Failed to contrast intercepts through emmeans in R

Question

I would like to test the simetry in the response of an observer to a contrast stimuli with different polarity, positive (white) and negative (black). I took the reaction time (RT) as dependent variable, along four different contrasts. It is known that the response time follows a Pieron curve whose asymptotas are placed (1) at observer threshold (Inf) and (2) at a base RT placed somewere between 250 and 450 msec. The knowledge allows us to linearize the relationship transforming the independent variable (effective contrast EC) as 1/EC^2 (tEC), so the equation linking RT to EC becomes:

RT = m * tEC + RT0

To test the symmetry I established the criteria: same slope and same intercept in the two polarities implies symmetry. To obtain the coefficients I made a linear model with interaction (coding trough a dummy variable for Polarity: Positive or Negative). The output of lm is clear to me, but some colegues prefer somthing more similar to an ANOVA output. So I decided to use emmeans to make the contrasts. With the slope is all right, but when computing the interceps starts the problem. The intercepts computed by lm are very different from the output of emmeans, and the conclusions are also different. In what follows I reproduce the example. The question is two fold: It is possible to use emmeans to solve my problem? If not, it is possible to make the contrasts through other packages (which one)?

Data

RT1000	EC	tEC	Polarity
596.3564	-25	0.001600	Negative
648.2471	-20	0.002500	Negative
770.7602	-17	0.003460	Negative
831.2971	-15	0.004444	Negative
1311.3331	15	0.004444	Positive
1173.8942	17	0.003460	Positive
1113.7240	20	0.002500	Positive
869.3635	25	0.001600	Positive

Code

# Model
model <- lm(RT1000 ~ tEC * Polarity, data = Data)

# emmeans
library(emmeans)
# Slopes
m.slopes <- lstrends(model, "Polarity", var="tEC")
# Intercepts
m.intercept <- lsmeans(model, "Polarity")

# Contrasts
pairs(m.slopes)
pairs(m.intercept)

Outputs

Modelo

term	estimate	std.error	statistic	p.value
(Intercept)	449.948	66.829	6.733	0.003
tEC	87205.179	20992.976	4.154	0.014
PolarityPositive	230.946	94.511	2.444	0.071
tEC:PolarityPositive	58133.172	29688.551	1.958	0.122

Slopes (it is all right)

Polarity	tEC.trend	SE	df	lower.CL	upper.CL
Negative	87205.18	20992.98	4	28919.33	145491.0
Positive	145338.35	20992.98	4	87052.51	203624.2

contrast	estimate	SE	df	t.ratio	p.value
Negative - Positive	-58133.17	29688.55	4	-1.958101	0.12182

Intercepts (problem)

Polarity	lsmean	SE	df	lower.CL	upper.CL
Negative	711.6652	22.2867	4	649.7874	773.543
Positive	1117.0787	22.2867	4	1055.2009	1178.957

contrast	estimate	SE	df	t.ratio	p.value
Negative - Positive	-405.4135	31.51816	4	-12.86285	0.000211

Computed intercepts through emmeans differs from the ones computed by lm. I think the problem is that the model is not defined for EC = 0. But I'm not sure.

Answer 1

What you are calling the intercepts are not; they are the model predictions at the mean value of tEC. If you want the intercepts, use instead:

m.intercept <- lsmeans(model, "Polarity", at = list(tEC = 0))

You can tell what reference levels are being used via

ref_grid(model)     # or   str(m.intercept)

Please note that the model fitted here consists of two lines with different slopes; hence the difference between the predictions changes depending on the value of tEC. Thus, I would strongly recommend against testing the comparison of the intercepts; those are predictions at a tEC value that, as you say, can't even occur. Instead, try to be less of a mathematician and do the comparisons at a few representative values of tEC, e.g.,

LSMs <- lsmeans(model, "Polarity", at = list(tEC = c(0.001, 0.003, 0.005)))
pairs(LSMs, by = tEC)

You can also easily visualize the fitted lines:

emmip(model, Polarity ~ tEC, cov.reduce = range)