an exercise in regression. i borrow thermometers from the science teachers and bring a large...
TRANSCRIPT
HOT CHOCOLATE COOLINGAn Exercise in Regression
TIME VS. TEMPERATURE
FIRST, COLLECT THE DATA…I borrow thermometers from the science teachers and bring a LARGE container of hot water. Reserve the hot chocolate mix for drinking, not temperature taking…
Hot Chocolate Cooling
Time Tempre... <new>
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0 147.2
1 144.7
2 142.2
3 140.2
4 138.0
5 136.2
6 133.9
7 131.9
8 130.3
9 128.8
10 127.4
11 125.8
12 124.7
13 123.3
14 122.0
15 120.6
16 119.3
17 118.4
18 117.2
19 115.9
20 115.0
NEXT, PLOT THE DATA…I have students use fathom or, more often, calculators to graph the data & calculate the r value & the regression equation.
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Time
0 2 4 6 8 10 12 14 16 18 20 22
Temperature = -1.57Time + 144.4; r2 = 0.98
Hot Chocolate Cooling Scatter Plot
JUST A HINT – THIS WATER WAS MUCH TOO HOT!You probably notice a
curve in the scatterplot – that curve is MUCH less noticeable if you start the temperature a bit cooler. Somewhere around 100º F is about right…
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Time
0 2 4 6 8 10 12 14 16 18 20 22
Temperature = -1.57Time + 144.4; r2 = 0.98
Hot Chocolate Cooling Scatter Plot
THEN, TALK ABOUT THE RESULTS…r2 = 0.98 – what does this mean?
What is r ?We can extrapolate!The equation from this model predicts the temperature after 2 hours to be -44º F. Is this a problem?
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Time
0 2 4 6 8 10 12 14 16 18 20 22
Temperature = -1.57Time + 144.4; r2 = 0.98
Hot Chocolate Cooling Scatter Plot
BUT YOU MUST INCLUDE…Even if you start at a lower temperature where the curve isn’t noticeable, and even with the high r value (r=-0.9899), the context and the residual plot clearly shows a non-linear pattern!
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Time
0 2 4 6 8 10 12 14 16 18 20 22
Temperature = -1.57Time + 144.4; r2 = 0.98
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Time
Hot Chocolate Cooling Scatter Plot
TRANSFORMATIONS ANYONE?