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1

It doesn't make much sense to compare coefficients to p-values like you do. This has nothing to do with logistic regression in particular, the same applies to linear regression, so let us use that as an example. If the model is $Y_i = \beta_0 + \beta_1 x_i + \epsilon_i$, but then you transform the predictor linearly to $Y_i = \beta_0^* + \beta_1^* (...


1

Bonferroni correction is not about sample size, but about the number of tests. Look at the rationale for applying the Bonferroni correction: If I have a single test, then the chance of having a false positive is $\alpha$. In other words, even if the H0 is true, you will reject it in a ratio of $\alpha$ cases. If you now use two tests, e.g. with different ...


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Yes (although I would not call it "concealing"). The point is that if an interaction is relevant, the main effect may or may not be significant in itself. Thus, it is generally considered a bad idea to start with an all-additive, all-linear model, since effects may be non-linear and non-additive. There are literally tons of posts on this. I recommend those ...


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What you apparently want to do is to start by evaluating the relationship between the 3 classes and each of your features individually. For each continuous feature you are proposing a one-way ANOVA of that feature against the known classification (3 classes). ANOVA is not appropriate for a similar evaluation of the binary features, as interpretation of ANOVA ...


2

Notice that the last three cases don't give exactly equal answers when properly rounded (e.g. in the last one, t-test gives 64.0% vs 64.1% in Bayes). If you have two iid. normally distributed groups of equal size, then the p-value is going to be equal to the Bayesian result, because they are the same function. The only difference lies in the interpretation. ...


1

The paired-sample t-test is used to measure the probability that we observe a difference in sample-means between two data sets at least as large as the one we observed, given that they are both generated by a Normal distribution of unknown variance and equal mean to each other. That is, that data set 1 is generated by distribution $\mathcal{N}(\mu, \sigma_1^...


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The short answer is that you have very strong evidence from the first dataset that the two populations have indeed different mean responses and strong evidence from the second. The way to combine results from multiple studies is by conducting a meta analysis. However in the case of just two studies it's irrelevant. Now, different scientific fields use ...


0

I don't know about the Python function in question. But if a t test is used to determine the p value --- which it probably is for at least moderately large sample sizes ---, the p value for a one-sided test follows the usual rules. That is, the p value for the one-sided test is either one-half the p value of the two-sided test, or 1 minus one-half the p ...


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What happens is that the regression forces a line through the heavily weighted data points. The solution is to scale the weights to mitigate their effect. Analysing the data with scaled weights yields reliable p-values.


3

The test statistic ($t$ in your example) and all calculations to reach that point depend only on the null hypothesis $H_0$ and nothing else. The p-value is affected by the alternative hypothesis $H_1$ as the $H_1$ identifies which values are considered as "extreme" values and the p-value calculates the proximity of the final result (your $t$) to those ...


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The reason that results in the wrong direction don’t give small p-values is because they provide terrible evidence in favor of the alternative. Imagine a null hypothesis of a fair coin and an alternative of a bias towards heads. You then flip the coin 100 times and get 99 tails. You have terrible evidence in favor of your alternative hypothesis. This can ...


1

Seems to me that the MCQ question cannot be answered with any singular value. Both of your justifications are true, and there is no way to know whether the null is true or false.


2

If you do as was suggested in a comment by Michael Lew and use log(nutrient) in the plot, you see a reasonably linear trend for each temperature: lattice::xyplot(length ~ log(nutrient) | temperature, data = mydata) Accordingly, I tried the following model, which fits quite well: mylm = lm(length ~ factor(temperature) * log(nutrient), data = mydata) The ...


1

When I read this question:"Does frequency of occurrence (FO) of pieces eaten differ between species or year?", I do not immediately think of using a 2-factor interaction model, but rather will have thought of testing two different models, one with "year" and one with "species". Furthermore the description of the variable "frequency of occurrence" makes me ...


3

Fitting 10K GLMs will be slow. A 2x2 Chi-squared test for each comparison of two event counts will be faster and will work well enough provided your event counts don't get too extreme, e.g. your event counts range from [5, 495] for your sample size of 500. You can work with the event counts directly so you won't need to create the vectors of zeroes and ones. ...


1

Some Machine Learning techniques are based on p values, e.g., ANOVA feature selection. https://scikit-learn.org/stable/modules/generated/sklearn.feature_selection.f_classif.html f_classif (ANOVA) returns a p-value for each feature. As Tim writes the p-value is used for hypothesis testing. ANOVA tests whether means of two or more samples are equal. A low p-...


4

$p$-values are used for hypothesis testing. In machine learning you don't have any hypothesis to test, & you don't care about it. You care about making accurate predictions and hypothesis testing has nothing to do with it. You can check the The Two Cultures: statistics vs. machine learning? thread for some related discussion.


2

Often with data that look like this, the goal is to determine the x value beyond which there is no further (statistical) increase in the y variable. And also to determine the plateau y value. This might be done with linear-plateau or quadratic-plateau models. But eyeballing your data, a Cate-Nelson approach might be more useful to determine these values. ...


1

The first step usually taken for non-linear relationships like this is by adding a polynomial term. A linear relationship is y ~ x, but you can use a quadratic approximation by doing y ~ x + x ^ 2. This is like x interacting with itself: the relationship between x and y depends on the value of x. It is still a linear model, since you are adding up ...


7

There is no intercept term in this model, so the best-fit line must go through the origin. A best-fit line that goes through the origin will clearly have a positive slope for the data you're showing. It seems including x_ran = sm.add_constant(x_ran) will add the constant term. You should then find that your intercept is significantly diffrent from zero, ...


5

Whatever X1 is, the p-value you are referring to is so low because the null hypothesis implicitly made by your modeling software/package is that X1 coefficient is equal to zero. You estimate a coef of 1746 $\pm$ SE 135... Which is definitely different from zero.


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