---
title: "Chi-square: intelligence vs economic condition (1072)"
subject: "Probability and Statistics"
chapter: 7
key: "semester_5/probability_and_statistics:chi-square-intelligence-economic-1072"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-intelligence-economic-1072
site: BCT Engineering
---

# Chi-square: intelligence vs economic condition (1072)

Probability and Statistics · Chapter 7: Inference concerning proportion

- Asked in: 2075 Chaitra, 6 marks; 2071 Chaitra, 6 marks (asked 2 times)
- Video: [Chi Square Test](https://www.youtube.com/watch?v=HKDqlYSLt68) — 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/chi-square-intelligence-economic-1072
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7

## Question

In a recent survey 1,072 Engineers were classified according to their intelligence (GPA in Bachelor) and economic conditions after graduation. Test whether there is any association between intelligence and economic condition. [χ² value for 3 d.f. = 5.991 as printed]
| Economic Condition after graduation | Excellent | Good | Mediocre | Dull |
|---|---|---|---|---|
| Good | 48 | 199 | 181 | 82 |
| Not good | 81 | 185 | 190 | 106 |

## Short answer

A chi-square test of independence gives χ²=9.735 with 3 degrees of freedom, against critical value 7.815 at 5% level. Since 9.735>7.815, reject H0; there is significant association between intelligence and economic condition.

## Answer

A chi-square test of independence is used to test association between intelligence (GPA) and economic condition. Total = 1072.

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
               Excellent      Good  Mediocre      Dull     Total
Good                  48       199       181        82       510
Not good              81       185       190       106       562
Total                129       384       371       188      1072
```

Expected frequencies:

```text
               Excellent      Good  Mediocre      Dull
Good               61.37    182.69    176.50     89.44
Not good           67.63    201.31    194.50     98.56
```

Terms (O − E)²/E:

```text
               Excellent      Good  Mediocre      Dull
Good               2.913     1.457     0.115     0.619
Not good           2.644     1.322     0.104     0.562
```

χ² = Σ (O − E)²/E = **9.735**

4. Degrees of freedom = (r − 1)(c − 1) = (2 − 1)(4 − 1) = **3**.
   Critical value: χ²(0.05; 3) = **7.815**. Reject H0 if χ² > 7.815.
5. Decision: calculated χ² > tabulated χ² (9.735 > 7.815, p-value = 0.0210), so **reject H0**.

**Conclusion:** there is a significant association between the two attributes; they are not independent.

Note: the value 5.991 printed in the question is the χ² value for 2 d.f. For a 2 × 4 table, d.f. = (2−1)(4−1) = 3 and the correct table value at 5% is 7.815. The decision is the same with either value because the calculated χ² is larger than both.

## Check yourself

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

1. How is each expected frequency E computed?
   A. Grand total/number of cells
   B. Row total/number of columns
   C. Observed/2
   D. (row total × column total)/grand total
   Answer: D. Under independence E = (row total × column total)/grand total. (Key point: Expected frequency)

2. What are the d.f. for the 2 × 4 table?
   A. 3
   B. 2
   C. 8
   D. 5
   Answer: A. d.f. = (2 − 1)(4 − 1) = 3. (Key point: Degrees of freedom)

3. What is the correct critical value at 5% with 3 d.f.?
   A. 5.991
   B. 7.815
   C. 9.488
   D. 3.841
   Answer: B. 5.991 is for 2 d.f. as the note says; for 3 d.f. it is 7.815. (Key point: Table value)

4. Calculated chi-square is 9.735. What is concluded?
   A. Independent; fail to reject H0
   B. Fewer d.f. are needed
   C. Reject H0; intelligence and economic condition are associated
   D. Accept H0 because p = 0.021
   Answer: C. 9.735 > 7.815, so H0 of independence is rejected. (Key point: Decision)

## Close variants set in other exams

- [Chi-square: intelligence vs economic condition (1650)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Achi-square-intelligence-economic-1650) — Probability and Statistics, chapter 7, asked 3×

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