---
title: "Paired test: safety program worker hours"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:paired-before-after-safety-program"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/paired-before-after-safety-program
site: BCT Engineering
---

# Paired test: safety program worker hours

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2073 Shrawan, 5 marks; 2071 Shrawan, 6 marks; 2070 Chaitra, 6 marks (asked 3 times)
- Video: [Matched or Paired Samples T-Test - Hypothesis Testing](https://www.youtube.com/watch?v=JiQR0lHLe74) — The Organic Chemistry Tutor
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/paired-before-after-safety-program
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

The following are the average weekly losses of worker hours due to accidents in 10 industrial plants before and after a certain safety program was put into operation:
| Before | 45 | 73 | 46 | 124 | 33 | 57 | 83 | 34 | 26 | 17 |
|---|---|---|---|---|---|---|---|---|---|---|
| After | 36 | 60 | 44 | 119 | 35 | 51 | 77 | 29 | 24 | 11 |
Use the 0.05 level of significance to test whether the safety program is effective.

## Short answer

Differences d = Before−After: d̄ = 5.20, s_d = 4.077. t = 5.20/(4.077/√10) = 4.033 > t_crit (one‑tailed, df=9, α=0.05) = 1.833, so reject H₀; the safety program significantly reduces lost hours.

## Answer

Paired (dependent) samples: the same plants are measured before and after. Let d = Before − After, the reduction in lost worker hours. μ_d is the mean reduction.

| Before | 45 | 73 | 46 | 124 | 33 | 57 | 83 | 34 | 26 | 17 |
|---|---|---|---|---|---|---|---|---|---|---|
| After | 36 | 60 | 44 | 119 | 35 | 51 | 77 | 29 | 24 | 11 |
| d | 9 | 13 | 2 | 5 | -2 | 6 | 6 | 5 | 2 | 6 |

Σd = 52, Σd² = 420, n = 10

d̄ = Σd/n = 5.20
s_d = √[(Σd² − n·d̄²)/(n − 1)] = √[(420 − 10×27.04)/9] = 4.0770

**Hypotheses**

- H₀: μ_d = 0 (the safety program is not effective)
- H₁: μ_d > 0 (the program reduces lost hours; one-tailed)

**Level of significance:** α = 0.05

**Test statistic:** t = d̄/(s_d/√n) = 5.20/(4.0770/√10) = 5.20/1.2893 = **4.033**

**Critical value:** df = n − 1 = 9, one-tailed t₀.₀₅ = 1.833. Reject H₀ if t > 1.833.

**Decision:** calculated t = 4.033 > 1.833 (p-value ≈ 0.0015), so H₀ is rejected.

**Conclusion:** at the 5% level there is sufficient evidence that the safety program is effective in reducing the weekly loss of worker hours.

## Check yourself

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

1. Why is a paired t-test used?
   A. Same plants measured before and after
   B. Samples are large
   C. Variances are known
   D. Samples are independent
   Answer: A. Measurements on the same plants are dependent pairs. (Key point: Test type)

2. What is H1?
   A. μd = 0
   B. μd < 0
   C. μd ≠ 0
   D. μd > 0 (d = Before − After)
   Answer: D. Effectiveness means a reduction, so a one-tailed test μd > 0. (Key point: Alternative)

3. What is t?
   A. d̄/(s_d/√n) ≈ 4.03
   B. d̄/s_d
   C. s_d/d̄
   D. d̄·√n
   Answer: A. t = 5.2/(4.077/√10) = 4.033. (Key point: Statistic)

4. Why reject H0?
   A. 4.033 > 1.833 with df = 9
   B. 4.033 < 1.833
   C. df = 10
   D. t is negative
   Answer: A. Calculated t exceeds the one-tailed critical value, so the program is effective. (Key point: Decision)

## Related in Probability and Statistics

- [Paired test: music and worker productivity](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apaired-t-test-music-productivity) — Probability and Statistics, chapter 6
- [CI: weekly study hours of students](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Astudy-hours-mean-ci-99-95) — Probability and Statistics, chapter 6
- [Paired t-test: advertisement effectiveness](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apaired-t-test-advertisement-sales) — Probability and Statistics, chapter 6
- [Paired t-test: dietary product weight loss](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apaired-t-test-dietary-weight-loss) — Probability and Statistics, chapter 6
- [Sample mean, variance, CV: patients' sleep hours](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Ahospital-patients-sleep-mean-cv) — Probability and Statistics, chapter 1, asked 3×
- [Poisson process; intersection accidents](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apoisson-process-accidents-intersection) — Probability and Statistics, chapter 2

## Related in other subjects

- [Power output with 46.85 kV sending, 33 kV receiving](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-output-46-85kv-33kv) — Power System Analysis I, chapter 6
- [N-R two iterations, y12 = 10∠−73.74°](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Anr-two-bus-y12-10) — Power System Analysis II, chapter 2
- [Gumbel design flood, factor of safety 1.3, 45-year life](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-25yr-life45-fos1-3) — Engineering Hydrology, chapter 6
- [Actual load estimation in industrial plants](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aindustrial-load-estimation-method) — Industrial Power Distribution and Illumination, chapter 1, asked 2×