---
title: "Regression equation for demand of fish"
subject: "Engineering Economics"
chapter: 1
key: "semester_6/engineering_economics:regression-demand-fish-price-60"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/regression-demand-fish-price-60
site: BCT Engineering
---

# Regression equation for demand of fish

Engineering Economics · Chapter 1: Introduction

- Asked in: 2065 Chaitra, 6 marks
- Video: [Demand Forecasting Regression method problem](https://www.youtube.com/watch?v=A6LWDQGnL70) — EDUCARE WITH CK
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/regression-demand-fish-price-60
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-1

## Question

The following data shows the demands for fish when the prices are as shown. Calculate the hypothesized regression equation. What shall be the demand if the price is set to be Rs. 60 per kg?
| S.N | Price per kg. | Quantity (tones) |
|---|---|---|
| 1 | 64 | 65 |
| 2 | 53 | 75 |
| 3 | 67 | 56 |
| 4 | 52 | 69 |
| 5 | 82 | 51 |
| 6 | 59 | 65 |
| 7 | 67 | 63 |
| 8 | 71 | 55 |

## Short answer

The regression line is Q = 110.20 − 0.743P, fitted by least squares. For a price of Rs 60 per kg, forecast demand is 65.6 tonnes.

## Answer

#### Method
Fit the hypothesised linear demand Q = a + bP by least squares, where P = price (Rs/kg) and Q = quantity (tonnes). n = 8.
b = (nΣPQ − ΣPΣQ) / (nΣP² − (ΣP)²), a = (ΣQ − bΣP) / n

| S.N | P | Q | PQ | P² |
|---|---|---|---|---|
| 1 | 64 | 65 | 4,160 | 4,096 |
| 2 | 53 | 75 | 3,975 | 2,809 |
| 3 | 67 | 56 | 3,752 | 4,489 |
| 4 | 52 | 69 | 3,588 | 2,704 |
| 5 | 82 | 51 | 4,182 | 6,724 |
| 6 | 59 | 65 | 3,835 | 3,481 |
| 7 | 67 | 63 | 4,221 | 4,489 |
| 8 | 71 | 55 | 3,905 | 5,041 |
| Σ | 515 | 499 | 31,618 | 33,833 |

b = (8 × 31,618 − 515 × 499) / (8 × 33,833 − 515²)
 = (252,944 − 256,985) / (270,664 − 265,225)
 = −4,041 / 5,439 = **−0.7430**

a = (499 − (−0.7430)(515)) / 8 = (499 + 382.65) / 8 = **110.20**

**Regression equation: Q = 110.20 − 0.743 P**

#### Demand at Rs 60 per kg
Q = 110.20 − 0.743 × 60 = 110.20 − 44.58 = **65.6 tonnes**

**Answer: Q = 110.20 − 0.743P; at Rs 60/kg demand is about 65.6 tonnes.** The negative slope means demand falls by about 0.74 tonne for every Rs 1 rise in price, as expected from the law of demand.

## Check yourself

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

1. Which formula gives the slope b of Q = a + bP?
   A. b = (nSigmaPQ - SigmaP SigmaQ) / (nSigmaP^2 - (SigmaP)^2)
   B. b = (SigmaPQ - SigmaP SigmaQ) / SigmaP^2
   C. b = (nSigmaPQ + SigmaP SigmaQ) / (nSigmaP^2 + (SigmaP)^2)
   D. b = (SigmaQ - a n) / SigmaP^2
   Answer: A. The least-squares slope uses n times the cross-product sum minus the product of sums, over n times the sum of squares minus the square of the sum. (Key point: Least squares formula)

2. The slope is -0.743. What does it mean?
   A. Demand rises 0.74 tonne for each Re 1 rise in price
   B. Demand falls about 0.74 tonne for each Rs 1 rise in price, consistent with the law of demand
   C. Demand falls by 74% when price rises
   D. Price falls 0.743 Rs for each tonne of demand
   Answer: B. A negative slope shows quantity moves opposite to price per unit of price. (Key point: Sign of slope)

3. Which is the demand at Rs 60 per kg?
   A. About 154.8 tonnes (110.20 + 0.743 x 60)
   B. About 44.6 tonnes
   C. About 65.6 tonnes (110.20 - 0.743 x 60)
   D. About 110.2 tonnes
   Answer: C. Substitute P = 60 into Q = 110.20 - 0.743P to get 110.20 - 44.58 = 65.6. (Key point: Prediction)

4. How is the intercept a computed once b is known?
   A. a = (SigmaQ + SigmaP)/n
   B. a = b x SigmaP / n
   C. a = SigmaQ / SigmaP
   D. a = (SigmaQ - b SigmaP)/n = (499 + 382.65)/8 = 110.20
   Answer: D. The line passes through the mean point, so a = mean Q - b x mean P. (Key point: Intercept)

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