Motivator: Novel metabolic role for BDNF in pancreatic β-cell insulin secretion
I’ll finish this some day…
knitr::opts_chunk$set(echo = TRUE, message=FALSE) library(tidyverse) library(data.table) library(mvtnorm) library(lmerTest) normal response niter <- 2000 n <- 9 treatment_levels <- c("cn", "high", "high_bdnf") insulin <- data.table(treatment = rep(treatment_levels, each=n)) X <- model.matrix(~ treatment, data=insulin) beta <- c(0,0,0) # no effects # the three responses are taken from the same cluster of cells and so have expected # correlation rho.
A skeletal response to a twitter question:
“ANOVA (time point x group) or ANCOVA (group with time point as a covariate) for intervention designs? Discuss.”
follow-up “Only 2 time points in this case (pre- and post-intervention), and would wanna basically answer the question of whether out of the 3 intervention groups, some improve on measure X more than others after the intervention”
Here I compare five methods using fake pre-post data, including
Setup Import Models as nested using “tank” nested within “room” as two random intercepts (using lme4 to create the combinations) A safer (lme4) way to create the combinations of “room” and “tank”: as two random intercepts using “tank2” Don’t do this This is a skeletal post to show the equivalency of different ways of thinking about “nested” factors in a mixed model. The data are measures of life history traits in lice that infect salmon.
Background Comparing marginal effects to main effect terms in an ANOVA table First, some fake data Comparison of marginal effects vs. “main” effects term of ANOVA table when data are balanced Comparison of marginal effects vs. “main” effects term of ANOVA table when data are unbalanced When to estimate marginal effects keywords: estimation, ANOVA, factorial, model simplification, conditional effects, marginal effects
Background I recently read a paper from a very good ecology journal that communicated the results of an ANOVA like that below (Table 1) using a statement similar to “The removal of crabs strongly decreased algae cover (\(F_{1,36} = 17.
This doodle was motivated Jake Westfall’s answer to a Cross-Validated question.
The short answer is yes but most R scripts that I’ve found on the web are unsatisfying because only the t-value reproduces, not the df and p-value. Jake notes the reason for this in his answer on Cross-Validated.
To get the adjusted df, and the p-value associated with this, one can use the emmeans package by Russell Lenth, as he notes here.
Fig 1C of the Replication Study: Melanoma exosomes educate bone marrow progenitor cells toward a pro-metastatic phenotype through MET uses an odd (to me) three stage normalization procedure for the quantified western blots. The authors compared blot values between a treatment (shMet cells) and a control (shScr cells) using GAPDH to normalize the values. The three stages of the normalization are
first, the value for the Antibody levels were normalized by the value of a reference (GAPDH) for each Set.
Motivation: https://pubpeer.com/publications/8DF6E66FEFAA2C3C7D5BD9C3FC45A2#2 and https://twitter.com/CGATist/status/1175015246282539009
tl;dr: Given the transformation done by the authors, for any response in day_0 that is unusually small, there is automatically a response in day_14 that is unusually big and vice-versa. Consequently, if the mean for day_0 is unusually small, the mean for day_14 is automatically unusually big, hence the elevated type I error with an unpaired t-test. The transformation is necessary and sufficient to produce the result (meaning even in conditions where a paired t-test isn’t needed, the transformation still produces elevated Type I error).
