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Assuming that the samples are not biased in favour of any of the drugs, you can try to calculate the 95% confidence interval for each of the new drugs and check whether or not they intersect with the …

Does this one do the job? mid_point <- function(long1, lat1, long2, lat2) { if(abs(long1-long2 < 180)) { long <- (long1+long2)/2 } else { long <- 180 + (long1+long2)/2 } if(long > 18 …

It would be good to first understand the output of the simpler linear regression model (your glm is just an adaptation of that model to a classification problem) Check my answer to this question Begin …

Welcome to Cross Validated! Actually, there is no reason why you shouldn't perform a mean comparison test. Equal sample size and varaince are not a requirement. You may want to check this to see how …

If you have enough evidence to believe that the patterns in 2020 and 2021 data are different (i.e: they're not just two samples of the same population), then you should not expect the same model to wo …

You can relabel your classes as 1 and 0 or TRUE and FALSE, now the model will do it the way you expect. If those are not the labels, well, I would try to guess it. You can try to see the prediction f …

You can do it on R but you'll have to write the code yourself (there's no explicit method for the lm class that will do it for you) If your model is $Y:=\beta_0+\beta_1a+\beta_2b+\beta_3ab+\beta_4a^2+\ … beta_5b^2+\epsilon$, then: $\frac{\delta Y}{\delta a} = \beta_1 + b\beta_3 +2a\beta_4$ R will provide the $\beta_k$ coefficients with coefficients(R2), returning a numeric. …

The idea of principal component analysis is to transform your original variables into new uncorrelated or "perpendicular" ones. So if your original variables were already pretty much uncorrelated, it …

With only 1 observation, it is quite easy. Look how many standard deviations out of the mean your new data point is (this is, subtract the "data" mean and divide by the "data" standard deviation) Now …

My interpretation would be that the middle values of the sample are close to what you'd expect from normally distributed data, as it follows the straight line from the diagram closely. However, it se …

Those two results are exactly the same: zero! Try to see if the problem persists when using a more "normal" number, i.e: predict(logitMod, testdata[1]) in a range like $(-2,2)$ or $(-3, 3)$

Please remember this is a statistics site, not a code one. However, a quick look at the emmeans documentation package reveals the following: Using cov.reduce cov.reduce may be a function, logical va …

I'm seeing some "inconsistencies" on how R calculates the Akaike Information Criterion (AIC) for linear regression models. I'd like to get its expression so I can calculate it myself. …

Assuming you're using the base R lm/glm functions: If you createthe dummy variables yourself, the step process will treat them as separate variables, so you may get some of them removed while the others … You can also feed them into the model as categorical and R will do the dummification for you. …

That final peak is due to overfitting. Your model is way too complex for the data you are trying to fit. It is just finding the degree-4 polynomial that intersects the x value you have at the mean of …