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Paired t-test: advertisement effectiveness

Short answer

Differences d = After−Before: d̄ = 1.833, s_d = 5.307, t = 0.846 (df = 5). At 5%, given two‑tailed t = 2.571, fail to reject H₀; no evidence the advertisement increased sales.

The sales figure of an item in six shops before and after and advertisement is given as:

Before532831485052
After582930555645
Test whether the advertisement was effective at 5% level of significance? (t-value for 5 degree of freedom = 2.571)

Paired samples (same shops before and after). Let d = After − Before.

Shop123456
Before532831485052
After582930555645
d51-176-7

Σd = 11, Σd² = 161, n = 6

d̄ = 11/6 = 1.833 s_d = √[(Σd² − n·d̄²)/(n − 1)] = √[(161 − 6×3.361)/5] = 5.307

Hypotheses

  • H₀: μ_d = 0 (advertisement not effective)
  • H₁: μ_d > 0 (sales increased; one-tailed)

Level of significance: α = 0.05

Test statistic: t = d̄/(s_d/√n) = 1.833/(5.307/√6) = 1.833/2.167 = 0.846

Critical value: df = 5. The given tabulated value is t = 2.571 (the two-tailed 5% value; the one-tailed 5% value would be 2.015). Either way the critical value is far above the calculated t.

Decision: calculated t = 0.846 < 2.571 (and < 2.015), so we fail to reject H₀ (p ≈ 0.218).

Conclusion: at the 5% level there is no sufficient evidence that the advertisement increased sales; it was not effective.

✓ Answer checked by Claude Sonnet 5.5

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