ISS 2016 Statistics Paper-1 Solution: Question 50 (Choosing Between Two Fitted Models)

This post continues our ongoing question-by-question walkthrough of the ISS (Indian Statistical Service) Statistics Paper-1 previous year papers. Today we take up Question 50 from the 2016 paper, which tests a very practical idea from curve fitting: when you have two different models fitted to the same data, how do you decide which one to trust for a prediction?

Quick Summary

• Topic:Principle of least squares — choosing between two fitted models

• Question Reference:ISS 2016, Statistics Paper-1, Question 50

• Correct Answer:Option (c) — put X = X0 in the model for which the E value obtained is the least

The Question As Asked

For a given data (Xi, Yi); i = 1, 2, 3, ..., n; using the principle of least-squares, two models, say M1 and M2, are fitted. Let M1 be Y = f1(X) and let M2 be Y = f2(X). Suppose X0 is a value of X for which the predicted value of Y is to be determined.

Define E1 = Σ[Yi − f1(Xi)]2 and E2 = Σ[Yi − f2(Xi)]2. Then which one of the following is the criterion for choosing the estimated value of Y for X = X0?

• (a) Put X = X0 in the model for which the Y value obtained is the least.

• (b) Put X = X0 in the model for which the Y value obtained is the greatest.

• (c) Put X = X0 in the model for which the E value obtained is the least.

• (d) Put X = X0 in the model for which the E value obtained is the greatest.

Step 1: Understand What M1 and M2 Actually Are

We are told that both M1 and M2 are obtained by fitting curves to the same set of data points (Xi, Yi) using the principle of least squares. That principle says: given a chosen form of the curve (say, a straight line for M1 and a quadratic for M2, or any two different functional forms), the “best” curve of that form is the one whose parameters minimise the sum of squared vertical deviations between the observed Yi and the fitted value f(Xi).

So M1 is the best-fitting curve of its own type, and M2 is the best-fitting curve of its own (possibly different) type. Nothing in the question tells us the two models have the same functional form — they are simply two competing ways of describing the relationship between X and Y.

Step 2: What Do E1 and E2 Represent?

E1 = Σ[Yi − f1(Xi)]2 is the residual sum of squares for model M1 — it adds up, over every data point, the squared gap between what was actually observed (Yi) and what the model predicted (f1(Xi)). E2 is the exact same quantity computed for model M2.

A small E value means the curve passes very close to the observed data points overall — the model explains the data well. A large E value means the curve, on average, misses the observed points by a wide margin — the model fits the data poorly, even if it happens to have been “optimally” fitted within its own form.

Step 3: Why the Predicted Y Value at X0 Is the Wrong Thing to Compare

Options (a) and (b) tempt you to compare the predicted Y values — f1(X0) versus f2(X0) — and simply pick whichever is smaller or larger. But think about what that number actually tells you: nothing about accuracy. A model could predict a very small or very large Y at X0 purely because of the shape of the curve it belongs to, regardless of whether that curve actually tracks the real data well elsewhere. There is no reason at all that “smallest predicted value” or “largest predicted value” should correspond to “most trustworthy prediction.” These two options are essentially a distraction — they describe quantities with no connection to model accuracy.

Step 4: Why the Error Sum of Squares Is the Right Criterion

The entire logic of least-squares fitting is built around minimising the residual sum of squares as the measure of how well a model fits the data. If we are choosing between two already-fitted competing models to decide which one's prediction at X0 we should trust, the natural and statistically sound approach is to ask: which of the two models fits the overall data better? And “fits the data better” is measured precisely by a smaller E value.

So the correct rule is: compare E1 and E2. Whichever model has the smaller residual sum of squares is the better-fitting model overall, and that is the model whose predicted value at X = X0 should be used as the estimate for Y.

Pro Tip: Whenever a question gives you two competing fitted models along with their respective error sums (E1, E2), your first instinct should be to treat E as a scoreboard for goodness of fit, not the predicted Y values themselves. This is exactly the kind of conceptual trap — “does a bigger predicted number mean a better model?” — that aspirants are trained to catch instantly through repeated drilling on past papers, the way it's typically taught in a structured ISS coaching environment: always ask “which quantity actually measures fit quality?” before comparing numbers across two models.

Step 5: Final Answer and Verification Against the Official Key

Based on the reasoning above, the correct choice is: put X = X0 in the model for which the E value obtained is the least. This corresponds exactly to option (c).

Checking against the official ISS 2016 Statistics Paper-1 answer key (Series A booklet), Question 50 is listed as option (c)as well. Our independent derivation matches the official key — no discrepancy here.

Why This Question Matters

This question is not really about computation at all — there is no arithmetic to perform. It is a pure concept-check on model selection, which sits at the heart of applied statistics and data analysis. Students who only memorise the mechanics of fitting a least-squares line or curve often trip up the moment a question asks them to reason about comparing two already-fitted models, because the exam is now testing judgment rather than formula recall. Recognising that the residual (error) sum of squares — not the predicted value itself — is the correct yardstick for model adequacy is a habit that pays off repeatedly, both in curve-fitting questions and later in regression and model-selection topics such as adjusted R-squared, AIC, and BIC.

Frequently Asked Questions

Why can't we just pick the model that predicts the value closer to the trend of nearby data points?

Because the question gives you no information about “nearby data points” at X0 — it only gives you summary statistics over the whole dataset (E1 and E2). The whole point of using E as the criterion is that it summarises overall fit without needing you to inspect individual points near X0.

Does a smaller E value always mean a better model in every situation?

Within the scope of this question, yes — among two already-fitted models being compared on the same dataset, the one with the smaller residual sum of squares fits the observed data better. In more advanced courses you would also worry about overfitting and model complexity, but that refinement is outside what this MCQ is testing.

Is this the same idea as R-squared in regression?

Yes, conceptually. R-squared is derived directly from the residual sum of squares (it measures the proportion of total variation explained by the model), so a model with a smaller E will generally have a higher R-squared. The underlying logic — smaller residual error means a better-fitting model — is identical.

What if both E1 and E2 are equal?

If E1 = E2, both models fit the data equally well by this criterion, and the question as posed would not give you a unique way to prefer one over the other based on E alone — you would need additional criteria such as model simplicity (parsimony) to break the tie.

Why do the options even mention the predicted Y value at X0?

They are there as plausible-looking distractors. Exam setters often include an option built from a quantity that sounds relevant (the actual value you are trying to estimate) but is statistically meaningless for judging which model to trust. Learning to tell apart “the quantity you want to estimate” from “the quantity that tells you how trustworthy the estimate is” is exactly the skill this question is testing.

Is this topic likely to repeat in future ISS papers?

Model comparison via error sums of squares is a recurring theme across ISS Paper-1 — you will see it resurface in simple linear regression, curve fitting, and numerical analysis sections, so it's worth understanding the underlying logic rather than memorising just this one answer.

If anything in this derivation felt unclear, drop your doubt in the comments below — and if you found this walkthrough useful, do share it with a fellow ISS aspirant who is working through the same paper.