---
title: "Test: heat capacity of coal from two mines"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:coal-two-mines-heat-capacity-test"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/coal-two-mines-heat-capacity-test
site: BCT Engineering
---

# Test: heat capacity of coal from two mines

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2070 Asar, 6 marks
- Video: [t- test - problem 7| The following random samples are measurements of the heat|Testing of hypothesis](https://www.youtube.com/watch?v=Bvt_a1r0DNc) — 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/coal-two-mines-heat-capacity-test
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

The following random samples are measurement of the heat producing capacity (in millions of calories per ton) of specimens of coal from two mines:
| Mine 1 | 8260 | 8130 | 8350 | 8070 | 8340 | - |
|---|---|---|---|---|---|---|
| Mine 2 | 7950 | 7890 | 7900 | 8140 | 7920 | 7840 |
Use the 0.01 level of significance to test whether the difference between means of these two samples is significant.

## Short answer

Pooled t-test assuming equal variances: sp²=13066.67, t=4.19, df=9, critical t=±3.250. |t| > 3.250, reject H0; difference in mean heat capacities is significant at 1% level.

## Answer

Samples are small and independent, so a pooled t-test is used (populations assumed normal with equal variances).

Sample statistics:

```text
n₁ = 5, Σx₁ = 41150, x̄₁ = 8230.0000, s₁² = 15750.0000
n₂ = 6, Σx₂ = 47640, x̄₂ = 7940.0000, s₂² = 10920.0000
```

1. H0: μ₁ = μ₂ (mean heat capacities are equal)
   H1: μ₁ ≠ μ₂ (two-tailed)
2. Level of significance: α = 0.01
3. Test statistic (small samples, equal variances assumed, pooled t-test):

```text
Sp² = [(n₁−1)s₁² + (n₂−1)s₂²] / (n₁+n₂−2)
    = [4×15750.0000 + 5×10920.0000] / 9 = 13066.6667
Sp  = 114.3095
t   = (x̄₁ − x̄₂) / [Sp √(1/n₁ + 1/n₂)]
    = (8230.0 − 7940.0) / [114.3095 × √(1/5 + 1/6)]
    = 290.0000 / 69.2179 = 4.190
```

4. Critical value: t at α = 0.01, df = n₁+n₂−2 = 9: ±3.250. Reject H0 if |t| > 3.250.
5. Decision: the calculated value falls in the rejection region (p-value = 0.0023), so **reject H0**.

**Conclusion:** the difference between the mean heat-producing capacities of the two mines is significant at the 1% level.

## Check yourself

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

1. Why a pooled t-test?
   A. Paired
   B. Small independent samples with equal variances assumed
   C. Large samples
   D. Known sigma
   Answer: B. n1 = 5 and n2 = 6 are small and independent. (Key point: Choice of test)

2. What is df?
   A. 9
   B. 11
   C. 5
   D. 10
   Answer: A. 5 + 6 - 2 = 9, critical value +-3.250 at 1%. (Key point: Degrees of freedom)

3. What is Sp^2?
   A. 10920
   B. 13066.67
   C. 15750 x 10920
   D. (15750 + 10920)/2
   Answer: B. [4 x 15750 + 5 x 10920]/9. (Key point: Pooled variance)

4. What is the conclusion?
   A. Fail to reject
   B. t = 4.19 > 3.25, difference significant
   C. t = 290
   D. t = 4.19 < 3.25, not significant
   Answer: B. H0 is rejected at 1%. (Key point: Decision)

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