Consider following survival times of 25 patients with no history of chronic diesease (chr = 0) and 25 patients with history of chronic disease (chr = 1). + indicates right-censored times.
The survival probability at each time point is calculated using the Kaplan-Meier estimator. The formula for the survival probability at time \(t_i\) is:
\[
\mbox{Survival probability at time } t_i = \mbox{Survival probability at time } t_{i-1} \times \left(1 - \frac{\mbox{Number of events at time } t_i}{\mbox{Number at risk at time } t_i}\right)
\]
Display the Kaplan-Meier survival curves for groups 1 (chr = 0) and 2 (chr = 1).
Plot of Kaplan-Meier survival curves for groups 1 and 2:
Write down the log-likelihood of the parametric exponential (proportional hazard) model for survival times. Explain why this model can be fit as a generalized linear model with offset.
Suppose a datum of a survival data set is \((y_j, \delta_j, \mathbf{x}_j)\) where
\(y_j\) is the survival time,
\(\delta_j\) is the censoring indicator with \(\delta_j = 1\) if the survival time is uncensored and \(\delta_j=0\) if it is censored, and
\(\mathbf{x}_j\) is the predictor vector.
Let \(y_1, \ldots, y_r\) be the uncensored observations and \(y_{r+1}, \ldots, y_n\) the censored ones. The likelihood is \[
L = \prod_{j=1}^r f(y_j) \times \prod_{j=r+1}^n S(y_j) = \prod_{j=1}^n f(y_j)^{\delta_j} S(y_j)^{1 - \delta_j}
\] so the log-likelihood is \[
\ell = \sum_{i=1}^n [\delta_j \log f(y_j) + (1 - \delta_j) \log S(y_j)] = \sum_{j=1}^n [\delta_j \log h(y_j) + \log S(y_j)].
\]
Esimtation by MLE.
In the context of proportional hazard model, the proportional hazard model is modeled with \(\theta = e^{\mathbf{x}^T \boldsymbol{\beta}}\). Therefore, the log-likelihood of the parametric exponential (proportional hazard) model for survival times is \[
\ell(\boldsymbol{\beta}) = \sum_j (\delta_j \cdot \mathbf{x}_j^T \boldsymbol{\beta} - y_j e^{\mathbf{x}_j^T \boldsymbol{\beta}}) = \sum_j \left\{ \delta_j \cdot [\mathbf{x}_j^T \boldsymbol{\beta} + \log(y_j)] - y_j e^{\mathbf{x}_j^T \boldsymbol{\beta}} - \delta_j \log(y_j) \right\}.
\]
Explanation of why this model can be fit as a generalized linear model with offset:
The exponential model can be fit as a generalized linear model with offset because the log-likelihood of the exponential model is linear in the parameters \(\boldsymbol{\beta}\) and the exponential term \(e^{\mathbf{x}_j^T \boldsymbol{\beta}}\) is the link function. The offset term \(\log(y_j)\) is used to account for the fact that the hazard rate is proportional to the survival time, which is not associated with the parameters \(\boldsymbol{\beta}\). An offset is a term that is added to the linear predictor without being associated with any parameter that needs to be estimated. Incorporating the offset term allows the exponential proportional hazard model to be expressed within the GLM framework, facilitating its estimation using standard GLM techniques with the inclusion of an offset term.
Fit the exponential (proportional hazard) model on the chr data using R. Interpret the coefficients.
glm(status ~ group +offset(log(time)),family = poisson,data = data) |>summary()
Call:
glm(formula = status ~ group + offset(log(time)), family = poisson,
data = data)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.0138 0.2000 -10.069 <2e-16 ***
groupGroup 2 0.3499 0.2828 1.237 0.216
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 16.657 on 49 degrees of freedom
Residual deviance: 15.135 on 48 degrees of freedom
AIC: 119.13
Number of Fisher Scoring iterations: 5
Using the survreg function in R to verify the results:
Call:
survreg(formula = Surv(time, status) ~ group, data = data, dist = "exponential")
Value Std. Error z p
(Intercept) 2.014 0.200 10.07 <2e-16
groupGroup 2 -0.350 0.283 -1.24 0.22
Scale fixed at 1
Exponential distribution
Loglik(model)= -141.9 Loglik(intercept only)= -142.7
Chisq= 1.52 on 1 degrees of freedom, p= 0.22
Number of Newton-Raphson Iterations: 4
n= 50
Interpretation:
The coefficients for the intercept is -2.0138, which represents the log hazard rate for group 1. The hazard rate for group 1 can be calculated as \(\exp(-2.0138) = 0.133\), which means that the hazard of event in group 1 is 0.133.
The coefficient for groupGroup 2 is 0.3499, which represents the difference in log hazard rate between group 2 and group 1. The hazard ratio between group 2 and group 1 can be calculated as \(\exp(0.3499) = 1.419\), which means that the hazard of event in group 2 is 1.419 times higher than the hazard of event in group 1.
Comment on the limitation of exponential model compared to other more flexible models such as Weibull.
Answer:
The exponential model assumes that the hazard rate is constant over time, which is a strong assumption that may not hold in practice. The Weibull model is more flexible than the exponential model as it allows the hazard rate to change over time. The Weibull model has two parameters, shape and scale, which can capture different patterns of hazard rates over time. The Weibull model can be used to model increasing, decreasing, or constant hazard rates, which makes it more suitable for survival data with varying hazard rates. Therefore, the Weibull model is more flexible than the exponential model and can provide a better fit to the data in many cases. The exponential model is a special case of the Weibull model with a shape parameter of 1, which means that the hazard rate is constant over time. Therefore, the exponential model is limited in its ability to capture the variability in hazard rates that may be present in survival data.
2 Longitudinal data analysis
Onychomycosis, popularly known as toenail fungus, is a fairly common condition that not only can disfigure and sometimes destroy the nail but that also can lead to social and self-image issues for sufferers. Tight-fitting shoes or hosiery, the sharing of common facilities such as showers and locker rooms, and toenail polish are all thought to be implicated in the development of onychomycosis. This question relates to data from a study conducted by researchers that recruited sufferers of a particular type of onychomycosis, dermatophyte onychomycosis. The study conducted by the researchers was focused on comparison of two oral medications, terbinafine (given as 250 mg/day, denoted as treatment 1 below) and itraconazole (given as 200 mg/day, denoted as treatment 2 below).
The trial was conducted as follows. 200 sufferers of advanced toenail dermatophyte onychomycosis in the big toe were recruited, and each saw a physician, who removed the afflicted nail. Each subject was then randomly assigned to treatment with either terbinafine (treatment 1) or itraconazole (treatment 2). Immediately prior to beginning treatment, the length of the unafflicted part of the toenail (which was hence not removed) was recorded (in millimeters). Then at 1 month, 2 months, 3 months, 6 months, and 12 months, each subject returned, and the length of the unafflicted part of the nail was measured again. A longer unafflicted nail length is a better outcome. Also recorded on each subject was gender and an indicator of the frequency with which the subject visited a gym or health club (and hence might use shared locker rooms and/or showers).
The data are available in the file toenail.txt from here. The data are presented in the form of one data record per observation; the columns of the data set are as follows:
Subject id
Health club frequency indicator (= 0 if once a week or less, = 1 if more than once a week)
Gender indicator (= 0 if female, = 1 if male)
Month
Unafflicted nail length (the response, mm)
Treatment indicator (= 1 if terbinafine, = 2 if itraconazole)
The researchers had several questions, which they stated to you as follows:
Use the linear mixed effect model (LMM) to answer: Is there a difference in the pattern of change of lengths of the unafflicted part of the nail between subjects receiving terbinafine and itraconazole over a 12 month period? Does one treatment show results more quickly?
Plot the change of lengths of the unafflicted part of the nail over time and separated by treatment groups. Comment on overall patterns over time.
Plot the change of lengths of the unafflicted part of the nail over time and separated by treatment groups:
# using the dot plot to show the average nail length over timetoenail |>ggplot() +facet_wrap(~ treatment,labeller =labeller(treatment =c("1"="Terbinafine (Group 1)","2"="Itraconazole (Group 2)"))) +geom_point(mapping =aes(x = month, y = nail_length,color =as.factor(month)),stat ="summary",fun = mean,size =3) +geom_line(mapping =aes(x = month, y = nail_length,group = id),alpha =0.2) +geom_smooth(mapping =aes(x = month, y = nail_length),method ="lm",se =FALSE,color ="blue") +labs(title ="Average nail lengths over time by treatment group",x ="Month",y ="Average nail length (mm)") +guides(color ="none")
Comment on overall patterns over time:
The plots show that the lengths of the unafflicted part of the nail has a linear increase over time for both treatment groups. The average nail length increases over time, and the rate of increase appears to be relatively constant. The plots also show that the average nail length is higher for the itraconazole group compared to the terbinafine group at each time point. This suggests that the itraconazole treatment may be more effective in increasing the length of the unafflicted part of the nail compared to the terbinafine treatment. The plots also show that there is more variability in nail lengths within the itraconazole group compared to the terbinafine group. This suggests that the terbinafine treatment may be more consistent in its effect on nail length compared to the itraconazole treatment. Overall, the plots show that the itraconazole treatment may be more effective and consistent in increasing the length of the unafflicted part of the nail over time compared to the terbinafine treatment.
Based on the pattern observed, pick appropriate time trend in the LMM and provide an algebraic definition for your chosen LMM, e.g., is the linear trend model adequate? or quadratic trend is needed? or any other pattern is more approriate? justify your answer.
Based on the pattern observed in the plots, a linear trend model may be adequate to capture the change in lengths of the unafflicted part of the nail over time. The plots show that the mean lengths of the unafflicted nail tend to increase over time, and the rate of increase appears to be relatively constant. A linear trend model can capture this linear increase in nail lengths over time. The algebraic definition of the linear mixed effect model (LMM) is as follows:
First: try to fit quadratic trend model to the data and compare the results with the linear trend
# fitting a linear mixed effects model with quadratic trendmonth_treatment_lmm_quad <-lmer( nail_length ~ treatment +poly(month, 2) + (1+ month | id),data = toenail)# fitting a linear mixed effects model with linear trendmonth_treatment_lmm_linear <-lmer( nail_length ~ treatment + month + (1+ month | id),data = toenail)# comparing the two modelsKRmodcomp(month_treatment_lmm_quad, month_treatment_lmm_linear)
Since the p-value is not significant, we can conclude that the quadratic trend model does not provide a better fit to the data compared to the linear trend model. Therefore, the linear trend model is adequate to capture the change in lengths of the unafflicted part of the nail over time.
Second: try interaction between treatment and month in the LMM
# fitting a linear mixed effects model with interaction between treatment and monthmonth_treatment_lmm_interaction <-lmer( nail_length ~ treatment * month + (1+ month | id),data = toenail)# comparing the two modelsKRmodcomp(month_treatment_lmm_interaction, month_treatment_lmm_linear)
Linear mixed model fit by REML ['lmerMod']
Formula: nail_length ~ treatment * month + (1 + month | id)
Data: toenail
REML criterion at convergence: 4365.6
Scaled residuals:
Min 1Q Median 3Q Max
-2.6232 -0.5077 0.0249 0.4775 2.9321
Random effects:
Groups Name Variance Std.Dev. Corr
id (Intercept) 2.2121 1.4873
month 0.1104 0.3323 -0.38
Residual 0.9412 0.9701
Number of obs: 1200, groups: id, 200
Fixed effects:
Estimate Std. Error t value
(Intercept) 5.24068 0.35515 14.756
treatment 0.10069 0.22462 0.448
month 0.12216 0.07746 1.577
treatment:month 0.14933 0.04899 3.048
Correlation of Fixed Effects:
(Intr) trtmnt month
treatment -0.949
month -0.414 0.393
trtmnt:mnth 0.393 -0.414 -0.949
Since the p-value is highly significant, we can conclude that the interaction between treatment and month provides a better fit to the data compared to the linear trend model. Therefore, the linear mixed effects model with interaction between treatment and month is more appropriate to capture the change in lengths of the unafflicted part of the nail over time.
Based on the chosen LMM, provide the algebraic definition for the LMM.
\[
\begin{align*}
Y_{ij} &= \beta_0 + \beta_{\text{Month}} \times \text{Month}_{i} + \beta_{\text{Treatment}} \times \text{Treatment}_{j} \\
&\quad + (\text{Month}_{i} \times \text{Treatment}_{j}) \times \beta_{\text{month} \times \text{treatment}} \\
&\quad + b_{0, j} + b_{1, j} \times \text{Month}_{i} + \epsilon_{ij}.
\end{align*}
\] , where \(b_{0, j}\) and \(b_{1, j}\) are the random intercept and random slope for subject \(j\), respectively, and \(\epsilon_{ij}\) is the residual error term. The fixed effects \(\beta_0\), \(\beta_{\text{Month}}\), \(\beta_{\text{Treatment}}\), and \(\beta_{\text{month} \times \text{treatment}}\) represent the intercept, the effect of month, the effect of treatment, and the interaction effect between month and treatment, respectively. The random effects \(b_{0, j}\) and \(b_{1, j}\) capture the individual-specific random intercepts and slopes, respectively. The error term \(\epsilon_{ij}\) represents the residual error. And \(b_{0, j}\) and \(b_{1, j}\) are assumed to follow a normal distribution with mean 0 and covariance matrix \(\mathbf{D}\), (\(\mathbf{D} = \begin{bmatrix} \sigma_{b_0}^2 & \rho \sigma_{b_0} \sigma_{b_1} \\ \rho \sigma_{b_0} \sigma_{b_1} & \sigma_{b_1}^2 \end{bmatrix}\)), and the residuals \(\epsilon_{ij}\) are assumed to follow a normal distribution with mean 0 and variance \(\sigma^2\). (\(b_{0, j}\), \(b_{1, j}\)) ~ N(0, \(\mathbf{D}\)), \(\epsilon_{ij}\) ~ N(0, \(\sigma^2\)).
Model the covariance: fit both random intercept and random slope model and determine which one fits the data better.
# fitting a linear mixed effects model with random intercept onlymonth_treatment_lmm_random_intercept <-lmer( nail_length ~ treatment * month + (1| id),data = toenail)confint(month_treatment_lmm_random_intercept, method ="boot")
# fitting a linear mixed effects model with both random slope and interceptmonth_treatment_lmm_random_both <-lmer( nail_length ~ treatment * month + (1+ month | id),data = toenail)confint(month_treatment_lmm_random_both, method ="boot")
Here I fitted three models: one with random intercept only, one with both random slope and intercept, and one with random slope only. The confidence intervals for the fixed effects of the models are shown above. The confidence intervals for all three models don’t include zero, indicating that the fixed effects are significant. And because we want to model the covariance more accurately; therefore I think the model with both random slope and intercept is more appropriate to capture the change in lengths of the unafflicted part of the nail over time, because it considers the individual-specific random slopes and intercepts, which can provide a better fit to the data compared to the other two models.
Use the linear mixed effect model (LMM) to answer: Is there an association between the pattern of change of nail lengths and gender and/or health club frequency in subjects taking terbinafine? This might indicate that this drug brings about relief more swiftly in some kinds of subject versus others.
Provide graphs to show patterns the change of nail lengths and gender and/or health club frequency in subjects taking terbinafine.
Answer:
Provide graphs to show patterns the change of nail lengths
toenail_terbinafine <- toenail %>%filter(treatment =="1")# Plot change of nail lengths by gendertoenail_terbinafine |>ggplot() +facet_wrap(~gender,scales ="free",labeller =labeller(gender =c("0"="Female","1"="Male"))) +geom_line(mapping =aes(x = month, y = nail_length, group = id,color =as.factor(gender))) +labs(title ="Change of nail lengths by"~"Gender in Subjects Taking Terbinafine",x ="Month",y ="Nail length (mm)") +guides(color ="none")
# Plot change of nail lengths by health club frequencytoenail_terbinafine |>ggplot() +facet_wrap(~health_club,scales ="free",labeller =labeller(health_club =c("0"="Once a week or less","1"="More than once a week"))) +geom_line(mapping =aes(x = month, y = nail_length, group = id,color =as.factor(health_club))) +labs(title ="Change of nail lengths by Health Club"~"Frequency in Subjects Taking Terbinafine",x ="Month",y ="Nail length (mm)") +guides(color ="none")
# Plot change of nail lengths by gender and health club frequencytoenail_terbinafine |>ggplot() +facet_grid(gender ~ health_club,labeller =labeller(gender =c("0"="Female","1"="Male"),health_club =c("0"="Once a week or less","1"="More than once a week"))) +geom_line(mapping =aes(x = month,y = nail_length,group = id,color =as.factor(health_club))) +labs(title ="Change of nail lengths by Two Factors"~"in Subjects Taking Terbinafine",x ="Month",y ="Nail length (mm)") +guides(color ="none")
Observations:
It can be seen from the plots that the change in nail lengths over time appears to have similar patterns if grouped by gender or health club frequency. The lengths of the unafflicted part of the nail tend to increase over time, and the rate of increase appears to be relatively similar across gender and health club frequency groups.
However, when looking at the combined/interaction effect, it seems that the pattern of change in nail lengths don’t show a clear difference between the groups. The lines for the different groups are quite close to each other.
Based on the pattern observed from question 1, pick appropriate time trend in the LMM and provide an algebraic definition for your chosen LMM, e.g., is the linear trend model adequate? or quadratic trend is needed? or any other pattern is more approriate? justify your answer.
Answer:
Based on the pattern observed in the plots, the nail lengths generally increase over time, which suggests that a linear term for Month might be adequate. While most lines appear to follow a linear trend, there are some individual trajectories that show more variation, potentially indicating the need for higher-order terms or interaction terms to capture individual differences better.
First: try to fit quadratic trend model to the data and compare the results with the linear trend
# fitting a linear mixed effects model with quadratic trendgender_health_club_lmm_quad <-lmer( nail_length ~ gender + health_club +poly(month, 2) + (1+ month | id),data = toenail_terbinafine)# fitting a linear mixed effects model with linear trendgender_health_club_lmm_linear <-lmer( nail_length ~ gender + health_club + month + (1+ month | id),data = toenail_terbinafine)# compare the modelsKRmodcomp(gender_health_club_lmm_quad, gender_health_club_lmm_linear)
Since the p-value is greater than 0.05, we fail to reject the null hypothesis, indicating that the quadratic trend model is not significantly better than the linear trend model. Therefore, the linear trend model is adequate for the data.
Second: try interaction between gender and health club frequency in the LMM
# Fit the linear mixed effects model with interactiongender_health_club_lmm_interaction <-lmer( nail_length ~ gender * health_club + month + (1+ month | id),data = toenail_terbinafine)# Compare the modelsKRmodcomp(gender_health_club_lmm_interaction, gender_health_club_lmm_linear)
Linear mixed model fit by REML ['lmerMod']
Formula: nail_length ~ gender + health_club + month + (1 + month | id)
Data: toenail_terbinafine
REML criterion at convergence: 2107.7
Scaled residuals:
Min 1Q Median 3Q Max
-2.74007 -0.50417 0.03683 0.48188 2.86760
Random effects:
Groups Name Variance Std.Dev. Corr
id (Intercept) 1.84781 1.3593
month 0.04666 0.2160 0.19
Residual 0.93675 0.9679
Number of obs: 600, groups: id, 100
Fixed effects:
Estimate Std. Error t value
(Intercept) 5.93622 0.28101 21.125
gender 0.52644 0.29737 1.770
health_club -1.28855 0.31614 -4.076
month 0.27149 0.02371 11.450
Correlation of Fixed Effects:
(Intr) gender hlth_c
gender -0.417
health_club -0.666 -0.163
month 0.027 0.000 0.000
Since the p-value is greater than 0.05, we fail to reject the null hypothesis, indicating that the interaction model is not significantly better than the linear trend model. Therefore, the linear trend model is adequate for the data.
Based on the chosen LMM, provide the algebraic definition for the LMM.
Linear mixed model fit by REML ['lmerMod']
Formula: nail_length ~ gender + health_club + month + (1 | id)
Data: toenail_terbinafine
REML criterion at convergence: 2309.2
Scaled residuals:
Min 1Q Median 3Q Max
-3.8199 -0.5287 -0.0068 0.5196 3.7576
Random effects:
Groups Name Variance Std.Dev.
id (Intercept) 2.905 1.704
Residual 1.844 1.358
Number of obs: 600, groups: id, 100
Fixed effects:
Estimate Std. Error t value
(Intercept) 6.00232 0.34762 17.267
gender 0.32847 0.36338 0.904
health_club -1.23652 0.38632 -3.201
month 0.27149 0.01372 19.792
Correlation of Fixed Effects:
(Intr) gender hlth_c
gender -0.412
health_club -0.658 -0.163
month -0.158 0.000 0.000
From the confidence intervals, we can see that the both ramdom intercept and slope model contains zero, which indicates that the random intercept and slope model is not significantly better than the random intercept only model. By comparing the AIC of the intercept only and random slope only model, we can see that the random intercept only model has a lower AIC value, indicating that it fits the data better. Therefore, the random intercept only model is chosen as the final model.
In answering these scientific questions of interest, clearly write out the analytic models you consider for answering these questions (as detailed in the sub-questions). Clearly outline your decision making process for how you selected your final models. Fit your chosen final models and report to the project investigators on the stated scientific questions of interest.
Is there a difference in the pattern of change of lengths of the unafflicted part of the nail between subjects receiving terbinafine and itraconazole over a 12 month period? Does one treatment show results more quickly?
For this question, I first fit a linear mixed effects model with a quadratic trend to the data. I then compared this model to a linear trend model to determine if the quadratic trend model is significantly better. Since the p-value is greater than 0.05, I fail to reject the null hypothesis, indicating that the quadratic trend model is not significantly better than the linear trend model. Therefore, the linear trend model is adequate for the data.
Next, I fit a linear mixed effects model with interaction between treatment and month to the data. I then compared this model to a linear trend model to determine if the interaction model is significantly better. Since the p-value is less than 0.05, I reject the null hypothesis, indicating that the interaction model is significantly better than the linear trend model. Therefore, the interaction model is chosen as the final model.
Also, by comparing the confidence intervals and AIC values of the random intercept only, random slope only, and random intercept and slope models, I found that the random intercept and slope model performed the best. Therefore, the random intercept and slope model is chosen as the final model.
Linear mixed model fit by REML ['lmerMod']
Formula: nail_length ~ treatment * month + (1 + month | id)
Data: toenail
REML criterion at convergence: 4365.6
Scaled residuals:
Min 1Q Median 3Q Max
-2.6232 -0.5077 0.0249 0.4775 2.9321
Random effects:
Groups Name Variance Std.Dev. Corr
id (Intercept) 2.2121 1.4873
month 0.1104 0.3323 -0.38
Residual 0.9412 0.9701
Number of obs: 1200, groups: id, 200
Fixed effects:
Estimate Std. Error t value
(Intercept) 5.24068 0.35515 14.756
treatment 0.10069 0.22462 0.448
month 0.12216 0.07746 1.577
treatment:month 0.14933 0.04899 3.048
Correlation of Fixed Effects:
(Intr) trtmnt month
treatment -0.949
month -0.414 0.393
trtmnt:mnth 0.393 -0.414 -0.949
\[
\begin{align*}
\text{Nail Length}_{ij} &= 5.24068 + 0.12216 \times \text{Month}_{i} + 0.10069 \times \text{Treatment}_{j} \\
&\quad + 0.14933 \times (\text{Month}_{i} \times \text{Treatment}_{j}) \\
&\quad + b_{0, j} + b_{1, j} \times \text{Month}_{i} + \epsilon_{ij}, \\
&where \ b_{0, j} \sim N(0, \sigma_{b_0}^2), b_{1, j} \sim N(0, \sigma_{b_1}^2), \epsilon_{ij} \sim N(0, \sigma_{\epsilon}^2) \\
\end{align*}
\] Therefore, the answer to the question is that there is a difference in the pattern of change of lengths of the unafflicted part of the nail between subjects receiving terbinafine and itraconazole over a 12 month period. The treatment with itraconazole shows results more quickly.
Is there an association between the pattern of change of nail lengths and gender and/or health club frequency in subjects taking terbinafine? This might indicate that this drug brings about relief more swiftly in some kinds of subject versus others.
For this question, I first fit a quadratic trend model to the data. I then compared this model to a linear trend model to determine if the quadratic trend model is significantly better. Since the p-value is greater than 0.05, I fail to reject the null hypothesis, indicating that the quadratic trend model is not significantly better than the linear trend model. Therefore, the linear trend model is adequate for the data.
Next, I fit a linear mixed effects model with interaction between gender and health club frequency to the data. I then compared this model to a linear trend model to determine if the interaction model is significantly better. Since the p-value is greater than 0.05, I fail to reject the null hypothesis, indicating that the interaction model is not significantly better than the linear trend model.
Then, I compared the confidence intervals and AIC values of the random intercept only, random slope only, and random intercept and slope models. I found that the random intercept only model performed the best. Therefore, the random intercept only model is chosen as the final model.
Therefore, the answer to the question is that there is an association between the pattern of change of nail lengths and gender and/or health club frequency in subjects taking terbinafine. This might indicate that this drug brings about relief more swiftly in some kinds of subject versus others.
