14
Dichotomizing a continuous covariate is ill-advised, as has been noted by other users.
One strategy I employ is to rescale the predictor to something more reasonable. 1 mmHg may not be a very meaningful scale on which to interpret changes in BP. But, if you rescale the predictor so that a difference of 1 unit represents say a difference 10 mmHg then ...
12
Similar to EdM's answer, a marginal effects plot is a useful way to showcase the relationship between a clinical measurement and outcome while holding other variables constant. These plots are helpful because they show the relationship between the predictor and outcome, so if the outcome is nonlinear, physicians can easily see this and interpret it ...
9
For digestibility, use examples from representative situations: in your example, maybe comparing risks at BP of 160 versus BP of 120.
For an approach that can take into account the multiple predictors typically important in clinical studies, use a nomogram. It provides a graphical tool to show how predictor values affect outcomes. The rms package in R ...
8
Another issue is that the relationship between the IV and the DV (here, BP and heart attack risk) might not be linear. I would think this sort of nonlinearity would be quite common in medical fields. Indeed, this is sometimes given as a reason for categorizing the continuous variable (albeit into more than two categories). But this isn't good. A better ...
answered Jan 14 at 12:53
6
I have many questions about the proposed approach, mostly because it is not clearly described. It's unlikely this question can be edited to give us much insight into the setting and rationale for this approach. Nonetheless, a few points can be addressed.
Sums of variances obtained by randomly partitioning a dataset 30-fold cannot be called "robustness". A ...
6
Those symbols do not have any specific meaning, what they mean should follow from context. Usually, in statistics we use Greek letters to denote parameters, but even this is not a general rule followed by everyone. $\beta$ and $\theta$ are popular choices by convention. Also by convention $\beta$ is popular choice for regression parameters.
5
$\theta$ is the Greek letter used in the most generality for some parameter (could be a vector), which could include regression coefficients.
$\beta$ would be specific to regression coefficients.
Unless something specifically deals with regression coefficients, I think the preference would be to use $\theta$. Even in a context where the parameter could be ...
4
A danger in having powerful tools available in standard software programs is that users don't always understand the underlying hidden assumptions.
Even if you will be using Stata for routine work, I recommend getting a copy of An Introduction to Statistical Learning and working through the examples in Chapter 6 of LASSO and ridge regression, with the code ...
4
This is a standard least squares problem and, as such, is solved by finding parameters that minimize the sum of squared residuals.
A general approach is first to fit the model
$$E[Y] = \beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k.$$
Replace the $y_i$ by their residuals
$$e_i = y_i - (\hat\beta_0 + \hat\beta_1 x_1 + \cdots + \hat\beta_k x_k).$$
The ...
4
Maximum Likelihood just like minimizing the square of the sum of residuals has no penalty factor to reduce coefficients to zero as LASSO does. So by definition Maximum Likelihood does not shrink coefficients to zero and LASSO does.
However, you can't automatically deduce that the LASSO model will generate a better fit and prediction of what you are ...
4
I'll add this: That the LASSO (and associated with it, ridge regression) have Bayesian interpretations is a glimmer of light in the problem of penalization and inference. However, it isn't as easy as "pick priors so that some covariates are shrunk to zero". In order for valid inferences to be made from a bayesian linear model, your priors have to actually ...
3
First of all, the LRT is not really a good measure for variable selection. It is intended to compare nested models for evaluation of a hypothesis. If you use the LRT to do variable selection for 10 variables, then you are almost certainly going to make a false positive.
If you want to do a form of variable selection, LASSO seems to be the way to go and ...
3
Here is how to do this for one cultivar:
plot(shoot ~ P, data = subset(DF, cultivar == "Dinninup"))
fit1 <- nls(shoot ~ ifelse(P < bp, m * P + c, m * bp + c),
data = subset(DF, cultivar == "Dinninup"),
start = list(c = 1, m = 0.05, bp = 25), na.action = na.omit)
summary(fit1)
#Formula: shoot ~ ifelse(P < bp, m * P + c, m * ...
3
A search of my electronic copy of ISLR shows the symbol $\beta$ as appearing on 99 pages. A quick look at the context in which that symbol appears on those pages indicates that all uses of that symbol represent coefficients in linear forms, in contexts of: standard linear regression, logistic regression, penalized regressions, and support vector machines. ...
2
The package mcp was made just for scenarios like this. See below how I structured your data as df later.
Fit a change point model
First, let's define a slope followed by a joined plateau. We add varying (random) change point locations (the left-hand side of the equation):
model = list(
shoot ~ 1 + P, # intercept and slope
1 + (1|cultivar) ~ 0 # ...
2
Many standard statistical tests can be reframed as simple linear models. So in your example, there is no difference between a two sample T-test testing for differences in X and Y versus a linear regression of Y and X against a grouping indicator. Your intuition that regression requires more data is not correct.
Also, as whuber mentions in his comment, you ...
2
The output is a linear combination of the inputs. Form of the input features does not matter; for example, $y=\beta_0+\beta_1\sin e^x$ can still be a linear regression model. We can always call our features with different names, e.g. let $u_1=xt,u_2=x^2$, then your model becomes:
$$y=\beta_0+\beta_1u_1+\beta_2u_2=u^T\beta$$
which is linear.
Edit: An example ...
2
If I understand you correctly, you are asking about testing the significance of the difference between two regressions on the same data set. This is not a standard significance testing problem. A standard significance testing problem states a null hypothesis ($H_0$) about how the data were generated, and uses the distribution assuming this $H_0$ of a certain ...
1
You appear to be misunderstanding the output from your model.
In your code, the line:
abline(fit.mice$coefficients[1:2], col="skyblue3")
plots a line with the correct intercept, but the wrong slope. fit.mice$coefficients[1] is the intercept, but fit.mice$coefficients[2] is the estimate for treat in the Mice added group, hence this is an offset to the ...
1
An ANCOVA model would be one way. It would look something like
ElephantDensity ~ Control * PlantDensity
This will fit fixed effects for Control and PlantDensity, along with an interaction term between them. The interaction will quantify the extent to which the association between PlantDensityand ElephantDensity varies at the different levels of treatment.
...
1
Penalized regression does not do an initial regression and then downweight points "outlying" from this initial regression. $L_1$ and $L_2$-penalized regressions do not weight observations at all (de facto). In fact, $L_1$-penalized regression (LASSO) is more prone to bias d/t outlier because it tends to select regressors based on large coefficient values. ...
1
The key thing for a statistical test is that the distribution of the test statistic can be computed under the null hypothesis. The null hypothesis for a likelihood ratio test is typically the less complex model (meaning fewer free parameters). If the less complex model holds (let's say that's model A), the likelihood from model B should normally be very ...
1
I'm going to hopefully clarify some things about the likelihood ratio test. If you still disagree, comment below.
First, a nested model is (pragmatically) a model for which we assume the value(s) of the parameter(s). Consider a vector of parameters partitioned into two disjoint sets $\mathbf{\Theta} = [\theta; \theta_0]$ . In this case, we are free to ...
1
I read a paper a while back that used an L2-norm to do sort of a "temporal smoothing" process across the predictions, which I thought was an interesting way to deal with time-varying components. I unfortunately can't find it but here's a similar paper I found
Learning Time-Varying Graphs using Temporally Smoothed L1-Regularized Logistic Regression
http://...
1
1(a) Anova() can be easier to understand in terms of evaluating the significance of a predictor in your model, even though there is nothing wrong with the output from summary().
The usual R summary() function reports something that can appear quite different from Anova(). A summary() function typically reports whether the estimated value for each ...
1
The interaction term you propose would give you a predictor that is 0 for values of $x$ less than or equal to 0.5 and equal to $x$ for values above that. In principle there's nothing wrong with that formulation of a predictor variable in your regression.
That wouldn't, however, test your hypothesis that "$x$ is a significant predictor of $y$ but only when $...
1
If you were to perform this as a linear regression, in R, you would receive the intercept term (the point at which the line touches the y-axis, which is equivalent to the value of y when the x-variable is 0). If you are using the ggplot2 package to graph the least squares line through the data points, ggplot can use a stat_smooth(method = "lm"), to help draw ...
1
Great question! As you've pointed out, they are not, and sometimes refer to wildly different things (i.e., precision is a property of the model in classification, and refers to a measure of variance in regression). Unfortunately, in statistics and i'm sure other disciplines, we tend to abuse notation and use the same word to denote different things. You've ...
1
I'll try to explain how I see it from statistical point of view.
I don't think MSE and MSR are the same thing (however most people don't differentiate between those two I guess).
Let's say that you do a simulation of data that can be described using regression model. Let's say that you generate the data "randomly" around parabola curve and you have the ...
1
Estimating a regression without a constant term is essentially saying "I know the intercept of my model is equal to zero. I don't need to estimate it because I know with certainty what it is." That's a pretty bold claim to make. If you're right, then you can estimate the slope with slightly increased precision. If you're wrong, your estimate of the slope ...
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