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Paired test: music and worker productivity

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%.

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 music219205226198209216
Week with music235186240203221205

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):
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
  1. Critical value: t at α = 0.01, df = n−1 = 5: ±4.032. Reject H0 if |t| > 4.032.
  2. 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.

✓ Answer checked by Claude Sonnet 5.5

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