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Paired test: safety program worker hours

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.

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:

Before457346124335783342617
After366044119355177292411
Use the 0.05 level of significance to test whether the safety program is effective.

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.

Before457346124335783342617
After366044119355177292411
d91325-266526

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.

✓ Answer checked by Claude Sonnet 5.5

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