---
title: "Chi-square: brand-region association (1%)"
subject: "Probability and Statistics"
chapter: 7
key: "semester_5/probability_and_statistics:chi-square-brand-region-1-percent"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-brand-region-1-percent
site: BCT Engineering
---

# Chi-square: brand-region association (1%)

Probability and Statistics · Chapter 7: Inference concerning proportion

- Asked in: 2069 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-1-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 using Chi-square test at α = 1%?
| | 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 1% is 9.210. Since 4.688 < 9.210, 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 = 0.01
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(0.01; 2) = **9.210**. Reject H0 if χ² > 9.210.
5. Decision: calculated χ² < tabulated χ² (4.688 < 9.210, 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 χ²(0.01; 2)?
   A. 5.991
   B. 7.815
   C. 13.277
   D. 9.210
   Answer: D. At 1% with 2 d.f. the table value is 9.210. (Key point: Critical value)

2. Why are d.f. 2?
   A. (2 − 1)(3 − 1) = 2
   B. 3 − 1
   C. 300 − 1
   D. 6 − 1
   Answer: A. d.f. = (r − 1)(c − 1). (Key point: Degrees of freedom)

3. 4.688 vs 9.210 means
   A. Reject H0
   B. Association
   C. Fail to reject H0; independent
   D. Use t
   Answer: C. 4.688 < 9.210, so no association is shown. (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 purchase and region (5%)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Achi-square-brand-region-5-percent) — Probability and Statistics, chapter 7, asked 4×

## Related in other subjects

- [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