Class Notes for 13^th and 15^th March

Reminder: Homework 4 on Design due next Wednesday (22^nd).

Monday 13^th Class

· Nonlinear models often result from compartmental models (biological “common sense”), and the parameters are usually very impt and interpretable (as compared with linear models)

· Need to give starting values, and that requires understanding the model function and sometimes some ingenuity

· Iterate to a solution using e.g. MGN method – results in parameter estimation, and then interpretation or prediction

· Rival model functions exist for the same dataset – e.g., SE2, MM2 (Michaelis Menton), and Lansky model functions all look very similar

· CI’s: two types – Wald (estimate +/- t*SE) is based on a parabolic approximation to the SSE or likelihood, and Likelihood-based. PLCI’s are often asymmetric, which makes more sense since usually our information about a parameter is asymmetric. Best to use PLCI’s, but they are a pain to find. The difference between WCI’s (Wald) and PLCI’s depends upon “curvature” – more on this later.

Wednesday 15^th Class

· Better understanding of MM2 model function parms, and how to give good starting values

· Ex. 2.4 – downward SE2 doesn’t fit (see residuals), but SE2 with a lag (“knot”) does fit: 95% WCI for knot is (25.16,46.19). Will PLCI (not given) be shifted R or shifted L?

· Ex. 2.6 – Fitting a (modified) LL4 model function for May and one for June; wish to test H₀: q_1M = q_1J, q_2M = q_2J, and q_3M = q_1J; tested using Full-and-Reduced F statistic,

F_3,24 = [(0.0206-0.0179)/3] / [0.0179/24] = 1.20,

which carries a p-value of 32.9%. We retain the claim of common upper and lower asymptotes and slopes for M and J.

· All our models so far are homoskedastic normal NLINs, but data in Ex. 2.7 show non-constant variance. Letting “rhs” denote the (mean) model function, we propose that VAR = s²*rhs^r, where r is an additional parameter to be estimated. The case where r = 0 is then constant variances across X. To test H₀: r = 0, we use Wald or LR. Wald gives t₅₅ = 1.4707/0.4699 = 3.13 and p = 0.0028. More reliable is the LR test c² = 254.0 – 245.3 = 8.7 and p = 0.0032. That Wald gives a similar p-value means quadratic approx. is good here. Regardless, we reject the null, and accept heteroskedasticity. One of the ramifications is that the SE for the LD50 drops from 0.3805 to 0.3297 (drops 13.4%).

· Ex. 2.8. Gerig data and Finney synergy model function.

z = x₁ + q₄*x₂ + q₅*SQRT{q₄*x₁*x₂}

is the effective dose, and this is related to Y = AGR using a dose-response function. q₄ is the relative potency of x₁ to x₂, and q₅ – which is the key parameter here – is the coefficient of synergy. q₅ < 0 indicates antagonism (between x₁ and x₂), q₅ = 0 indicates independent action, and q₅ > 0 indicates synergy. Wald test of H₀: q₅ = 0 gives t₁₁ = -0.8266/0.1830 = -4.52 and p = 0.00087. More reliable is the LR test F_1,11 = 6.68 and p = 0.0254. These data and this analysis suggest moderate evidence of significant antagonism between these two phenolic acids.