ISS 2016 Statistics Paper-2 Solution: Question 3 (Two-Way ANOVA F-Statistic)

Continuing our question-by-question walk through the ISS (Indian Statistical Service) Statistics Paper-2 previous year papers, we now arrive at Question 3 of the 2016 paper. This one sits squarely in the Design of Experiments portion of the syllabus, and it is a favourite examiner's trick because it looks like it needs three ANOVA tables when really it only needs one.

Quick Summary

• Topic: Two-Way Analysis of Variance (ANOVA) without replication

• Question Reference: ISS 2016, Statistics Paper-2, Question 3

• Correct Answer: Option (b) — F ≈ 1.24

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

Question:A certain task can be accomplished in a factory by four different workers on five different types of machines. A sample study in context of a two-way design without repeated value is being made with two fold objectives of examining whether the four workers differ with respect to mean productivity and whether it is same for five machines. The researcher involved in this study reports gathered data as under:

Sum of squares for variance between machines = 35.2 Sum of squares for variance between workers = 53.8 Sum of squares for total variance = 174.2

The F-statistic between machines is equal to:

• (a) 0.95

• (b) 1.24

• (c) 1.48

• (d) 1.75

Step 1: Identify the Design

The phrase "two-way design without repeated value" is the key signal here. It tells us we have two factors being studied at once — workers and machines — and each combination of (worker, machine) has been observed exactly once. No replication means there is no separate "interaction" sum of squares that can be isolated; whatever variation is not explained by workers or by machines automatically becomes the error (residual) variation.

There are 4 workers and 5 machines, so the layout is a 4 × 5 table with a total of N = 4 × 5 = 20 observations (one per cell).

Step 2: Work Out the Degrees of Freedom

In any ANOVA, before you can compute a mean square, you need the degrees of freedom (df) associated with each source of variation. The rule is simple: df = (number of levels of that factor) − 1.

• Total df = N − 1 = 20 − 1 =19

• df for machines (the factor of interest here) = number of machines − 1 = 5 − 1 =4

• df for workers = number of workers − 1 = 4 − 1 =3

In a two-way layout without replication, the total df splits cleanly into three non-overlapping parts: df(machines) + df(workers) + df(error) = df(total). So the error df is found by subtraction:

df(error) = df(total) − df(machines) − df(workers) = 19 − 4 − 3 =12

Step 3: Find the Missing Sum of Squares (Error SS)

Just as the degrees of freedom split additively, the sums of squares also split additively in this design:

SS(total) = SS(machines) + SS(workers) + SS(error)

We are given SS(total) = 174.2, SS(machines) = 35.2, and SS(workers) = 53.8. Rearranging for the one piece we don't have yet:

SS(error) = SS(total) − SS(machines) − SS(workers) = 174.2 − 35.2 − 53.8 =85.2

Pro Tip: The single most common mistake on this type of question is forgetting that SS(error) has to be backed out by subtraction — students who haven't drilled this kind of two-way table enough will instead try to compute the F-ratio directly from SS(machines) and SS(workers), or divide by the wrong "total" figure entirely. The way this is usually taught is to write out the full skeleton ANOVA table first — every source, its df, and a blank for its SS and MS —

Step 4: Convert Sums of Squares into Mean Squares

A "mean square" (MS) is nothing but the sum of squares divided by its own degrees of freedom — it converts a raw total of squared deviations into an average, so that sources with different numbers of levels can be fairly compared.

MS(machines) = SS(machines) ÷ df(machines) = 35.2 ÷ 4 =8.8

MS(error) = SS(error) ÷ df(error) = 85.2 ÷ 12 =7.1

(We don't actually need MS(workers) for this particular question, since we're only asked for the F-statistic "between machines," but for completeness: MS(workers) = 53.8 ÷ 3 ≈ 17.93.)

Step 5: Compute the F-Statistic for Machines

The F-statistic for any factor in ANOVA is always the ratio of that factor's mean square to the error mean square — it measures how much variation the factor explains relative to the "background noise" that no factor explains.

F(machines) = MS(machines) ÷ MS(error) = 8.8 ÷ 7.1 = 1.2394…

Rounded to two decimal places, F(machines) ≈1.24, which is option (b).

This matches the official answer key, which also records the answer to Question 3 as (b). Our independently derived value and the key are in full agreement.

Why This Question Matters

Two-way ANOVA without replication is a recurring theme across ISS Statistics Paper-2 — it shows up not just as direct numerical questions like this one, but also folded into questions on randomized block designs, Latin square designs, and general linear model theory. The core skill being tested isn't really arithmetic; it's whether you can correctly partition total variation into its additive components using only the degrees-of-freedom logic, even when one of the sums of squares is deliberately withheld so you have to derive it. Get comfortable with that skeleton-table approach and a whole family of design-of-experiments questions becomes routine rather than intimidating.

Final Answer

The F-statistic between machines is 1.24, confirming option (b)as correct — consistent with the official ISS 2016 Statistics Paper-2 answer key.

Frequently Asked Questions

Why is there no separate "interaction" term in this ANOVA?

Interaction can only be estimated when a design has repeated observations at each combination of factor levels. Since this is explicitly a "design without repeated value" (one observation per worker–machine cell), there is no way to separate interaction from pure random error, so both get lumped together into a single error term.

How do I know which degrees of freedom to subtract from which?

Always start from the total df, which is always (total number of observations − 1). Then subtract the df of every named factor in the design. Whatever remains is the error df. This additive relationship holds for any balanced design without replication, not just this specific worker–machine example.

Could I have found the F-statistic for workers using the same data?

Yes. You would use MS(workers) = 53.8 ÷ 3 ≈ 17.93 in the numerator, and the same MS(error) = 7.1 in the denominator, giving F(workers) ≈ 2.53. The question here only asks about machines, but the same error mean square is used as the denominator for testing either factor.

What does an F-value close to 1 actually tell us?

An F-value near 1 suggests that the variation attributed to that factor (machines, in this case) is roughly the same size as the random error variation, meaning there is no strong statistical evidence that machines significantly affect productivity. A much larger F-value, well beyond the relevant critical value from F-tables, would be needed to reject the null hypothesis of no machine effect.

Is the order of subtraction in SS(error) = SS(total) − SS(machines) − SS(workers) important to remember exactly?

The order of subtraction doesn't matter mathematically since it's just addition rearranged, but what matters is remembering that all three quantities are linked by one additive identity. As long as you know SS(total) and any two of the remaining three sources, you can always find the third by subtraction.

Where can I find similarly styled solved questions for practice?

This solution is part of an ongoing, question-by-question series covering every ISS Statistics Paper-2 question from 2016 onward, published here on Sunrise Classes' blog, so working through the earlier and later posts in the series in order is a good way to build the same design-of-experiments intuition systematically.

If anything in this derivation feels unclear, or if you spot a step you'd solve differently, drop a comment below — working through the disagreement is often the fastest way to actually learn the concept. And if this helped, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.