Resource Lesson
Linear Regression and Data Analysis Methods
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When you enter your data from an experiment in an EXCEL spreadsheet, EXCEL will provide you with a regression line. This is a line of best fit or trend line. It approximates the best linear representation for the data which you have entered. In some cases, none of your data points may actually fall on the line; while in others, the line will pass through practically every data point. A statistical method of determining the precision of your data is the correlation coefficient, R
^{2}
. The closer its value is to 1.000, the better the internal consistency within your data.
The equation of this regression line is
d = 2.5t + 6
The correlation coefficient is 0.9833 which is acceptable but would have been better had points 2 and 5 deviated less from the line.
The equation of this regression line is
P = 0.37T - 1
The correlation coefficient of 1.0000 is phenomenal! Notice that the extrapolated line almost passes through T = 0 K, the definition of Absolute Zero.
Formulas and trend line equations
Another aspect of regression lines is the relationship between their equations and the theoretical equations for the physical quantities being graphed.
In the case of the first graph,
d = 2.5t + 6
, the physics behind this behavior states that "distance" equals "rate" times "time;" that is,
d = rt
. Therefore, when the two equations are compared, the coefficient of
t
must represent the rate, or 2.5 m/sec.
d
=
r
t
d
=
2.5
t
+ 6
Thus, on the average, the object moved 2.5 meters each second. The significance of the y-axis intercept is the object's initial position. That is, when time t = 0 sec the object started at a position 6 meters above the "origin."
In the case of the second graph,
P = 0.37T - 1
. The physics behind this behavior is the ideal gas law which states that
PV = nRT
; where,
P
is the pressure in Pascals,
V
is the volume in m
^{3}
,
T
is the temperature in Kelvin, and
n
is the number of moles present in the sample. Solving for
P
reveals that
P = (nR/V) T
. Comparing this with the equation of our line shows that the numerical value of the slope equals the expression nR/V.
PV = nRT
P
= (
nR/V
)
T
P
=
0.37
T
- 1
Since our formula demands that pressure be measured in Pa, and our data is measured in kPa, we must convert 0.37 kPa to 0.37 x 10
^{3}
Pa or 370 Pa. Setting this equivalent numerical value for the slope equal to the expression nR/V yields the following expression for
V
.
370 = nR/V
370 V = nR
V = nR/370
If we now substitute in that 1 mole was present and R equals 8314 J/mole K, we can discover that the volume of gas present was
V = nR/370
V = (1)(8314)/(370)
V = 22.5 m
^{3}
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