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

# Paired t-test: dietary product weight loss

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2073 Chaitra, 5 marks
- Video: [Matched or Paired Samples T-Test - Hypothesis Testing](https://www.youtube.com/watch?v=JiQR0lHLe74) — The Organic Chemistry Tutor
- 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-dietary-weight-loss
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of at least 3 pounds. Eight subjects use the product for one month, and the resulting weight loss data are reported below. Do the data support the claim of the producer of the dietary product with the probability of a type I error set to 0.05?
| Subjects | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Weight before (lb) | 165 | 201 | 195 | 198 | 155 | 143 | 150 | 187 |
| Weight after (lb) | 161 | 195 | 192 | 193 | 150 | 141 | 146 | 183 |

## Short answer

Paired t-test on weight losses (before − after): d̄ = 4.125 lb, t = 2.553. At α = 0.05, critical value t(0.05,7) = −1.895; since t > −1.895, we fail to reject H0. The data support the claim of average weight loss of at least 3 lb.

## Answer

The same subjects are weighed before and after, so a paired t-test on the weight losses d = before − after is used. The claim (μ_d ≥ 3) is placed in H0, and rejection would mean the claim is not supported.

1. H0: μ_d ≥ 3 lb (average loss is at least 3 lb, the claim)
   H1: μ_d < 3 lb (average loss is less than 3 lb) (left-tailed)
2. Level of significance: α = 0.05
3. Differences d (before − after):

```text
d  = 4, 6, 3, 5, 5, 2, 4, 4
n  = 8, Σd = 33, d̄ = 4.1250
Σ(d−d̄)² = 10.8750, s_d² = 1.5536, s_d = 1.2464
t  = (d̄ − μ_d)/(s_d/√n) = (4.1250 − 3)/(1.2464/√8)
   = 1.1250/0.4407 = 2.553
```

4. Critical value: t at α = 0.05, df = n−1 = 7: −1.895. Reject H0 if t < −1.895.
5. Decision: the calculated value falls in the acceptance region (p-value = 0.9810), so **fail to reject H0**.

**Conclusion:** the data support the producer's claim that the average weight loss is at least 3 lb.

## Check yourself

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

1. Why a paired t-test?
   A. The same subjects are measured before and after
   B. Sample is large
   C. Variance known
   D. Two independent groups
   Answer: A. Before and after weights belong to the same eight subjects, so the differences d are analysed. (Key point: Choice of test)

2. How are the hypotheses set?
   A. H0: mu_d < 3; H1: mu_d >= 3
   B. H0: mu_d = 0; H1: mu_d != 0
   C. H0: mu_d >= 3 (the claim); H1: mu_d < 3
   D. H0: mu_d = 8
   Answer: C. The claim of at least 3 lb loss is placed in H0, giving a left-tailed test. (Key point: Hypotheses)

3. What is df and the critical value?
   A. 7; -1.895
   B. 7; +-2.365
   C. 8; -1.860
   D. 15; -1.753
   Answer: A. df = n-1 = 7, left-tailed at 5% gives -1.895. (Key point: Degrees of freedom)

4. What is the decision?
   A. Fail to reject H0 since t is not below -1.895; claim supported
   B. Reject because t = 2.553 > 1.895
   C. Test is inconclusive
   D. Reject H0; claim unsupported
   Answer: A. Positive t is not in the left rejection region, so the data support the claim. (Key point: Decision)

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