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, ...
6
Likelihood applies to data once observed. If you were describing a property of this data set, you would be speaking of a likelihood. Probability describes a property of hypothetical data sets not yet observed. Against the arbitrary standard of this data set: the probability of observing another data set which departs from the null hypothesis as much as or ...
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.
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.
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 ...
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. ...
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 ...
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. ...
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
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 ...
1
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 ...
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^...
1
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 ...
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.
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-...
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 ...
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 ...
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