ISS 2016 Statistics Paper-2 Solution: Question 10 (Testing β₁ = 0 in Regression)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, we now arrive at the tenth item of the 2016 paper. If you have been following along, you already know the drill: no answer is accepted on faith here, every result is derived from scratch so that you can reproduce the logic on any similar question in the exam.

Quick Summary

• Topic:Simple linear regression — distributional results needed to test H0 : β1 = 0

• Question Reference:ISS 2016, Statistics Paper-2, Question 10

• Correct Answer:Option (a) — Statements 1, 2 and 3

Here is the question exactly as it appears in the test booklet:

Q10.Which of the following statements are correct in order to obtain the hypothesis H0 : β1 = 0 for the model yi = β0 + β1xi + εi ?

1. yi is N(β0 + β1xi, σ²)

2. β̂1 is N(β1, σ²/Σ(xi − x̄)²)

3. (n − 2)σ̂²/σ² is χ²n−2

4. β̂1 and σ̂² are not independent

Select the correct answer using the code given below :

• (a) 1, 2 and 3

• (b) 1 and 2 only

• (c) 1, 2 and 4

• (d) 3 and 4

Step 1: Setting Up the Model and Its Assumptions

Before checking any of the four statements, write down exactly what the simple linear regression model assumes. We have n pairs of observations (x1, y1), (x2, y2), …, (xn, yn), and the model says

yi = β0 + β1xi + εi,   i = 1, 2, …, n

with the classical Gauss-Markov-plus-normality assumptions: the errors ε1, ε2, …, εn are independent, each has mean 0 and the same variance σ², and — this is the extra piece needed for hypothesis testing — each εi is normally distributed. In short, εi ~ N(0, σ²) independently across i. Everything in this question follows from applying basic normal-distribution facts to this one assumption.

Step 2: Checking Statement 1 — The Distribution of yi

Look at the model equation again: yi = β0 + β1xi + εi. For a fixed xi (in regression, the x-values are treated as fixed, known constants, not random), the quantity β0 + β1xi is just a constant number. So yi is literally "a constant plus a normal random variable."

A basic property of the normal distribution is that adding a constant to a normal random variable shifts its mean by that constant and leaves the variance unchanged. Since εi ~ N(0, σ²), we get

yi = (constant) + εi  ⟹  yi ~ N(β0 + β1xi, σ²)

which is exactly Statement 1. So Statement 1 is TRUE.

Step 3: Checking Statement 2 — The Distribution of β̂1

Recall the ordinary least squares (OLS) formula for the slope estimator:

β̂1 = Σ(xi − x̄)(yi − ȳ) / Σ(xi − x̄)²

It's a standard algebraic exercise (worth doing once on paper so you never forget it) to rewrite the numerator so that β̂1 becomes a weighted sum of the yi's alone:

β̂1 = Σ ciyi,   where ci = (xi − x̄) / Sxx, and Sxx = Σ(xi − x̄)²

This matters because a weighted sum of independent normal random variables is itself normal. Since each yi is normal (Step 2) and the yi's are independent (because the εi's are independent), β̂1 — being a linear combination Σciyi — must also be normal.

Now work out its mean and variance:

Mean:E(β̂1) = β1 (this is the standard unbiasedness result for OLS — you can verify it by substituting E(yi) = β0 + β1xi into ΣciE(yi) and simplifying, using the facts Σci = 0 and Σcixi = 1).

Variance:Since the yi's are independent, Var(Σciyi) = Σci²Var(yi) = σ²Σci². Working out Σci² = Σ(xi−x̄)²/Sxx² = Sxx/Sxx² = 1/Sxx, so Var(β̂1) = σ²/Sxx = σ²/Σ(xi − x̄)².

Putting the mean and variance together: β̂1 ~ N(β1, σ²/Σ(xi − x̄)²), which is exactly Statement 2. So Statement 2 is TRUE.

Step 4: Checking Statement 3 — The Chi-Square Result for σ̂²

Let SSE = Σ(yi − ŷi)² be the residual sum of squares, and recall that the unbiased estimator of σ² in simple linear regression is σ̂² = SSE/(n − 2). The divisor is n − 2 (not n) because two parameters, β0 and β1, were estimated from the data before the residuals could be computed — each estimated parameter "uses up" one degree of freedom.

A classical result (proved using the fact that SSE is a quadratic form in the yi's built from a projection matrix of rank n − 2) states that, under normality of the errors,

SSE / σ² ~ χ²n−2

Since σ̂² = SSE/(n − 2), multiplying both sides by (n − 2) gives (n − 2)σ̂² = SSE, so

(n − 2)σ̂² / σ² = SSE / σ² ~ χ²n−2

which is exactly Statement 3. So Statement 3 is TRUE.

Step 5: Checking Statement 4 — Independence of β̂1 and σ̂²

This is the statement most candidates get tripped up on, because it is phrased as a negative ("are not independent") and because the underlying reason is a slightly deeper geometric fact about normal random vectors rather than a plug-and-chug formula.

Pro Tip: whenever an ISS question splits a regression estimate (like β̂1) from a residual-based quantity (like σ̂² or SSE), your first instinct should be to check whether they come from orthogonal projections of the same normal vector y. If they do, they are independent — full stop, no further calculation needed. This "projection first, algebra later" habit is exactly how independence questions in the linear model are usually taught, and once it becomes automatic you'll answer this entire family of questions in seconds rather than minutes.

Here is the geometric picture. The fitted values ŷi live in the column space of the design matrix (a 2-dimensional subspace spanned by the constant vector and the x-vector), while the residuals ei = yi − ŷi live in the orthogonal complement of that subspace. Both β̂1 (which is a function of the fitted-value subspace) and σ̂² (which is built entirely from the residuals, i.e., the orthogonal subspace) are functions of the same underlying multivariate normal vector y. A key theorem about multivariate normal vectors says: if you project a normal vector onto two orthogonal subspaces, the two projections are statistically independent. Applying that here, β̂1 and σ̂² (equivalently, β̂1 and SSE) are independent of each other.

This is the exact regression analogue of a fact you may already know from basic statistics: for an iid normal sample, the sample mean and the sample variance are independent. The regression version simply replaces "sample mean" with β̂1 (from the fitted-value subspace) and "sample variance" with σ̂² (from the residual subspace).

Since β̂1 and σ̂² actually ARE independent, the claim in Statement 4 — that they are "not independent" — is false. So Statement 4 is FALSE.

Step 6: Why These Three Facts Matter for the Hypothesis Test

It's worth pausing to see why the question links these particular three statements to testing H0 : β1 = 0. To test this hypothesis you build a t-statistic:

t = (β̂1 − β1) / √(σ̂²/Sxx)

The numerator is standard normal after dividing by its true standard deviation (this comes from Statement 2). The denominator is built from σ̂², whose scaled version is chi-square (Statement 3). And the reason the ratio of "standard normal" to "square root of chi-square over its degrees of freedom" follows a clean tn−2 distribution is precisely because the numerator and denominator are independent — which is guaranteed by the fact that Statement 4, as worded, is false. Remove the independence and the neat t-distribution result would simply not hold. So Statements 1, 2 and 3 are the three ingredients you actually need; Statement 4's claim, if it were true, would break the whole construction.

Final Answer

Statements 1, 2 and 3 are correct; Statement 4 is incorrect. That matches option (a) — 1, 2 and 3, which is also the answer given in the official answer key (Series A, Question 10 → A). Our independent derivation and the official key agree.

Why This Question Matters

This question is really testing whether you understand *where* the t-test for a regression coefficient comes from, rather than whether you can recite the formula t = β̂1/SE(β̂1) from memory. In the actual exam, questions are frequently built exactly this way — by taking a standard, memorised result and breaking it into its component assumptions, then asking you to identify which pieces are genuinely necessary. Students who repeatedly work through this kind of decomposition — normality of y, distribution of the estimator, distribution of the error variance, and independence between the two — tend to find that the entire chapter on inference in linear regression stops feeling like a list of formulas and starts feeling like one connected argument.

Frequently Asked Questions

Why is the divisor (n − 2) used for σ̂² instead of n or (n − 1)?

Because two parameters, β0 and β1, are estimated from the data before the residuals can be computed. Each estimated parameter removes one degree of freedom from the n original observations, leaving n − 2 degrees of freedom for estimating the error variance. This is exactly analogous to using (n − 1) in the one-sample case, where only the mean is estimated.

Is β̂1 and σ̂² independence a special feature of simple linear regression, or does it hold more generally?

It holds for the general multiple linear regression model as well, as long as the errors are independent and normally distributed. The regression coefficient estimates always live in the column space of the design matrix, and σ̂² is always built from the residuals in the orthogonal complement, so the same orthogonal-projection argument applies regardless of how many predictors you have.

What would change if the errors were not assumed normal?

Without normality, β̂1 would still be unbiased with the same variance formula (by the Gauss-Markov theorem, which does not require normality), but you could no longer claim it is exactly normally distributed, and the exact chi-square and t-distribution results in Statements 3 and the hypothesis test would no longer hold exactly — you would instead need to rely on large-sample approximations.

How is this different from testing significance of the intercept β0?

The logic is identical — you would just replace β̂1 with β̂0 and use its own variance formula, σ²[1/n + x̄²/Sxx], in place of σ²/Sxx. The independence of β̂0 from σ̂² holds for exactly the same orthogonal-projection reason.

Why does the question say "in order to obtain the hypothesis" rather than simply "which statements are true"?

The phrasing is a hint about the purpose of the question: it wants you to identify the minimal correct set of distributional facts that are actually used in deriving the test, not just any true statement about the model. That's why a false statement like Statement 4 is deliberately included as a distractor — to check whether you understand the role each fact plays, not just whether you've memorised isolated results.

Where can I find worked examples with actual numbers for this kind of hypothesis test?

Any standard regression chapter that covers the t-test for a single coefficient will have numerical examples — the key is to always identify SSE, compute σ̂² = SSE/(n−2), compute Sxx, and then plug into t = (β̂1 − 0)/√(σ̂²/Sxx) before comparing against the tn−2 critical value.

If anything in this derivation still feels unclear, drop your doubt in the comments below — working through exactly where you got stuck is often the fastest way to fix it. And if you know a fellow ISS aspirant who's grinding through the same paper, sharing this with them costs you nothing and might save them the exact confusion this question is designed to create.