---
title: "Chi-square distribution; brand-region association"
subject: "Probability and Statistics"
chapter: 7
key: "semester_5/probability_and_statistics:chi-square-define-brand-region"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-define-brand-region
site: BCT Engineering
---

# Chi-square distribution; brand-region association

Probability and Statistics · Chapter 7: Inference concerning proportion

- Asked in: 2072 Chaitra, 5 marks; 2072 Kartik, 5 marks (asked 2 times)
- 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-define-brand-region
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7

## Question

Define chi-square distribution. 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 distribution is sum of squares of independent standard normals with k d.f., positively skewed, mean k, variance 2k. For independence test, χ²=4.688 with d.f.=2, critical 5.991 at 5%. Fail to reject H0, no significant association between brand purchase and region.

## Answer

**Chi-square distribution:** the distribution of the sum of squares of k independent standard normal variables, χ² = ΣZᵢ², with k degrees of freedom. It is positively skewed, takes only non-negative values, has mean k and variance 2k, and is used for tests of independence, goodness of fit and variance.

#### Test of association between brand purchase and region
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. Which is true of the chi-square distribution?
   A. Positively skewed, non-negative, mean k and variance 2k
   B. Takes negative values
   C. Symmetric about zero
   D. Mean 0 and variance 1
   Answer: A. It is right-skewed, non-negative, with mean k and variance 2k. (Key point: Properties)

2. What are the d.f. for a 2 × 3 table?
   A. 6
   B. 2
   C. 5
   D. 3
   Answer: B. (2 − 1)(3 − 1) = 2, so the 5% critical value is 5.991. (Key point: Degrees of freedom)

3. What is E for 'Purchase, Central'?
   A. 40
   B. 50
   C. 46.67
   D. 33.3
   Answer: C. E = 140 × 100/300 = 46.67. (Key point: Expected frequency)

4. Calculated chi-square = 4.688 vs 5.991. What is concluded?
   A. Reject H0; association exists
   B. Calculated value is too small to compute
   C. Use d.f. = 3
   D. Fail to reject H0; no significant association
   Answer: D. 4.688 < 5.991, so independence is not rejected. (Key point: Decision)

## Close variants set in other exams

- [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×
- [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×