14

That’s the logarithm of the odds ratio, not the odds ratio itself. An odds ratio less than zero is nonsense. Looking at the behavior of a logarithm function (the base could be 2, could be 10, could be $e$), the function achieves values less than zero when the argument is less than 1, so a negative log of the odds ratio means that the odds ratio is between 0 ...


11

Is it okay that the direction of the fixed effects is opposite to the intercept in my model? Yes. The interpretation of the intercept is that it is the log-odds of the event (DV = 1) for the those in the none group (so, being negative, it is a protective effect), while the estimates for the other 2 groups are the log-odds of the event in each of those ...


6

I think it's a typo. The derivative of the logistic curve with respect to $x$ is: $$ \frac{\beta\mathrm{e}^{\alpha + \beta x}}{\left(1 + \mathrm{e}^{\alpha + \beta x}\right)^{2}} $$ So for their example where $\alpha = -1.40, \beta = 0.33$ it is: $$ \frac{0.33\mathrm{e}^{-1.40 + 0.33 x}}{\left(1 + \mathrm{e}^{-1.40 + 0.33 x}\right)^{2}} $$ Evaluated at the ...


5

The problem with using the estimated coefficients for interpretation is that because of the link function used in Logistic regression their effect is nonlinear to $\bf{X}$. \begin{equation} log(\frac{\pi_i}{1-\pi_i}) = \alpha + \beta X_i + \epsilon_i \end{equation} Therefore, \begin{equation} \frac{\pi_i}{1-\pi_i} = exp(\alpha + \beta X_i) = e^\alpha e^{\...


4

For a continuous variable $x$, the marginal effect of $x$ in a logit model is $$\Lambda(\alpha + \beta x)\cdot \left[1-\Lambda(\alpha + \beta x)\right]\cdot\beta = p \cdot (1 - p) \cdot \beta,$$ where the inverse logit function $\Lambda$ is $$\Lambda(z)=\frac{\exp{z}}{1+\exp{z}}.$$ Here $p$ is a probability, so the factor $p\cdot (1-p)$ is maximized when $...


3

If $p$ is the probability of an event, $o$ the odds (the ratio of probabilities of the event happening and it not happening), and $\alpha$ the log-odds, then $\alpha = \log_e\left(o\right) = \log_e\left(\frac{p}{1-p}\right)$ $o = e^\alpha = \left(\frac{p}{1-p}\right)$ $p = \frac{o}{1+o} = \frac{e^\alpha}{1+e^\alpha}=\frac{1}{1+e^{-\alpha}}$ with the ...


3

That is a huge association. It goes from basically everybody below to everybody above who attend. Fitting the model: att <- c(0,1,0,1,0,1) exam <- factor(c(0,0,1,1,2,2)) w <- c(1482, 300, 1094, 2822, 57, 1422) f <- polr( exam ~ att, weights=w) gives Call: polr(formula = exam ~ att, weights = w) Coefficients: att 2.925251 Intercepts: ...


3

Quantifying non-response bias is a very difficult statistical exercise. You are absolutely right that it should affect the reported figures... but the standard software won't do it, so there is where most people stop and give up. Some responsible practices include: Creating and utilizing nonresponse adjusted weights, and utilizing them in your standard ...


2

The R help manual cites the Fisher letter to the Australian Journal of Statistics. In it he notes, by example: If the observations in a $2 \times 2$ table are distinctly out of proportion (and indeed in other cases also) we may wish to set limits to the true product ratio, e.g. the observed table $$ \begin{array}{cc} 10 & 3 \\ 2 & 15 \end{...


2

The problem with converting odds ratios into risk differences is that the risk difference for a 1-unit change in age depends on the level of age. This is because logistic regression is a nonlinear model. It is linear in the log odds but nonlinear in the probability. To get risk differences at each level of age, you need to use a marginal effects procedure (...


2

$$\hat Var(log(\frac p{1-p})) \approx \frac 1 {Np(1-p)}$$ $$\hat Var\left[log\left(\frac {p_1/(1-p_1)}{p_2/(1-p_2)}\right)\right] \approx \frac 1 a + \frac 1 b+ \frac 1 c +\frac 1 d$$ where $p = X/N$ is estimate of parameter $\pi$ in $X\sim Bin(\pi,n)$ First one is for log ODDS, and second one is for log ODDS RATIO. In the linked page, OP misused ODDS.


2

Your results differ because you are actually fitting two different models, one where you adjust for all predictors except for $x_1$, and another where you adjust for all predictors except $x_2$. What's happening is that you are trying to treat the vector of predictors $\{x_1, \ldots, x_k\}$ as dummy variables by assuming that one needs to be omitted from the ...


2

While it cannot create the table in exactly how you specified, you can calculate risk ratios (and other measures) using the zEpid library. This library supports both calculating from summary counts (details here) and directly from pandas DataFrame objects (details here). The library does not directly calculate p-values, but you can easily do this by a ...


2

Norm Breslow argued that we should prefer the odds ratio as an association measure precisely because it's possible to have an odds ratio of exactly $\infty$. Some exposures in reality have deterministic relationship with the outcomes, but relative risks have theoretical upper bounds that depend on the "design" (margins of the contingency table). If the ...


1

Let's write the risk ratio and the odds ratio in comparable forms with the numerators and denominators each representing a row in your table: $$RR = \dfrac{30/(30+70)}{20/(20+100)} =1.8\\ OR = \dfrac{30/70}{20/100}\approx 2.143 $$ so they are not the same, as you say. If one of them had been $1$ then the other would also be $1$, as for example in $$RR = \...


1

Note that with the command shown here, the use of the cumulative logit model renders the (REFERENCE=FIRST) specification irrelevant, and the default setting of (ORDER=ASCENDING) is applied, so the model is set up to predict the probability of the lower category (I'm assuming that's non-use). Thus the threshold (intercept) term is opposite in sign of what you'...


1

The table displayed in this question shows the choices of parameter values for a simulation study, not experimentally determined or calculated odds ratios. The simulation involved 12 binary covariates (the $\mathsf {x_i}$) each with a prevalence of 20% having the indicated associations with outcome without treatment ($\mathsf {OR_C}$). In one set of ...


1

I assume that you have scaled your percentages to values in [0,1] (or in the open interval (0,1) after you've added/subtracted 0.001; you might want to consider adjusting by smaller values, as values very close to 0/1 will be extreme on the log-odds or logit scale; e.g. logit(0.001) = -6.9, logit(0.01) = -4.59). I'll call these numbers "pleasure indices". ...


1

Yes, you could report it that way. The probability of the outcome when eat_hotdog17=0 is $$ p = \dfrac{1}{1+\exp(-0.814)} \approx 30\% $$ When eat_hotdog=1 $$ p = \dfrac{1}{1+\exp(-0.814 - 0.464)} \approx 21\% $$


1

This is an interesting problem and not one I've thought about before, but I think we can reason through this using the standard logic of regression. You're right in noticing that you can't include all the categories though it's because the design matrix isn't full rank, not because of collinearity (if there is an intercept in the model). In regression with ...


1

Okay, I think I've figured this one out. First some notation. We have an observation $i$ in group $j$ where the outcome of interest $Y$ follows a Bernoulli distribution and you'll be placing a linear predictor on it's $p$ parameter using the inverse logit link: \begin{align} Y_{ij} &\sim \text{B}(p_{ij})\\ p_{ij} &= \text{logit}^{-1}(\alpha + \beta ...


1

Odds ratio (g1 vs g2) = odds(Y=1|g1)/odds(Y=1|g2) = exp(log(odds(Y=1|g1))-log(odds(Y=1|g2))) = exp(a1-a2). First "=" comes from the definition of odds ratio. Second "=" comes from x/y = exp(log(a)-log(b)) Third "=" comes from the definition of logistic regression. For any 2 subjects A and B, let $X_A$ and $X_B$ are their values on covariates, then OR(A ...


1

Neither, maybe you are confused with odds and odds-ratios. But odds and odd ratios are two different things. The odds ratio is defined as the ratio of the odds of A in the presence of B and the odds of A in the absence of B. Or equivalently (due to symmetry), the ratio of the odds of B in the presence of A and the odds of B in the absence of A. Given ...


1

The ANOVA is Type II sum of squares and the method you used below is Type III sum of squares. In general, they don't agree. Models for independent data give the same results. For mixed models, they differ. Type II and Type III sum of squares handle the random effect differently. The Type III method assumes, (incorrectly in most cases) that if the "true" data ...


1

Here, because of the presence of interaction, the main effects don't have a direct interpretation. That's not true. The "Farm Small" effect is the log odds ratio for TB where there are 0 cattle (if it's a projection, shift "Cattle" to a useful center). The "cattle" effect is the odds ratio for "TB" for "one differing value of cattle" where farmsize is big. ...


1

@Bjorn is trying to get you to see that having odds ratios above 1 in the sample doesn't tell you whether those odds ratios would be above 1 in the population. It's entirely consistent for there to be an odds ratio less than 1 in the population (i.e., that walking and calisthenics do indeed reduce the odds of ADL disability) and for you to have observed the ...


1

To test whether the odds ratios are different, you could run a meta-analysis on all 9 RCTs and test drug type as a moderator, with drug 3 set as the reference group. If it's significant, you could say it has the greater effect relative to placebo. Personally, I would comment on heterogeneity for this single analysis. You can't conclude that these odds ratios ...


1

The issue with statistical model assumptions in general is that they are never fulfilled precisely, so we more or less always do analyses in which assumptions are violated. The important question is whether assumptions are violated in ways that will cause misleading results, which is not exactly the question that misspecification tests such as Kolmogorov-...


1

The problem with the approach you have described is that those who are exposed to Injury $I$ are not mutually exclusive of those who are exposed to Condition $C$. So there are presumably individuals who have condition $C$ who are also Injured ($I$). But you also could have individuals who have condition $C$ who are not injured ($I$). As a result simply ...


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