Validation - Statistics

[bartjoosen]

If you use a regression line through the points used in the excel file of JA, combining the 2 injections, plotted versus the concentration, without any weighing, you have a non significant intercept (t-test).

But in an ideal world, you first have to prove that the data is homoscedastic.
If your data is heteroscedastic, you have to use weighing or log, x^2, 1/x.... to make sure your data is homoscedastic.

If your intercept is significant, you can use a calibration curve instead of a single point calibration. If you use a log-log scale, your residuals reduce by a great amount.

[JI2002]

If y=ax^b and you use log on both sides of the equation, then you have

logy = loga + b logx

so the intercept in logy ~ logx curve (loga) is the log of the slope in y ~ x curve. The slope in log y ~ logx curve (b) should be close to 1.00(between 0.98 to 1.02) if the ralationship between y and x is linear. Otherwise y ~ x curve is considered quadratic. In your case, the slope was 0.9925.

If y = ax^b + c, which means y ~ x curve has a significant intercept,
then logy = log (ax^b + c), there is no linear ralationship between logy and logx.