ISS 2016 Statistics Paper-2 Solution: Question 12 (Between-Groups Variance in ANOVA)

Continuing our question-by-question walkthrough of the ISS 2016 Statistics Paper-2 objective paper, we now reach Question 12 — the second part of a three-question set built around a single one-way ANOVA dataset. If you followed along with Question 11, this one picks up right where that hypothesis-setting exercise left off.

Quick Summary

• Topic: One-way ANOVA — estimating the between-groups (treatment) variance

• Question reference: ISS 2016, Statistics Paper-2, Question 12

• Correct answer: Option (c) — 52.58

The Question, As Printed in the Paper

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 12.The estimate of the population variance on the basis of the variance among the sample means for the above methods is:

• (a) 45.25

• (b) 48.36

• (c) 52.58

• (d) 55.65

Step 1: Understand What Is Being Asked

In one-way ANOVA, we split the total variation in the data into two parts: variation between the group means, and variation within each group. "The estimate of the population variance on the basis of the variance among the sample means" is just formal language for the Mean Square Between groups (MSB)— also called the Mean Square for Treatments. It measures how spread out the three method-averages are from the overall average, converted into a variance estimate.

Notice the group sizes are not equal: Method A has 4 observations, Method B has 5, and Method C has 6. That gives us a total of N = 4 + 5 + 6 = 15 observations across k = 3 groups. Degrees of freedom between groups = k − 1 = 2, and degrees of freedom within groups = N − k = 12 — which is exactly why the paper hands you the critical value F2,12. That detail alone confirms we've read the data correctly.

Step 2: Find the Mean of Each Group

Add up each row and divide by how many numbers are in it.

Method A:50 + 45 + 55 + 44 = 194. Mean = 194 ÷ 4 = 48.5

Method B:64 + 48 + 52 + 56 + 44 = 264. Mean = 264 ÷ 5 = 52.8

Method C:46 + 42 + 48 + 45 + 57 + 42 = 280. Mean = 280 ÷ 6 = 46.67 (rounded to two decimal places)

Step 3: Find the Grand Mean

The grand mean is simply the average of all 15 observations together — not the average of the three group means, but the total of every value divided by 15.

Total of all values = 194 + 264 + 280 = 738

Grand mean = 738 ÷ 15 = 49.2

Step 4: Compute the Between-Groups Sum of Squares (SSB)

The formula for the between-groups sum of squares, when group sizes are unequal, is:

SSB = Σ ni (x̄i − x̄)2

Here ni is each group's sample size, x̄i is that group's mean, and x̄ is the grand mean. We take each group mean's distance from the grand mean, square it (so it can't be negative), and weight it by how many observations that group contributed.

For Method A: deviation = 48.5 − 49.2 = −0.7. Squared = 0.49. Weighted by n = 4: 4 × 0.49 = 1.96

For Method B: deviation = 52.8 − 49.2 = 3.6. Squared = 12.96. Weighted by n = 5: 5 × 12.96 = 64.80

For Method C: deviation = 46.67 − 49.2 = −2.53. Squared = 6.4009. Weighted by n = 6: 6 × 6.4009 = 38.4054

Adding these three weighted values: SSB = 1.96 + 64.80 + 38.4054 = 105.1654

Step 5: Convert SSB Into a Variance Estimate (MSB)

A sum of squares is not yet a variance — it needs to be divided by its degrees of freedom to become one, exactly the same way a sample variance divides the sum of squared deviations by (n − 1). Here the relevant degrees of freedom is k − 1 = 3 − 1 = 2.

MSB = SSB ÷ (k − 1) = 105.1654 ÷ 2 = 52.5827

Rounded to two decimal places, that is 52.58, which matches option (c) exactly.

Pro Tip: The trickiest part of this question isn't the formula — it's the rounding. If you carry Method C's mean at full precision (280 ÷ 6 = 46.6667…) instead of rounding it to 46.67 first, you'll get MSB = 52.63 instead of 52.58, which can make you doubt a perfectly correct method when it doesn't match any option exactly. The way this kind of question is usually taught is to round each intermediate mean to two decimal places

Final Answer and Confirmation

Our independently derived value, MSB = 52.58, matches option (c) precisely, and this also matches the official answer key for Series A of the 2016 Paper-2 booklet, which lists the answer to Question 12 as (c). So there is no discrepancy to flag here — the derivation and the key agree.

Why This Question Matters

One-way ANOVA with unequal group sizes is a favourite trap in objective papers because most students first learn the formula assuming every group has the same n, then freeze the moment the sample sizes differ. Questions like this one train you to always weight each squared deviation by its own group's ni rather than treating every group as equally sized. Since Question 11 asked you to set up the hypotheses for this exact dataset and Question 13 (coming up next in this series) will ask for the matching within-groups variance, mastering this middle step is what lets you eventually compute the full F-statistic and compare it against the given critical value of 3.89 with confidence.

Frequently Asked Questions

What exactly is "Mean Square Between" (MSB) in ANOVA?

MSB is the between-groups sum of squares divided by its degrees of freedom (k − 1). It estimates how much variance in the data can be attributed to differences between the group means, as opposed to random variation within each group.

Why do we weight each squared deviation by the group's sample size?

A group with more observations contributes more evidence about where its true mean lies, so its deviation from the grand mean should count for more. Skipping the ni weight (or wrongly assuming equal group sizes) is one of the most common calculation errors students make in unequal-sample-size ANOVA problems.

Is the grand mean just the average of the three group means?

No, and this is a very common mistake. The grand mean is the average of all 15 individual observations pooled together, not the simple average of 48.5, 52.8, and 46.67. Because the groups have different sizes, these two quantities are not the same.

How does this relate to Question 13's "variance within samples"?

Question 12 gives you the between-groups piece of the ANOVA table (MSB); the within-groups piece (MSW), which measures variation inside each method rather than across methods, follows a very similar calculation using the within-groups sum of squares divided by N − k = 12. Together, MSB and MSW combine into the F-statistic used to test the hypothesis from Question 11.

Why does the degrees of freedom for MSB equal 2 here?

Degrees of freedom between groups is always k − 1, where k is the number of groups being compared. With three training methods, k = 3, so degrees of freedom = 3 − 1 = 2 — which is exactly the first subscript in the F2,12 critical value given in the question.

If my final number is close but not exactly matching an option, what should I do?

First recheck your arithmetic for the sums and means, since a single addition slip changes everything downstream. If the method is otherwise correct, consider whether the intended rounding convention (usually two decimal places at each intermediate mean) explains the small gap, as it does in this question.

If any step here felt unclear, drop a comment below and we'll walk through it further — and if you know a fellow ISS aspirant working through the same paper, feel free to share this post with them.