ISS 2016 Statistics Paper-2 Solution: Question 13 (Within-Groups Variance in ANOVA)

Continuing our question-by-question walkthrough of the ISS Statistics Paper-2 archive, we're now up to Question 13 from the 2016 paper. This one sits inside a linked group of three questions built around a single dataset, so before we touch the arithmetic, we'll make sure the shared setup is completely clear.

Quick Summary

• Topic:One-Way ANOVA — estimating the within-groups (error) population variance

• Question:ISS 2016, Statistics Paper-2, Question 13

• Correct Answer:(d) 39.097

The Question

For the next three (03) items that follow:A company used three different methods to train its employees. The number of units of output produced by different employees trained by the three training methods is given below:

• Method A: 50, 45, 55, 44

• Method B: 64, 48, 52, 56, 44

• Method C: 46, 42, 48, 45, 57, 42

It has to be tested whether the three different methods are satisfactory at 5% level of significance (Given F2,12 at 0.05 = 3.89 as the critical value).

Question 13:The estimate of the population variance within the samples for the different methods is:

• (a) 25.67

• (b) 31.07

• (c) 38.646

• (d) 39.097

Step 1: Understand What "Variance Within Samples" Means

In one-way ANOVA, the total variation in the data is split into two independent parts:

• Between-groups variation — how much the group means differ from each other (this is what Question 12 in this same set asks about).

• Within-groups variation — how much the individual observations inside each group scatter around their own group's mean.

The within-groups estimate of the population variance is nothing but the pooled Mean Square Within (MSW), also called the error mean square. It tells us how noisy the data is inside each training method, independent of whether the methods themselves differ. We calculate it by adding up the sum of squared deviations of each observation from its own group's mean — across all three groups — and then dividing by the within-groups degrees of freedom.

Step 2: Find Each Group's Mean

First, add up each group and divide by how many employees are in it.

Method A (n = 4): 50 + 45 + 55 + 44 = 194, so mean = 194 ÷ 4 =48.5

Method B (n = 5): 64 + 48 + 52 + 56 + 44 = 264, so mean = 264 ÷ 5 =52.8

Method C (n = 6): 46 + 42 + 48 + 45 + 57 + 42 = 280, so mean = 280 ÷ 6 =46.667 (approximately)

Step 3: Find the Sum of Squares Within Each Group

For each group, subtract the group's own mean from every observation in that group, square the result, and add up all the squares. A quick shortcut that avoids fraction-heavy subtraction is the identity:

Sum of squared deviations = Σx² − (Σx)² ÷ n

Method A:

Σx² = 50² + 45² + 55² + 44² = 2500 + 2025 + 3025 + 1936 = 9486

(Σx)² ÷ n = 194² ÷ 4 = 37636 ÷ 4 = 9409

SSA = 9486 − 9409 =77.00

Method B:

Σx² = 64² + 48² + 52² + 56² + 44² = 4096 + 2304 + 2704 + 3136 + 1936 = 14176

(Σx)² ÷ n = 264² ÷ 5 = 69696 ÷ 5 = 13939.2

SSB = 14176 − 13939.2 =236.80

Method C:

Σx² = 46² + 42² + 48² + 45² + 57² + 42² = 2116 + 1764 + 2304 + 2025 + 3249 + 1764 = 13222

(Σx)² ÷ n = 280² ÷ 6 = 78400 ÷ 6 = 13066.67

SSC = 13222 − 13066.67 =155.33

Step 4: Add the Three Sums of Squares Together

The total Sum of Squares Within (SSW) is simply the sum of the three group-level sums of squares — there's no need to bring in the grand mean at all for this part:

SSW = SSA + SSB + SSC = 77.00 + 236.80 + 155.33 =469.13

Step 5: Divide by the Within-Groups Degrees of Freedom

The degrees of freedom within groups is the total number of observations minus the number of groups: N − k.

Total observations, N = 4 + 5 + 6 = 15. Number of groups, k = 3.

Degrees of freedom = N − k = 15 − 3 = 12 (this is exactly the "12" you see in the critical value F2,12 given in the question — a nice built-in check that we're on the right track).

Estimate of population variance within samples = SSW ÷ (N − k) = 469.13 ÷ 12 =39.09

Pro Tip: The single most common slip on this type of question is dividing by the wrong degrees of freedom — using N − 1 = 14 (as if there were only one group) instead of N − k = 12. Another frequent error is computing each squared deviation from the overall grand mean instead of each observation's own group mean. Within-groups variance is only ever about scatter

Final Answer and Verification Against the Official Key

Our independently derived value is approximately 39.09, which rounds, allowing for the small cascading rounding difference that comes from carrying Method C's mean to only three decimal places, to the option given as 39.097. This matches option (d), which is also what the official answer key lists for this item. So our derivation and the official key agree:the correct answer is (d) 39.097.

Why This Question Matters

This question tests whether a candidate genuinely understands the mechanics behind the F-test in one-way ANOVA, rather than just memorising "F = between variance ÷ within variance." Every F-test rests on being able to correctly separate and compute both pieces of variance from raw data — under real exam pressure, with no calculator shortcuts beyond the Σx² identity. Once you can compute the within-groups variance for a dataset like this one without hesitation, the rest of the ANOVA table (the F-ratio, and the accept/reject decision against the critical value) falls into place in a matter of seconds.Getting the degrees of freedom right is often what separates a correct answer from a near-miss on test day.

Frequently Asked Questions

What is the difference between within-groups variance and between-groups variance?

Within-groups variance measures how spread out the observations are inside each individual group, around that group's own mean. Between-groups variance measures how far the group means themselves are spread out from the overall grand mean. ANOVA compares these two quantities to decide whether the differences between group means are larger than what ordinary random scatter within groups would explain.

Why do we divide by N − k and not N − 1?

N − 1 would be the correct divisor only if all the data belonged to a single group. Because we are pooling deviations computed separately from three different group means, we "use up" one degree of freedom for each group mean estimated, leaving N − k degrees of freedom overall. Here that is 15 − 3 = 12, which is exactly the value paired with F2,12 in the question.

Is there a faster formula than computing Σx² for each group?

You can compute deviations directly (x − mean) and square them, but that usually produces messier decimals, especially when the mean isn't a whole number, as in Method C. The Σx² − (Σx)²/n identity avoids working with fractional deviations until the very last step, which keeps the arithmetic faster and less error-prone under time pressure.

How is this within-groups variance used later in the ANOVA test?

It becomes the denominator of the F-ratio: F = (Mean Square Between) ÷ (Mean Square Within). If this computed F-ratio exceeds the critical value — here, F2,12 at 0.05 = 3.89 — we reject the null hypothesis that all three training methods produce the same mean output.

Does a negative deviation ever cause problems in this calculation?

No — since every deviation is squared before being added up, negative and positive deviations both contribute positively to the sum of squares. The sign only matters if you're computing a deviation by hand and want to sanity-check that your arithmetic is consistent, not in the final SS value itself.

What happens if I accidentally pool all 15 observations into one combined variance instead?

You would get a completely different, and incorrect, number, because that single combined variance would include the between-groups spread as well as the within-groups spread, conflating the two sources of variation that ANOVA is specifically designed to separate.

If any step above didn't fully click, drop a comment below and we'll walk through it further — and if you found this useful, do share it with a fellow ISS aspirant who's working through the same paper.