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

# Paired t-test: advertisement effectiveness

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2076 Chaitra, 5 marks
- Video: [T Test for Paired Sample Means with Example Problem | Hypothesis Testing Statistics](https://www.youtube.com/watch?v=t3jV5WxRk68) — Shakehand with Life
- 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-advertisement-sales
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

The sales figure of an item in six shops before and after and advertisement is given as:
| Before | 53 | 28 | 31 | 48 | 50 | 52 |
|---|---|---|---|---|---|---|
| After | 58 | 29 | 30 | 55 | 56 | 45 |
Test whether the advertisement was effective at 5% level of significance? (t-value for 5 degree of freedom = 2.571)

## 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.

## Answer

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

| Shop | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Before | 53 | 28 | 31 | 48 | 50 | 52 |
| After | 58 | 29 | 30 | 55 | 56 | 45 |
| d | 5 | 1 | -1 | 7 | 6 | -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.

## Check yourself

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

1. Why is a paired t-test appropriate?
   A. Same shops before and after the advertisement
   B. Large samples
   C. Different shops
   D. Known σ
   Answer: A. The observations are dependent pairs. (Key point: Paired test)

2. What is H0?
   A. μd > 0
   B. μd = 0
   C. μd < 0
   D. μd ≠ 0 and effective
   Answer: B. H0: no change in mean sales. (Key point: Hypotheses)

3. What is t?
   A. 2.571
   B. 0.846
   C. 1.833
   D. 2.167
   Answer: B. t = 1.833/(5.307/√6) = 0.846. (Key point: Statistic)

4. What is the conclusion?
   A. Reject H0; advertisement effective
   B. Fail to reject H0; not shown effective
   C. Accept H1
   D. Reduce df
   Answer: B. 0.846 < 2.571, so there is insufficient evidence of effectiveness. (Key point: Decision)

## Related in Probability and Statistics

- [Paired test: safety program worker hours](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apaired-before-after-safety-program) — Probability and Statistics, chapter 6, asked 3×
- [Normal: 31% under 45, 8% over 64](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Anormal-31-percent-under-45) — Probability and Statistics, chapter 3
- [Corrected mean and SD after item error](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Acorrected-mean-sd-20-items) — Probability and Statistics, chapter 1
- [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
- [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
- [IOE entrance scores: estimator and consistency](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aioe-entrance-sixty-estimator-consistency) — Probability and Statistics, chapter 6

## Related in other subjects

- [7 and 28 day strength of 53 grade OPC](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Aopc-53-grade-strength) — Civil Engineering Materials, chapter 5
- [Chi-square uniformity test, 40 numbers](https://www.bctengineer.com/question.md?key=semester_8%2Fsimulation_and_modeling%3Achi-square-uniformity-table-2074) — Simulation and Modeling, chapter 6
- [Stability: HS 175/52, BP 150/48](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Astability-hs175-52-bp150-48) — Engineering Geology II, chapter 6
- [AVL tree for 10 20 30 50 45 40 8 5 3](https://www.bctengineer.com/question.md?key=semester_4%2Fdata_structure_and_algorithm%3Aavl-build-10-20-30) — Data Structure and Algorithm, chapter 6