---
title: "ANOVA: shear force of bolts at three positions"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:anova-bolts-three-positions"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-bolts-three-positions
site: BCT Engineering
---

# ANOVA: shear force of bolts at three positions

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2070 Chaitra, 6 marks
- Video: [As a part of the investigation collapse of the roof of a building a testing lab| ANOVA| MA3251|SNM|](https://www.youtube.com/watch?v=jImLMlSaek0) — Maths Tutor Sherly
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-bolts-three-positions
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

As a part of investigation of the collapse of the roof of a building, a testing laboratory is given all the available bolts that connected the steel structure at three different positions on the roof. The forces required to shear each of these bolts (coded values) are as follows:
| Position 1 | 90, 82, 79, 98, 83, 91 |
|---|---|
| Position 2 | 105, 89, 93, 104, 89, 95, 86 |
| Position 3 | 83, 89, 80, 94 |
Perform an ANOVA to test at the 0.05 level of significance whether the difference among the sample means at the three positions are significant.

## Short answer

One-way ANOVA: F=2.333, critical F(0.05;2,14)=3.739. Since 2.333 < 3.739, fail to reject H0; differences among the three position means are not significant at 5% level.

## Answer

One-way ANOVA with unequal group sizes (6, 7 and 4).

1. H0: μ₁ = μ₂ = μ₃ (mean shear force is the same at the three positions)
   H1: at least one position mean differs (F-test, right-tailed)
2. Level of significance: α = 0.05
3. Computations:

```text
Position 1   n=6  total=523  mean=87.1667
Position 2   n=7  total=661  mean=94.4286
Position 3   n=4  total=346  mean=86.5000

Grand total T = 1530,  N = 17
Correction factor CF = T²/N = 1530²/17 = 137700.0000
Total SS  = Σx² − CF = 138638.0000 − 137700.0000 = 938.0000
SS between = Σ(Ti²/ni) − CF = 137934.4524 − 137700.0000 = 234.4524
SS within  = Total SS − SS between = 703.5476
```

4. ANOVA table:

| Source | SS | d.f. | MS | F |
|---|---|---|---|---|
| Between positions | 234.4524 | 2 | 117.2262 | 2.333 |
| Within (error) | 703.5476 | 14 | 50.2534 | |
| Total | 938.0000 | 16 | | |

5. Critical value: F(0.05; 2, 14) = 3.739. Reject H0 if F > 3.739.
6. Decision: the calculated value falls in the acceptance region (p-value = 0.1335), so **fail to reject H0**.

**Conclusion:** the difference among the sample means is not significant at the 5% level.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How do SS between formulae handle unequal sizes?
   A. Sum of Ti^2/ni minus CF, using each group's own ni
   B. Average the group sizes
   C. Assume all ni = 17
   D. Use only the largest group
   Answer: A. Each group total is squared and divided by its own size. (Key point: Unequal groups)

2. What are the d.f.?
   A. (2, 17)
   B. (3, 14)
   C. (2, 14)
   D. (2, 16)
   Answer: C. Between: 3-1 = 2; within: 17-3 = 14. (Key point: Degrees of freedom)

3. What is F?
   A. 50.25/117.23
   B. 938/17
   C. 117.226/50.253 = 2.333
   D. 234.45/703.55
   Answer: C. MS between over MS within. (Key point: F ratio)

4. What is the conclusion?
   A. Reject H0
   B. F = 2.333 < 3.739, differences not significant
   C. Position 2 clearly differs
   D. F > 3.739, significant
   Answer: B. Fail to reject H0 at 5%. (Key point: Decision)

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