---
title: "Chi-square: brand purchase and region (5%)"
subject: "Probability and Statistics"
chapter: 7
key: "semester_5/probability_and_statistics:chi-square-brand-region-5-percent"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-brand-region-5-percent
site: BCT Engineering
---

# Chi-square: brand purchase and region (5%)

Probability and Statistics · Chapter 7: Inference concerning proportion

- Asked in: 2073 Chaitra, 5 marks
- Video: [Chi Square Test Explained with Example | Calculate χ² & Test the Null Hypothesis](https://www.youtube.com/watch?v=f53nXHoMXx4) — Vectors Academy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-brand-region-5-percent
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7

## Question

From the following data can you conclude that there is association between the purchase of brand and geographical region? Use 5% level of significance.
| | Central | Eastern | Western |
|---|---|---|---|
| Purchase brand | 40 | 55 | 45 |
| Do not purchase brand | 60 | 45 | 55 |

## Short answer

Chi-square test of independence gives chi-square = 4.688 with df = 2; critical value at 5% is 5.991. Since 4.688 < 5.991, fail to reject H0; no significant association between brand purchase and region.

## Answer

Chi-square test of independence between brand purchase and region. Grand total = 300.

1. H0: the two attributes are independent (no association).
   H1: the two attributes are associated (dependent).
2. Level of significance: α = 0.05
3. Test statistic: χ² = Σ (O − E)²/E, where E = (row total × column total)/grand total.

Observed frequencies with totals:

```text
               Central Eastern Western   Total
Purchase            40      55      45     140
No purchase         60      45      55     160
Total              100     100     100     300
```

Expected frequencies:

```text
               Central Eastern Western   
Purchase         46.67   46.67   46.67
No purchase      53.33   53.33   53.33
```

Terms (O − E)²/E:

```text
               Central Eastern Western   
Purchase         0.952   1.488   0.060
No purchase      0.833   1.302   0.052
```

χ² = Σ (O − E)²/E = **4.688**

4. Degrees of freedom = (r − 1)(c − 1) = (2 − 1)(3 − 1) = **2**.
   Critical value: χ²(0.05; 2) = **5.991**. Reject H0 if χ² > 5.991.
5. Decision: calculated χ² < tabulated χ² (4.688 < 5.991, p-value = 0.0960), so **fail to reject H0**.

**Conclusion:** there is no significant association; the two attributes may be taken as independent.

## Check yourself

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

1. What is H0?
   A. Brand purchase and region are associated
   B. Equal variance
   C. Equal means
   D. Brand purchase and region are independent
   Answer: D. H0: no association. (Key point: Hypotheses)

2. E for 'No purchase, Eastern' is
   A. 53.33
   B. 50
   C. 45
   D. 46.67
   Answer: A. 160 × 100/300 = 53.33. (Key point: Expected value)

3. d.f. for the 2 × 3 table are
   A. 3
   B. 2
   C. 5
   D. 6
   Answer: B. (2 − 1)(3 − 1) = 2. (Key point: Degrees of freedom)

4. 4.688 vs 5.991 gives
   A. Reject H0
   B. Association at 5%
   C. Fail to reject H0; no significant association
   D. Need more d.f.
   Answer: C. Calculated is below the table value. (Key point: Decision)

## Close variants set in other exams

- [Chi-square distribution; brand-region association](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Achi-square-define-brand-region) — Probability and Statistics, chapter 7, asked 4×
- [Chi-square: brand-region association (1%)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Achi-square-brand-region-1-percent) — Probability and Statistics, chapter 7, asked 4×