19

A p value is the probability of observing a test statistic as or more extreme than the researcher's own test statistic, assuming the null hypothesis, and an assumed distribution model are both true. So when you see frequentist statistical software provide regression output that includes something like: name estimate S.E. t (or z) p value cons 3.0 0.015 200 ...


15

If you're trying to generate data from logistic regression's assumed data generating mechanism, your code does not do that. Logistic regression's data generating mechanism looks like $$ \eta = X\beta$$ $$ p = \dfrac{1}{1+e^{-\eta}}$$ $$ y \sim \operatorname{Binomial}(p, n) $$ What it looks like you're trying to do is create a linear regression in the log ...


11

This is actually a good CV question because it is the pivotal moment when a trained statistician becomes a professional statistician. Learning moment 1: Never judge a person by their pedigree or training By you not responding seriously to a critique of your methods because the criticism came from a non-statistician, you are selling yourself short, very short ...


8

This post analyzes the situation; explains the sense in which linear relations are transitive; locates the problem in a lack of transitivity of the slope estimates; and characterizes all cases where transitivity of the estimates does hold. You did well by explicitly writing the error terms. Let's name them (because they are separate things) and combine the ...


6

Quite generally, insignificant variables do not have to be removed from regression models (this also holds for parameters belonging to terms such as $X^2$). Also, generally, standard regression tests are invalid in models that stem from data driven selection.


6

When you have a factor variable in a regression model, that is, a categorical variable with multiple levels, that variable should be treated as a whole. So you should mostly disregard the t-tests for each separate level, and test the variable as a whole. In a linear regression model that would be an F-test. Such multiple-df tests are often called chunk tests,...


5

I don't think an entire data presentation is needed to give some intuition behind this phenomenon. While we would expect that a logistic regression and OLS model will, on average, produce parameter estimates (slopes or log-odds ratios) that are similar sign, it's entirely possible they will disagree for a given dataset and analysis. The most likely issue is ...


4

The long algebraic manipulations involved in standard demonstrations are bothersome. There ought to be a demonstration that is simple, direct, and provides insight into what the terms in the formulas mean. By "simple" I will allow liberal use of substitution in formulas and application of the most straightforward algebraic reductions, along with ...


4

Yes. $$\begin{align}y_i & =\alpha+(\beta_1+\beta_2)x_1+\beta_2x_2+\epsilon_i\\ \\ & = \alpha+\beta_1x_1+\beta_2x_1+\beta_2x_2+\epsilon_i\\ \\ & =\alpha+\beta_1x_1+\beta_2(x_1+x_2)+\epsilon_i\end{align}$$


4

Your model seems to have perfect separation. Notice the large standard errors of the coefficient estimates, and the tell-tale warning messages: Warning messages: 1: glm.fit: algorithm did not converge 2: glm.fit: fitted probabilities numerically 0 or 1 occurred Your model never actually got fit by the software! As you provided a fixed cutoff of AHI > 5 ...


4

I would interpret these results as: every 1 unit increase in parenting_style is associated with a 0.06 increase in aggression every 1 unit increase in sibling_aggression is associated with a 0.09 increase in aggression the p-values indicate that if these associations were actually absent, the probability of observing these data (or data even more extreme) ...


4

Unlike maximum likelihood estimation for a well-chosen correlation structure, the cluster bootstrap doesn't really "know" about the correlation structure. The cluster bootstrap allows for a fairly general correlation structure (just as the robust cluster sandwich estimator does) but that's not necessarily enough to be optimal. The sandwich ...


4

Those models are similar, but the key different thing is we model $\log{E(Y)}$ for GLM and $E(\log Y)$ for LM. Thus, we can estimate $Y$ directly in GLM and $\log Y$ in LM. GLM: We can directly say $E(Y)=\exp(\beta_0+\beta_1X)$. In this case, $\beta_1$ catches the effect of a unite change in $X$ on $Y$ (acutually, log of ratio of $Y$). LM: we only say $E(\...


3

From the estimation perspective, pweights is internally used the same way as any other weight. in OLS: $$ min_{\beta} = \sum{(y-\beta X)^2 * w} $$ The only difference with other methods is during the estimation of standard errors. When you use pweights, is like requesting robust standard errors. (sandwich formula). For further details on how exactly weights ...


3

Dichotomozing and binning continuous variables is a poor poor idea. Not only are the splits more or less arbitrary, but splitting continuous variables leads to residual confounding and decreases the degrees of freedom in the model (maybe not a problem in your particular example, you likely have oodles of data). This can lead to poorer measures of model ...


3

This is because the estimated coefficients are in a model controlling for other variables, whereas the plot you made does not control for other variables. If you were to make a simple model just of the outcome regressed on the single scale you are looking at, the model coefficients would reproduce the marginal plot exactly. There are many reasons why this ...


3

If you want $\gamma$ to be exactly 0.5 you must add it as a parameter with assigned coefficient. Leaving it as an unspecified intercept will estimate the coefficient instead. Depending on the features and size of the data you're working with the model may find a value close to 0.5, but it will have extra variance compared to using a fixed coefficient. (Of ...


3

From JMP: Introduction to Statistics: Correlation: What is correlation? Correlation is a statistical measure that expresses the extent to which two variables are linearly related (meaning they change together at a constant rate). It’s a common tool for describing simple relationships without making a statement about cause and effect. Key points to ...


3

TL;DR: Yes, it does return biased coefficients. I believe this is incorrect, if we are using the word "bias" in the way that people talk about it with regard to statistical and machine learning. (EDIT: Actually it IS correct, just a little confusing, explained below in the edit. Leaving this here so others can track my thought process, since it is ...


3

The t-test is the right way to do it if we do not know the variance (which we pretty much never know in practice). Thus, the answer to your question is because the t-test is the correct method. So why use the z-test as your have described? It is a fine approximation of the t-test for large sample sizes. There is a sense in which the t-distribution converges ...


3

The basic setup is to use a robust algorithm for root finding and supply it with the function you defined. You'll need one that doesn't require a derivative (the function is not even continuous, but is monotonic, so methods like binary section should still work) A suitable algorithm will require you to find a good starting point, and/or will require you to ...


3

Rank regression is an informal procedure not based on a fully specified statistical model. So you can expect to have interpretation problems and especially have problems generalizing it in various ways such as adjusting for multiple covariates and incorporating repeated measures. It is better to conceptualize the problem as a semiparametric ordinal model. ...


3

This set of questions probes your understanding of the relationships among (a) the parameter estimates and (b) their variance-covariance matrix when the explanatory variables in an ordinary least squares (OLS) regression are transformed. Let's begin, then, by discussing this generally. And because this is a self-study question, I won't provide the ...


3

Let's consider an example. We have $50$ dogs, $50$ cats, and $50$ yaks. On some variable of interest, $Y$, dogs follow a $N(0, 1)$ distribution, cats follow a $N(1, 1)$ distribution, and yaks follow a $N(1, 1)$ distribution. Thus, yaks and cats have the same mean, but dogs have a different mean. When you ask if the means for all three species are the same, ...


2

You have given proper justification in your question. If you want to know the effect of one with the rest held constant, use a main effects model. The principle of marginality states that main effects do not exist in the presence of an interaction, so beware that this is only valid if there is indeed only a negligible interaction effect. Note that the ...


2

You are correct that your Model 2 makes the most sense if you wish to standardize your continuous predictor C. The confusion comes from what the intercept and the coefficient for the binary predictor B mean in Model 0 versus Model 2. I assume that Stata is using treatment coding of the predictors, and that by standardizing C you mean subtracting its mean and ...


2

+1 to Demetri's answer. Skip model coefficients altogether. (Better: put them in an appendix, in case someone asks for them, and to show you did your homework.) As you write, they are hard to interpret and can be gotten wrong. But more importantly, nobody cares about your coefficients, anyway! What people care about is the impact on the observable. So pick a ...


2

For the coefficients and standard errors (and $p$-values) to be identical you would need to have interactions between geneder and all the other variables, and also have separate values of the overdispersion parameter by gender Your interaction model has a separate group coefficient for male and female but has the same coefficients across gender for age, ...


2

The main effect of D (b) is the amount of change expected in Y for each unit change in D, controlling for X. The interaction of X and D allows this change to become larger or smaller based on the value of X.


2

Section 4.2 of the time-dependence vignette of the R survival package discusses the problem with your second model: The issue is that the above code does not actually create a time dependent covariate, rather it creates a time-static value for each subject based on their value for the covariate time; no differently than if we had constructed the variable ...


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