Ideal Straight Line Fit for a Temperature Sensor Output

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SUMMARY

The discussion focuses on calculating the ideal straight line fit for a temperature sensor output using linear regression. The slope (M) was calculated as 0.94 and the y-intercept (b) as 118.33, based on the provided data points for temperature and resistance. Participants emphasized the importance of using correct units for temperature and resistance, as well as the need for precise calculations without rounding errors. Additionally, online regression tools, such as the one available at socscistatistics.com, were recommended for verifying results.

PREREQUISITES
  • Understanding of linear regression concepts
  • Familiarity with statistical calculations such as sums and averages
  • Knowledge of temperature measurement units (ºC, ºF, K)
  • Experience with data visualization techniques, such as creating tables
NEXT STEPS
  • Explore linear regression analysis using Python libraries like NumPy and SciPy
  • Learn how to utilize online regression tools for statistical validation
  • Study the significance of units in scientific measurements and calculations
  • Investigate best practices for presenting data in tabular formats
USEFUL FOR

Data analysts, engineers working with temperature sensors, and anyone involved in statistical modeling and data presentation will benefit from this discussion.

Fenrir
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Homework Statement
I am doing a distance learning course and need to find the ideal straight line of a sensor output. The sensor is a temperature sensor with an OHM output.
From a temperature range of 0-250 it has an output of 120-364ohm.
0 = 120
50 = 178
00 = 201
150 = 249
200 = 303
250 = 364

I need to generate the ideal straight line equation and values.
Relevant Equations
y = Mx + b (Least squares)
X 0 50 100 150 200 250
Y 120 178 201 249 303 364
XY 0 8900 20100 37350 60600 91000
X^2 0 2500 10000 22500 40000 62500

∑X 750
∑Y 1415
∑XY 217950
∑X^2 137500

M = (6*217950)-(750*1415) / (6*137500) - (750)^2
M = 1643 / 1750
M = 0.9388571429
M = 0.94

b = 1415-0.94*750 / 6 = 118.333333
b = 118.33

Mx + b
0.94*0 + 118.33 = 118.33
0.94*50 +118.33 = 165.33
0.94*100 +118.33 = 212.33
0.94*150 +118.33 = 259.33
0.94*200 +118.33 = 306.33
0.94*250 +118.33 = 353.33

N = Resistance - ISL

120-118.33 = 1.67
178-165.33 = 12.67
201-212.33 = -11.33
249-259.33 = -10.33
303-306.33 = -3.33
364-353.33 = -10.67

Temp 0 50 100 150 200 250
Resistance 120 178 201 249 303 364
ISL 118.33 165.33 212.33 259.33 306.33 353.33
N 1.67 12.67 -11.33 -10.33 -3.33 10.67

I'm looking at my answers and i feel like i have gotten it wrong somewhere, is there a mistake in what i have done?
Rounding only done here to simplify.

Edited to correct ∑X^2
 
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That looks ok. However, I would not have rounded M for the calculation of b.
 
In addition to what @DrClaude said...

There is online software available, e.g. https://www.socscistatistics.com/tests/regression/default.aspx
You can compare your answer to that of the software.

Some other points:

Fenrir said:
... temperature sensor with an OHM output.
It's better to say that the output is the sensor's resistance. (The 'ohm' (##\Omega##) is the unit of resistance.)

Fenrir said:
From a temperature range of 0-250
Units are needed. For example, the temperature could be in units of ºC (degrees centigrade), ºF (degrees Farhenheit) or K (kelvin). Knowing the correct units is very important.

You may be expected to give M and b with their units.

Fenrir said:
0 = 120
.
250 = 364
It's not a good idea to write things like '0 = 120'. It's wrong. Ideally, use a table with suitable column headings.
 
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