ISS 2016 Statistics Paper-1 Solution: Question 27 (Chi-Square vs t-Distribution Relations)
Continuing our question-by-question walkthrough of the ISS Statistics Paper-1 archive, this post picks up the 2016 paper at Question 27 — a short conceptual check that quietly tests whether you really understand where the chi-square distribution and Student's t-distribution come from, rather than just memorising their formulas.
Quick Summary
Topic: Sampling distributions of the sample mean — deriving chi-square and t statistics from a normal population
Question Reference: ISS 2016, Statistics Paper-1, Question 27
Correct Answer: (b) 2 only
The Question As Asked
Consider the following relations (with usual notations) where X ~ N(μ, σ²):
χ2(2) = (X̄ − μ)2 / σ2
t = (X̄ − μ)√n / s
Which of the above is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Step 1: Recall How a Single Normal Variable Standardises
We are told X ~ N(μ, σ2). The very first tool you reach for with any normal variable is standardisation: subtract the mean, divide by the standard deviation.
If X ~ N(μ, σ2), then Z = (X − μ)/σ ~ N(0, 1).
This is the building block for everything else in this question, so it is worth being completely comfortable with it before moving on.
Step 2: Find the Distribution of the Sample Mean X̄
Relation 1 uses X̄, the sample mean of n independent observations drawn from N(μ, σ2), not a single observation X. This distinction matters enormously, and it is exactly where this question is testing you.
For n independent and identically distributed observations X1, X2, …, Xn from N(μ, σ2):
E(X̄) = μ (the sample mean is an unbiased estimator of μ)
Var(X̄) = σ2/n (variance shrinks as the sample size grows, because averaging cancels out random noise)
Since a linear combination of independent normal variables is itself normal, we get:
X̄ ~ N(μ, σ2/n)
Step 3: Standardise X̄ Correctly
Applying the same standardisation rule from Step 1, but now using the standard deviation of X̄ itself, which is σ/√n (not σ):
Z = (X̄ − μ) / (σ/√n) = (X̄ − μ)√n / σ ~ N(0, 1)
Notice the crucial √n term. It appears because we are standardising the sample mean, whose spread is smaller than that of a single observation by a factor of √n.
Step 4: Square It to Get a Chi-Square Variable
A standard result you should have memorised cold for this exam: if Z ~ N(0, 1), then Z2 follows a chi-square distribution with 1 degree of freedom, written χ2(1). More generally, the sum of squares of k independent standard normal variables follows χ2(k).
Squaring the result from Step 3:
Z2 = [(X̄ − μ)√n / σ]2 = n(X̄ − μ)2 / σ2 ~ χ2(1)
Now compare this to what relation 1 in the question actually claims: (X̄ − μ)2/σ2 ~ χ2(2). Two things are wrong with it:
The correct expression needs a factor of n multiplying (X̄ − μ)2/σ2; the version in the question is missing it entirely.
Even setting that aside, a single squared standard normal always gives 1 degree of freedom, never 2.
Pro Tip: Whenever you see a chi-square claim built from a sample mean, do a two-second sanity check before trusting the degrees of freedom: count how many independent standard normal variables are being squared and added. One squared normal is always
So relation 1 is false.
Step 5: Check Relation 2 — Where Does the t-Distribution Come From?
In Step 3 we standardised X̄ using the true population standard deviation σ. In practice, σ is almost never known — we only have the data, from which we estimate it using the sample standard deviation s, defined by:
s2 = (1/(n−1)) Σ (Xi − X̄)2
Replacing σ with the estimate s in the standardised formula gives exactly the statistic in relation 2:
t = (X̄ − μ)√n / s
Because s is itself a random quantity computed from the same sample (not a fixed constant like σ), this ratio is no longer exactly standard normal. Instead, William Gosset showed (publishing as "Student") that it follows a new distribution, the t-distribution, with n − 1 degrees of freedom:
t = (X̄ − μ)√n / s ~ t(n−1)
This result rests on two classical facts from normal-sample theory: X̄ and s2 are statistically independent when sampling from a normal population, and (n−1)s2/σ2 ~ χ2(n−1). Dividing the standard normal numerator by the square root of an independent chi-square (scaled by its degrees of freedom) is precisely the definition of a t-distributed variable. So relation 2 is true exactly as written — no missing terms, no wrong subscripts.
Step 6: Final Answer
Relation 1 is false (wrong scaling and wrong degrees of freedom), and relation 2 is true (the textbook one-sample t-statistic). That means only relation 2 is correct.
Correct option:(b) 2 only.
This matches the official ISS 2016 Statistics Paper-1 answer key, which lists B for Question 27 — so our independently derived answer and the official key are in full agreement.
Why This Question Matters
This question is a favourite type in the ISS exam because it doesn't ask you to compute anything — it asks whether you actually understand why the chi-square and t-distributions look the way they do, rather than treating them as formulas to plug numbers into. The moment you can derive Var(X̄) = σ2/n and connect it to "squared standard normal = chi-square with 1 df" on your own, questions like this stop being memory tests and start being two-line derivations you can knock out under time pressure.
Frequently Asked Questions
What exactly is the difference between the chi-square distribution and the t-distribution here?
The chi-square distribution in this question comes from squaring a single standard normal variable — it has no unknown parameters left to estimate, since σ is assumed known. The t-distribution appears specifically because we replaced the unknown σ with its sample estimate s, introducing extra randomness that fattens the tails compared to a standard normal.
Why does Var(X̄) equal σ²/n and not just σ²?
X̄ is an average of n independent observations, and for independent variables, variances add: Var(X̄) = Var((X1+...+Xn)/n) = (1/n2)(nσ2) = σ2/n. Averaging cancels out random fluctuations, so the sample mean is always less variable than a single observation, and more so as n grows.
Why do we divide by n − 1, not n, when computing s² for the t-statistic?
Dividing by n − 1 makes s2 an unbiased estimator of σ2, correcting for the fact that X̄ itself is estimated from the same data, which uses up one "degree of freedom." This is also exactly why the resulting t-statistic carries n − 1 degrees of freedom rather than n.
Is (X̄ − μ)²/σ² always chi-square with 1 degree of freedom?
Not as written in relation 1 — it is missing the factor of n. The correct statement is n(X̄ − μ)2/σ2 ~ χ2(1). Dropping the n changes the scale of the distribution entirely, so the "as is" expression in the question does not follow any standard chi-square distribution.
Why are X̄ and s² required to be independent for the t-distribution derivation to work?
The formal definition of a t-distributed variable is a standard normal divided by the square root of an independent chi-square variable (scaled by its degrees of freedom). If X̄ and s2 were not independent, the ratio (X̄ − μ)√n/s would not follow a clean t-distribution at all — this independence is a special property of sampling from a normal population.
How do I quickly decide between a z-test and a t-test in exam questions like this?
If the population standard deviation σ is given or assumed known, standardise using σ and you get a normal (z) statistic. The moment σ is unknown and replaced by the sample standard deviation s, the statistic becomes t-distributed with n − 1 degrees of freedom instead of standard normal.
If any step above felt unclear, drop a comment below and we'll walk through it further — and if this helped, consider sharing it with a fellow ISS aspirant working through the same paper.
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