---
title: "Paired test: music and worker productivity"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:paired-t-test-music-productivity"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/paired-t-test-music-productivity
site: BCT Engineering
---

# Paired test: music and worker productivity

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2069 Chaitra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/paired-t-test-music-productivity
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

In a manufacturing company the new modern manager is in a belief that music enhances the productivity of workers. He made observations on 6 workers for a week and recorded the production before and after the music was installed. From the data given below, can you conclude that the productivity has indeed changed due to music? (α = 1%)
| Week without music | 219 | 205 | 226 | 198 | 209 | 216 |
|---|---|---|---|---|---|---|
| Week with music | 235 | 186 | 240 | 203 | 221 | 205 |

## Short answer

Paired t-test: d=with−without, d̄=2.833, s_d=14.525, t=0.478, df=5, critical t=±4.032. |t| < 4.032, fail to reject H0; no significant productivity change at 1%.

## Answer

The same six workers are observed under both conditions, so a paired t-test is used with d = production with music − production without music. As the question asks whether productivity has *changed*, a two-tailed test is used.

1. H0: μ_d = 0 (music has no effect on productivity)
   H1: μ_d ≠ 0 (productivity has changed) (two-tailed)
2. Level of significance: α = 0.01
3. Differences d (with music − without music):

```text
d  = 16, -19, 14, 5, 12, -11
n  = 6, Σd = 17, d̄ = 2.8333
Σ(d−d̄)² = 1054.8333, s_d² = 210.9667, s_d = 14.5247
t  = (d̄ − μ_d)/(s_d/√n) = (2.8333 − 0)/(14.5247/√6)
   = 2.8333/5.9297 = 0.478
```

4. Critical value: t at α = 0.01, df = n−1 = 5: ±4.032. Reject H0 if |t| > 4.032.
5. Decision: the calculated value falls in the acceptance region (p-value = 0.6529), so **fail to reject H0**.

**Conclusion:** productivity has not changed significantly at the 1% level; the data do not show a real effect of music.

## Check yourself

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

1. Why paired?
   A. Large sample
   B. Same six workers measured twice
   C. Independent groups
   D. ANOVA
   Answer: B. The differences d are analysed. (Key point: Choice of test)

2. Why two-tailed?
   A. He wants a decrease
   B. The manager asks if productivity has changed
   C. He wants an increase
   D. Always two-tailed
   Answer: B. 'Changed' means H1: mu_d != 0. (Key point: Tails)

3. What is df and critical value at 1%?
   A. 6; +-3.707
   B. 5; +-2.571
   C. 11; +-3.106
   D. 5; +-4.032
   Answer: D. df = n-1 = 5. (Key point: Degrees of freedom)

4. What is the conclusion?
   A. t = 2.83, reject
   B. t = 0.478, reject
   C. t = 0.478, fail to reject; no significant change
   D. Music clearly raises output
   Answer: C. |t| is far below 4.032. (Key point: Decision)

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