---
title: "Regression: chemical yield vs temperature"
subject: "Probability and Statistics"
chapter: 5
key: "semester_5/probability_and_statistics:yield-temperature-regression-200c"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/yield-temperature-regression-200c
site: BCT Engineering
---

# Regression: chemical yield vs temperature

Probability and Statistics · Chapter 5: Correlation and regression

- Asked in: 2070 Chaitra, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/yield-temperature-regression-200c
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5

## Question

Observation on the yield of a chemical reaction taken at various temperatures was recorded as follows:
| X (°C) | 150 | 150 | 200 | 250 | 250 | 300 | 150 |
|---|---|---|---|---|---|---|---|
| Y% | 75.4 | 81.2 | 85.5 | 89 | 90.5 | 96.7 | 75.4 |
Fit a simple linear regression and estimate value of yield at 200°C.

## Short answer

ŷ = 58.581 + 0.1266X. At X=200°C, estimated yield ≈ 83.9%.

## Answer

Let X = temperature (°C), Y = yield (%), n = 7. Fit Ŷ = a + bX.

Sums (n = 7):

```text
        x        y       xy       x²       y²
      150     75.4    11310    22500  5685.16
      150     81.2    12180    22500  6593.44
      200     85.5    17100    40000  7310.25
      250       89    22250    62500     7921
      250     90.5    22625    62500  8190.25
      300     96.7    29010    90000  9350.89
      150     75.4    11310    22500  5685.16
---------------------------------------------
     1450    593.7   125785   322500 50736.15
```

Σx = 1450, Σy = 593.7, Σxy = 125785, Σx² = 322500, Σy² = 50736.15

Means: x̄ = 207.1429, ȳ = 84.8143

Sxx = Σx² − (Σx)²/n = 22142.8571
Sxy = Σxy − (Σx)(Σy)/n = 2804.2857
Syy = Σy² − (Σy)²/n = 381.9086

Slope b = Sxy/Sxx = 2804.2857/22142.8571 = 0.1266
Intercept a = ȳ − bx̄ = 84.8143 − (0.1266)(207.1429) = 58.5806

**Fitted line: Ŷ = 58.581 + 0.1266 X**   (r = 0.964)

Estimate at X = 200°C:

Ŷ = 58.5806 + 0.12665 × 200 = 58.5806 + 25.3290 = **83.91%**

**Answer: Ŷ = 58.58 + 0.1266X; the estimated yield at 200°C is about 83.9%.**

## Check yourself

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

1. Which is the model fitted?
   A. Temperature on yield
   B. Yield Y on temperature X: Y = a + bX
   C. Y = bX with no intercept
   D. Yield squared on X
   Answer: B. Yield depends on temperature so Ŷ = a + bX. (Key point: Model)

2. How is the slope obtained?
   A. b = Sxy/Sxx = 2804.29/22142.86 ≈ 0.1266
   B. b = Sxx/Sxy
   C. b = Sxy/Syy
   D. b = ȳ/x̄
   Answer: A. b = Sxy/Sxx ≈ 0.1266 % per °C. (Key point: Slope)

3. What is the estimated yield at 200°C?
   A. About 83.9%
   B. About 89.6%
   C. About 58.6%
   D. About 84.8%
   Answer: A. Ŷ = 58.58 + 0.1266(200) ≈ 83.9%. (Key point: Estimate)

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