesis tests for a population variance, and it also underlies the derivation of the t-distribution (which is built from a standard normal divided by the square root of an independent chi-square variable, scaled by its degrees of freedom). If you understand why the degrees of freedom drop from n to (n − 1) when you replace μ with X̄, you will find a whole cluster of related ISS questions — on t-distributions, F-distributions, and interval estimation for σ² — become far easier to reason through rather than memorise.

Frequently Asked Questions

Why is the sample variance defined with a divisor of (n − 1) instead of n?

Dividing by (n − 1) rather than n makes S² an unbiased estimator of the population variance σ². If we divided by n instead, the estimator would be biased downward on average, because the sample mean X̄ is always at least as close to the data points as the true (unknown) population mean μ would be.

What exactly are degrees of freedom in this context?

Degrees of freedom count how many values in a set of numbers are free to vary once certain constraints are imposed. Here, the n deviations (Xi − X̄) must sum to zero, so only (n − 1) of them are independently free, which is why the resulting chi-square distribution has (n − 1) degrees of freedom rather than n.

How is this result different if we used the true population mean μ instead of X̄?

If you compute ∑ (Xi − μ)²/σ² using the true population mean μ, there is no constraint linking the deviations, so all n of them are independent, and the sum follows a χ² distribution with n degrees of freedom instead of (n − 1).

Is S² independent of X̄ in this setup?

Yes, and this is a remarkable and important property of sampling from a normal population: when the population is normal, the sample mean X̄ and the sample variance S² are statistically independent of each other. This independence is what makes the t-distribution derivation possible.

Where else does this χ² with (n − 1) d.f. result get used in the ISS syllabus?

It is directly used to construct confidence intervals and hypothesis tests for a single population variance σ², and it is one of the two building blocks (along with an independent standard normal) used to derive the Student’s t-distribution for inference about the population mean when σ is unknown.

Could option (a), χ² with (n − 2) d.f., ever be correct for a similar-looking question?

Yes — a (n − 2) degrees of freedom result typically shows up in simple linear regression, where the residual sum of squares divided by σ² follows a χ² distribution with (n − 2) degrees of freedom, because two parameters (the slope and intercept) are estimated from the data instead of just one mean. It’s a different scenario from this question, but the same principle of degrees of freedom lost is at play.

Have a doubt about this derivation, or spotted a step you’d like explained differently? Drop a comment below, and feel free to share this with a fellow ISS aspirant who might be revising sampling distributions this week.