I have data of fixed effects: genotypes = C, E, K, M; age = 30, 45, 60, 75, 90 days; random effects: block = 1, 2, 3; and variable = weight_DM.
The file is in: https://drive.google.com/open?id=1_H6YZbdesK7pk5H23mZtp5KhVRKz0Ozl
I have the slopes linear and quadratic of ages for each genotype, but I do not have the intercepts and standard errors. The R code is:
library(nlme)
library(lme4)
library(car)
library(carData)
library(emmeans)
library(ggplot2)
library(Matrix)
library(multcompView)
datos_weight <- read.csv2("D:/investigacion/publicaciones/articulos-escribiendo/pennisetum/pennisetum-agronomicas/data_weight.csv",header=T, sep = ";", dec = ",")
parte_fija_3 <- formula(weight_DM
~ Genotypes
+ Age
+ I(Age^2)
+ Genotypes*Age
+ Genotypes*I(Age^2))
heterocedasticidad_5 <- varComb(varExp(form = ~fitted(.)))
correlacion_4 <- corCompSymm(form = ~ 1|Block/Genotypes)
modelo_43 <- gls(parte_fija_3,
weights = heterocedasticidad_5,
correlation = correlacion_4,
na.action = na.omit,
data = datos_weight)
anova(modelo_43)
#response
Denom. DF: 48
numDF F-value p-value
(Intercept) 1 597.3828 <.0001
Genotypes 3 2.9416 0.0424
Age 1 471.6933 <.0001
I(Age^2) 1 22.7748 <.0001
Genotypes:Age 3 5.9425 0.0016
Genotypes:I(Age^2) 3 0.7544 0.5253
#################################
#test whether the linear age slopes of each genotype is equal to zero
################################
(tendencias_em_lin <- emtrends(modelo_43,
"Genotypes",
var = "Age"))
#response
Genotypes Age.trend SE df lower.CL upper.CL
C 1.693331 0.2218320 48 1.247308 2.139354
E 1.459517 0.2135037 48 1.030239 1.888795
K 2.001097 0.2818587 48 1.434382 2.567811
M 1.050767 0.1301906 48 0.789001 1.312532
Confidence level used: 0.95
(tendencias_em_lin_prueba <- update(tendencias_em_lin,
infer = c(TRUE,TRUE),
null = 0))
#response
Genotypes Age.trend SE df lower.CL upper.CL t.ratio p.value
C 1.693331 0.2218320 48 1.247308 2.139354 7.633 <.0001
E 1.459517 0.2135037 48 1.030239 1.888795 6.836 <.0001
K 2.001097 0.2818587 48 1.434382 2.567811 7.100 <.0001
M 1.050767 0.1301906 48 0.789001 1.312532 8.071 <.0001
Confidence level used: 0.95
########################################
#test differences between slope of the age linear for each genotype
########################################
CLD(tendencias_em_lin,
adjust = "bonferroni",
alpha = 0.05)
#response
Genotypes Age.trend SE df lower.CL upper.CL .group
M 1.050767 0.1301906 48 0.7128801 1.388653 1
E 1.459517 0.2135037 48 0.9054057 2.013628 12
C 1.693331 0.2218320 48 1.1176055 2.269057 12
K 2.001097 0.2818587 48 1.2695822 2.732611 2
Confidence level used: 0.95
Conf-level adjustment: bonferroni method for 4 estimates
P value adjustment: bonferroni method for 6 tests
significance level used: alpha = 0.05
Questions
How to test whether the age intercepts of each genotype is equal to zero?
How to test differences between intercepts of the age for each genotype?
Which are the standard errors of the intercepts of each genotype?
Thanks for your help.
You can answer these questions by using emmeans() in similar ways to what you did with emtrends().
Also look at the documentation for summary.emmGrid, and note that you can choose whether to do CIs, tests, or both. e.g.,
emm <- emmeans(...)
summary(emm, infer = c(TRUE,TRUE))
summary(emm, infer = c(TRUE,FALSE)) # or confint(emm)
summary(emm, infer = c(FALSE,TRUE)) # or test(emm)
True intercepts
If in fact you want the actual y intercepts, you can do that using
emm <- emmeans(..., at = list(age = 0))
The, predictions are made at age 0, which are the intercepts in the regression equations for each set of conditions. However, I would like to try to dissuade you from doing that because this because (a) these predictions are huge extrapolations, hence their standard errors are huge as well; and (b) it makes no practical sense to predict responses at age 0. For that reason, I think question #1 is basically meaningless.
If you leave that at part out, then emmeans() makes predictions at the mean age in the dataset. Those predictions will have a much smaller standard error than the intercepts do. Since you have interactions involving age in the model, the predictions compare differently at each age. I suggest it would be useful to put
emm <- emmeans(..., cov.reduce = FALSE, by = "age")
which is equivalent to using at to specify the set of age values that occur in the dataset, and doing separate comparisons at each of those age values.
Related
Does anyone know how to extract the parameter estimates of random term when using the (1 | …) syntax in a glmer model (including se, t ratio and p value)? I’m only able to access the average variance and std. deviance with the summary function.
Some background: I used cohort and period random terms (both factorized), where period = each survey year, and cohort = 8 birth cohorts. My model empty model looks like this :
glmer(pid ~ age + age2 + (1 | cohort) + (1| period)
There's a bit of a conceptual problem with what you are doing. The random effects do not have the same standing in statistical theory as the fixed effects. You are not really supposed to be making inferences on their estimates since you don't have a random sampling from their overall population. Hence you need to make some unteseted assumptions on their distribution. That said, there are apparently times when you might want to do it but with care that you are not making unsupportable claims. See: https://stats.stackexchange.com/questions/392314/interpretation-of-fixed-effect-coefficients-from-glms-and-glmms .
Dimitris Rizopoulosthen responded to a request for the possibility of getting "an average" of the random effects conditional on the fixed effects (rather the flipped version of mixed models inference). He offered a function in his GLMM package:
https://drizopoulos.github.io/GLMMadaptive/articles/Methods_MixMod.html#marginalized-coefficients
This is his example ......
install.packages("GLMMadaptive"); library(GLMMadaptive)
set.seed(1234)
n <- 100 # number of subjects
K <- 8 # number of measurements per subject
t_max <- 15 # maximum follow-up time
# we constuct a data frame with the design:
# everyone has a baseline measurment, and then measurements at random follow-up times
DF <- data.frame(id = rep(seq_len(n), each = K),
time = c(replicate(n, c(0, sort(runif(K - 1, 0, t_max))))),
sex = rep(gl(2, n/2, labels = c("male", "female")), each = K))
# design matrices for the fixed and random effects
X <- model.matrix(~ sex * time, data = DF)
Z <- model.matrix(~ time, data = DF)
betas <- c(-2.13, -0.25, 0.24, -0.05) # fixed effects coefficients
D11 <- 0.48 # variance of random intercepts
D22 <- 0.1 # variance of random slopes
# we simulate random effects
b <- cbind(rnorm(n, sd = sqrt(D11)), rnorm(n, sd = sqrt(D22)))
# linear predictor
eta_y <- as.vector(X %*% betas + rowSums(Z * b[DF$id, ]))
# we simulate binary longitudinal data
DF$y <- rbinom(n * K, 1, plogis(eta_y))
#We continue by fitting the mixed effects logistic regression for y assuming random intercepts and random slopes for the random-effects part.
fm <- mixed_model(fixed = y ~ sex * time, random = ~ time | id, data = DF,
family = binomial())
.... and then the call to his marginal_coefs function.
marginal_coefs(fm, std_errors=TRUE)
Estimate Std.Err z-value p-value
(Intercept) -1.6025 0.2906 -5.5154 < 1e-04
sexfemale -1.0975 0.3676 -2.9859 0.0028277
time 0.1766 0.0337 5.2346 < 1e-04
sexfemale:time 0.0508 0.0366 1.3864 0.1656167
I'm running a hurdle lognormal model using the GLMMadaptive package in R. Both the continuous part as well as the zero-part have categorical variables defined in the fixed effects. I would like to run an ANOVA on these categorical variables to detect if there is a main effect.
I've seen that using the glmmTMB package you are able to separately run an ANOVA on the conditional model and the zero-part model separately, as is demonstrated here.
Is there a similar strategy available for the GLMMadaptive package? (The glmmTMB does not support hurdle lognormal models as far as I understood). Perhaps using the joint_tests function from the emmeans package? If so, how do you define that you want to test the zero-part model? As emmeans::joint_tests(hurdlemodel) only gives the F-tests for the conditional part of the model.
Or as an alternative method, could you compare the fit of the models where you exclude the variable of interest against a the full model, as is demonstrated for the relevance of random effects in this vignette?
Many thanks!
The suggestion by Russ Lenth in the comments are implemented below, using the data and model in the GLMMadaptive two-part model vignette:
library(GLMMadaptive)
library(emmeans)
# data generating code from the vignette:
{
set.seed(1234)
n <- 100 # number of subjects
K <- 8 # number of measurements per subject
t_max <- 5 # maximum follow-up time
# we construct a data frame with the design:
# everyone has a baseline measurement, and then measurements at random follow-up times
DF <- data.frame(id = rep(seq_len(n), each = K),
time = c(replicate(n, c(0, sort(runif(K - 1, 0, t_max))))),
sex = rep(gl(2, n/2, labels = c("male", "female")), each = K))
# design matrices for the fixed and random effects non-zero part
X <- model.matrix(~ sex * time, data = DF)
Z <- model.matrix(~ 1, data = DF)
# design matrices for the fixed and random effects zero part
X_zi <- model.matrix(~ sex, data = DF)
Z_zi <- model.matrix(~ 1, data = DF)
betas <- c(1.5, 0.05, 0.05, -0.03) # fixed effects coefficients non-zero part
shape <- 2 # shape/size parameter of the negative binomial distribution
gammas <- c(-1.5, 0.5) # fixed effects coefficients zero part
D11 <- 0.5 # variance of random intercepts non-zero part
D22 <- 0.4 # variance of random intercepts zero part
# we simulate random effects
b <- cbind(rnorm(n, sd = sqrt(D11)), rnorm(n, sd = sqrt(D22)))
# linear predictor non-zero part
eta_y <- as.vector(X %*% betas + rowSums(Z * b[DF$id, 1, drop = FALSE]))
# linear predictor zero part
eta_zi <- as.vector(X_zi %*% gammas + rowSums(Z_zi * b[DF$id, 2, drop = FALSE]))
# we simulate negative binomial longitudinal data
DF$y <- rnbinom(n * K, size = shape, mu = exp(eta_y))
# we set the extra zeros
DF$y[as.logical(rbinom(n * K, size = 1, prob = plogis(eta_zi)))] <- 0
}
#create categorical time variable
DF$time_categorical[DF$time<2.5] <- "early"
DF$time_categorical[DF$time>=2.5] <- "late"
DF$time_categorical <- as.factor(DF$time_categorical)
#model with interaction in fixed effects zero part and adding nesting in zero part as in model above
km3 <- mixed_model(y ~ sex * time_categorical, random = ~ 1 | id, data = DF,
family = hurdle.lognormal(), n_phis = 1,
zi_fixed = ~ sex * time_categorical, zi_random = ~ 1 | id)
#### ATTEMPT at QDRG function in emmeans ####
coef_zero_part <- fixef(km3, sub_model = "zero_part")
vcov_zero_part <- vcov(km3)[9:12,9:12]
qd_km3 <- emmeans::qdrg(formula = ~ sex * time_categorical, data = DF,
coef = coef_zero_part, vcov = vcov_zero_part)
Output:
> joint_tests(qd_km3)
model term df1 df2 F.ratio p.value
sex 1 Inf 11.509 0.0007
time_categorical 1 Inf 0.488 0.4848
sex:time_categorical 1 Inf 1.077 0.2993
> emmeans(qd_km3, pairwise ~ sex|time_categorical)
$emmeans
time_categorical = early:
sex emmean SE df asymp.LCL asymp.UCL
male -1.592 0.201 Inf -1.99 -1.198
female -1.035 0.187 Inf -1.40 -0.669
time_categorical = late:
sex emmean SE df asymp.LCL asymp.UCL
male -1.914 0.247 Inf -2.40 -1.429
female -0.972 0.188 Inf -1.34 -0.605
Confidence level used: 0.95
$contrasts
time_categorical = early:
contrast estimate SE df z.ratio p.value
male - female -0.557 0.270 Inf -2.064 0.0390
time_categorical = late:
contrast estimate SE df z.ratio p.value
male - female -0.942 0.306 Inf -3.079 0.0021
Checking if contrasts correspond with zero-part fixed effects:
> fixef(km3, sub_model = "zero_part")
(Intercept) sexfemale time_categoricallate sexfemale:time_categoricallate
-1.5920415 0.5568072 -0.3220390 0.3849780
> (-1.5920) - (-1.5920 + 0.5568)
[1] -0.5568 #matches contrast within "early" level of "time_categorical"
> (-1.5920 + -0.3220) - (-1.5920 + -0.3220 + 0.5568 + 0.3850)
[1] -0.9418 #matches contrast within "late" level of "time_categorical"
The function emmeans::qdrg() can sometimes be used to create the needed object for a model not directly supported by emmeans. See its documentation. In very simple models (e.g., inheriting from lm, it may be enough to supply the object and data arguments.
That usually does not work for more sophisticated models, in which case
you will need to specify data, the fixed-effects formula for the conditional or zero part of the model, and the associated regression coefficients (coef) and variance-covariance matrix (vcov) for the part of the model in question. Often with models like this with multiple components, you likely will have to pick a subset of the coefficients and covariance matrix. These all must conform: the length of coef must equal the number of rows and columns of vcov and the number of columns in the model matrix generated by formula [which may be checked via model.matrix(formula, data = data)].
qdrg() will not work for a multivariate model -- or at least it's tricky -- because the implied model involves other factor(s) that delineate the levels of the multivariate response. If there are special provisions for, say, spline smoothing, that is another instance where qdrg() probably can't be made to work.
Once qdrg() actually runs and produces results, it is a good idea to use it to estimate some contrasts that are estimated by the model parameterization. For example, suppose that the model was fitted with the default contr.treatment contrasts. Then the regression coefficients are interpretable as a comparison with the first level as a reference level. Accordingly, if we computedrg <- qdrg(...), and one of the factors is "treat", look at contrast(rg, "trt.vs.ctrl1", simple = "treat"), and check to see if the first set of estimated contrasts matches the main-effect estimates for treat.
I will illustrate all of this with a simple lm model, ignoring the fact that it is already supported by emmeans.
> warp.lm <- lm(breaks ~ wool * tension, data = warpbreaks)
Here is the reference grid
> rg <- qdrg(~ wool * tension, coef = coef(warp.lm), vcov = vcov(warp.lm),
+ df = df.residual(warp.lm), data = warpbreaks)
Here is a sanity check -- First, look at the model summary:
> summary(warp.lm)$coef
Estimate Std. Error t value Pr(>|t|)
(Intercept) 44.55556 3.646761 12.217842 2.425903e-16
woolB -16.33333 5.157299 -3.167032 2.676803e-03
tensionM -20.55556 5.157299 -3.985721 2.280796e-04
tensionH -20.00000 5.157299 -3.877999 3.199282e-04
woolB:tensionM 21.11111 7.293523 2.894501 5.698287e-03
woolB:tensionH 10.55556 7.293523 1.447251 1.543266e-01
Second, look at selected contrasts:
> contrast(rg, "trt.vs.ctrl1", simple = "wool")
tension = L:
contrast estimate SE df t.ratio p.value
B - A -16.33 5.16 48 -3.167 0.0027
tension = M:
contrast estimate SE df t.ratio p.value
B - A 4.78 5.16 48 0.926 0.3589
tension = H:
contrast estimate SE df t.ratio p.value
B - A -5.78 5.16 48 -1.120 0.2682
> contrast(rg, "trt.vs.ctrl1", simple = "tension")
wool = A:
contrast estimate SE df t.ratio p.value
M - L -20.556 5.16 48 -3.986 0.0005
H - L -20.000 5.16 48 -3.878 0.0006
wool = B:
contrast estimate SE df t.ratio p.value
M - L 0.556 5.16 48 0.108 0.9863
H - L -9.444 5.16 48 -1.831 0.1338
P value adjustment: dunnettx method for 2 tests
Comparing with the regression coefficients, we do confirm that the first contrast for wool is estimated as -16.33, matching the regression coefficient for woolB. Also, the first set of contrasts for tension are estimated as -20.556 and -20.0, matching the regression coefficients for tensionM and tensionH. The SEs and t ratios match as well. (The P values for the second set do not match due to the multiplicity adjustment.)
I built a linear mixed model and did a post hoc test for it. Fixed factors are the phase numbers (time) and the group.
statistic_of_comp <- function (x, df) {
x.full.1 <- lmer(x ~ phase_num + group + (1|mouse), data=df, REML = FALSE)
x_phase.null.1 <- lmer(x ~ group + (1|mouse), data=df, REML = FALSE)
print(anova (x.full.1, x_phase.null.1))
summary(glht(x.full.1, linfct=mcp(phase_num="Tukey")))
}
Now my problem is, that I want to do a post hoc test with more than one fixed factor. I found the following
linfct=mcp(phase_num="Tukey", group="Tukey)
but that doesn't give the result I want. At the moment I get the comparison for the groups with Tukey (every group with every other group) and the comparison between the two phases.
What I want is a comparison of the phase_numbers for every group.
e.g. group1 phase1-phase2 ..., group2 phase1-phase2 etc.
I'm sure you can do this with multcomp, but let me illustrate how to do it with the emmeans package. I'm going to use a regular linear model (since you haven't given a reproducible example), but the recipe below should work just as well with a mixed model.
Linear model from ?emmeans (using a built-in data set):
warp.lm <- lm(breaks ~ wool * tension, data = warpbreaks)
Apply emmeans(), followed by the pairs() function:
pairs(emmeans(warp.lm , ~tension|wool))
wool = A:
contrast estimate SE df t.ratio p.value
L - M 20.556 5.16 48 3.986 0.0007
L - H 20.000 5.16 48 3.878 0.0009
M - H -0.556 5.16 48 -0.108 0.9936
wool = B:
contrast estimate SE df t.ratio p.value
L - M -0.556 5.16 48 -0.108 0.9936
L - H 9.444 5.16 48 1.831 0.1704
M - H 10.000 5.16 48 1.939 0.1389
For more information, see ?pairs.emmGrid or vignette("comparisons",package="emmeans") (which clarifies that these tests do indeed use Tukey comparisons by default).
I'm doing logistic regression on Boston data with a column high.medv (yes/no) which indicates if the median house pricing given by column medv is either more than 25 or not.
Below is my code for logistic regression.
high.medv <- ifelse(Boston$medv>25, "Y", "N") # Applying the desired
`condition to medv and storing the results into a new variable called "medv.high"
ourBoston <- data.frame (Boston, high.medv)
ourBoston$high.medv <- as.factor(ourBoston$high.medv)
attach(Boston)
# 70% of data <- Train
train2<- subset(ourBoston,sample==TRUE)
# 30% will be Test
test2<- subset(ourBoston, sample==FALSE)
glm.fit <- glm (high.medv ~ lstat,data = train2, family = binomial)
summary(glm.fit)
The output is as follows:
Deviance Residuals:
[1] 0
Coefficients: (1 not defined because of singularities)
Estimate Std. Error z value Pr(>|z|)
(Intercept) -22.57 48196.14 0 1
lstat NA NA NA NA
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 0.0000e+00 on 0 degrees of freedom
Residual deviance: 3.1675e-10 on 0 degrees of freedom
AIC: 2
Number of Fisher Scoring iterations: 21
Also i need:
Now I'm required to use the misclassification rate as the measure of error for the two cases:
using lstat as the predictor, and
using all predictors except high.medv and medv.
but i am stuck at the regression itself
With every classification algorithm, the art relies on choosing the threshold upon which you will determine whether the the result is positive or negative.
When you predict your outcomes in the test data set you estimate probabilities of the response variable being either 1 or 0. Therefore, you need to the tell where you are gonna cut, the threshold, at which the prediction becomes 1 or 0.
A high threshold is more conservative about labeling a case as positive, which makes it less likely to produce false positives and more likely to produce false negatives. The opposite happens for low thresholds.
The usual procedure is to plot the rates that interests you, e.g., true positives and false positives against each other, and then choose what is the best rate for you.
set.seed(666)
# simulation of logistic data
x1 = rnorm(1000) # some continuous variables
z = 1 + 2*x1 # linear combination with a bias
pr = 1/(1 + exp(-z)) # pass through an inv-logit function
y = rbinom(1000, 1, pr)
df = data.frame(y = y, x1 = x1)
df$train = 0
df$train[sample(1:(2*nrow(df)/3))] = 1
df$new_y = NA
# modelling the response variable
mod = glm(y ~ x1, data = df[df$train == 1,], family = "binomial")
df$new_y[df$train == 0] = predict(mod, newdata = df[df$train == 0,], type = 'response') # predicted probabilities
dat = df[df$train==0,] # test data
To use missclassification error to evaluate your model, first you need to set up a threshold. For that, you can use the roc function from pROC package, which calculates the rates and provides the corresponding thresholds:
library(pROC)
rates =roc(dat$y, dat$new_y)
plot(rates) # visualize the trade-off
rates$specificity # shows the ratio of true negative over overall negatives
rates$thresholds # shows you the corresponding thresholds
dat$jj = as.numeric(dat$new_y>0.7) # using 0.7 as a threshold to indicate that we predict y = 1
table(dat$y, dat$jj) # provides the miss classifications given 0.7 threshold
0 1
0 86 20
1 64 164
The accuracy of your model can be computed as the ratio of the number of observations you got right against the size of your sample.
In R, I am searching for a way to estimate confidence intervals for linear contrasts for lmer models that use either kenward-rogers or satterthwaite degrees of freedom and SE.
For example, I can compute a CI for a fixed effect parameter in a mixed model like SAS with R, using the t-value (with df from KR) and SE.
mod<-lmerTest::lmer(y~time1+treatment+time1:treatment+(1|PersonID),data=data)
lmerTest::summary(mod,ddf = "Kenward-Roger")
This output:
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 49.0768 1.0435 56.4700 47.029 < 2e-16 ***
time1 5.8224 0.5963 48.0000 9.764 5.51e-13 ***
treatment 1.6819 1.4758 56.4700 1.140 0.2592
time1:treatment 2.0425 0.8433 48.0000 2.422 0.0193 *
Allows a CI for time1 like:
5.8224+abs(qt(0.05/2, 48))*0.5963 #7.021342
5.8224-abs(qt(0.05/2, 48))*0.5963 #4.623458
I would like to do this same thing for a linear contrast of the fixed coefficients. This is the p-value but there is no SE output.
pbkrtest::KRmodcomp(mod,matrix(c(0,0,1,0),nrow = 1))
stat ndf ddf F.scaling p.value
Ftest 1.2989 1.0000 56.4670 1 0.2592
Is there anyway to get a SE or a CI from lmer linear contrasts that uses this type of df?
For this, you have at least two options: using the lsmeans package, or doing it manually (using functions vcovAdj.lmerMod and pbkrtest::get_Lb_ddf). Personally, I go with the later if the contrast to be tested is not very "simple", because I find the syntax in lsmeans a bit complicated.
To exemplify, take the following model:
library(pbkrtest)
library(lme4)
library(nlme) # for the 'Orthodont' data
# 'age' is a numeric variable, while 'Sex' and 'Subject' are factors
model <- lmer(distance ~ age : Sex + (1 | Subject), data = Orthodont)
Linear mixed model fit by REML ['lmerMod']
Formula: distance ~ age:Sex + (1 | Subject)
…
Fixed Effects:
(Intercept) age:SexMale age:SexFemale
16.7611 0.7555 0.5215
from which we would like to obtain stats on the difference between the coefficients for age in males and females (i.e., age:SexMale - age:SexFemale).
Using lsmeans:
library(lsmeans)
# Evaluate the contrast at a value of 'age' set to 1,
# so that the resulting value is equal to the regression coefficient
lsm = lsmeans(model, pairwise ~ age : Sex, at = list(age = 1))$contrast
produces:
contrast estimate SE df t.ratio p.value
1,Male - 1,Female 0.2340135 0.06113276 42.64 3.828 0.0004
Alternatively, doing the calculation manually:
# Specify the contrasts: age:SexMale - age:SexFemale
# Must have the same order as the fixed effects in the model
K = c("(Intercept)" = 0, "age:SexMale" = 1, "age:SexFemale" = -1)
# Retrieve the adjusted variance-covariance matrix, to calculate the SE
V = pbkrtest::vcovAdj.lmerMod(model, 0)
# Point estimate, SE and df
point_est = sum(K * fixef(model))
SE = sqrt(sum(K * (V %*% K)))
df = pbkrtest::get_Lb_ddf(model, K)
alpha = 0.05 # significance level
# Calculate confidence interval for the difference between the 'age' coefficients for males and females
Delta_age_CI = point_est + SE * qt(c(0.5 * alpha, 1 - 0.5 * alpha), df)
will result in a point estimate equal to 0.2340135, SE 0.06113276, df 42.63844, and confidence interval [0.1106973, 0.3573297]