# Tag Info

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Your R code gives the following prop.test(x=c(55,56),n=c(100,216), alternative = "less") 2-sample test for equality of proportions with continuity correction data: c(55, 56) out of c(100, 216) X-squared = 24.096, df = 1, p-value = 1 alternative hypothesis: less 95 percent confidence interval: -1.0000000 0.3934584 sample ...

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Bootstrapping done properly will not inflate your metrics. Bootstrapping allows you obtain the estimate of the probability distribution of the test statistic. Getting the p-values from this distribution involves using the actual sample size. Here's an example of incorrect application of bootstrapping. A scientist was asked to obtain 99.9th quantile of the ...

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ANOVA assumes all group distributions under consideration to be normal with the same variance. Consequently, the only way they can differ is in their means. However, you allude to the fact that the test could pick up on other differences, and if you want to use ANOVA to test something else, it is common in statistics to use a surrogate test. The best example ...

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low (vs. high) has a restricted range (i.e., only 2 values). Number of children presumably has a much wider range. The narrower the range of an $X$ variable, the less power there is to test it. That's a general principle. Specific to this example, it sounds like the underlying variable (income) was subjected to a median split, or something similar. If ...

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Significance is not purely about the magnitude of an effect. For instance, if the parameter/effect is $1$ light-year or $9.46×10^{12}$ km, then it is different figures but the same distance. (You might argue that this example is bad because I changed the units, but how are you gonna make sure that the units match when you compare 'pets per unit of income' ...

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See how you get the message "waiting for profiling to be done..." when you run confint? That is because confint returns profile likelihood confidence intervals, which inverts the likelihood ratio test statistic to find regions where the LRT would fail to be rejected. The p values in summary come from a Wald test. The two methods are almost the same,...

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The answer by @Dave2e is fine (+1), but I wanted to give an Answer based mainly on a specific example and showing computations of P-values. Consider the following fictitious data: set.seed(2022) x1 = rnorm(30, 350, 50) x2 = rnorm(30, 300, 70) summary(x1); length(x1); sd(x1) Min. 1st Qu. Median Mean 3rd Qu. Max. 205.0 309.6 346.7 344.2 ...

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This is a common way to express scientific notation. $$A e B = A \times 10^{B}$$ We do this to save space. If your case, it would be possible to write the number as $0.000000000000465$, but that is a lot of writing, and it’s easy for our eyes to glaze over all of the digits. At a glance, can you tell if that number is bigger or smaller than $0.... 2 I think, "assume within the null hypothesis" is not a valid statement. We do not assume nothing in null hypothesis. We check if hull hypothesis can be rejected. And to be able to check it, we make some assumptions. These assumptions help us to develop formulas for test statistic and p-value. So: if we also assume that standard deviation in all the ... 2 First note that a bootstrap sample (draws with replacement) must contain exactly the same number of observations (independent draws) like the original data. I wonder however how bootstrap could be useful in this case. As an alternative to the parametric paired t-test, there is a Monte Carlo test, but it is not based on bootstrapping, but on label ... 2 The two comments from years ago get at why stepwise regression might not be so helpful for your task. However, it is legitimate to wonder why variables are "insignificant" if the stepwise procedure removes "insignificant" variables. A potential culprit is multicollinearity of your features. In fact, with multicollinearity, we can have a ... 2 Let us define$\mu_1= 383$and$\mu_2= 168$with reasonable small standard deviations. The differences between the means are over 4 standard deviations apart. By eye one can judge, with a large difference, that it is very very unlikely to occur by chance alone. With regard to a discussion of the formal analysis: The right tail t-test is comparing if$\mu_1(...

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Your idea does work. This can be viewed as a type of Monte Carlo simulation where you randomly pick a set of points from the target distribution and regress to them, and you get a distribution of MSE. you can then check if the MSE you get for the real target distribution is significantly lower than from a randomly assigned distribution. The common way of ...

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You cannot just divide the p-value from a 2-sided hypothesis by 2 to get a p-value from a 1-sided hypothesis. You must first check which side of the t distribution your test statistic falls on and whether that is consistent with or not consistent with your null hypothesis. A t-test will use a test statistic (we’ll call it x). Under the null hypothesis, x ...

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p-value < 0.0509 Incredible, this is the closest I've seen someone get to the nominal value. Please keep in mind that The Difference Between “Significant” and “Not Significant” is not Itself Statistically Significant. Though you have failed to reject the null (by a hair's worth) I would look at the confidence interval for the effect to see what sorts ...

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This question or questions very like it come up a lot. Your issue is that there are two distinct and incompatible ways of testing hypotheses using p-values, and many textbooks contain a confusing mix of the two. There is already an excellent explanation of the two approaches in this answer, but I'll summarise it (badly) here: In Fisher's approach, you ...

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Both methods yield the exact same point estimate; in a sense, the one-step regression method is using the two-step method implicitly. You are correct that the naive standard errors are incorrect for the two-step method, and it indeed should not be used. The authors of the paper do not propose using the naive standard errors from such a regression. Any tests ...

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There are some issues with the type of used cost functions. For these issues see a previous second version of this answer. The current question is edited. Your first fit already determined an effect that the mean is dependent on the features. Or at least expressing the mean as a linear function of the regressors gives a lower mean squared error than ...

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This means $4.65 \times 10^{-13}$. Note that this is almost certainly not an exact number, but rather a statement of machine precision - basically it's closer to zero than your computer can measure.

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It sounds like what you are after is a meta-analysis. This is the frequentist analog to a Bayesian analysis using an informative prior. A meta-analysis can come in many forms. One option is to combine the subject-level observations into a single data set and run a single model to produce p-values and confidence intervals. This is the ideal approach. ...

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The F-statistic on which the computation of a p-value is based can be expressed in terms of $R^2$ (and also the degrees of freedom). See also Does $r$-squared have a $p$-value? The relationship between $R^2$ and the F-statistic is for simple linear regression with a sample of size $n$ the following $$F = \frac{R^2}{1-R^2} (n-2)$$ So yes, there is a relation ...

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You should use the output from margins (though possibly modified or augmented). This really depends on the question you are trying to answer by running this probit. Your question lacks that info, so I will tell you what Stata calculated and give some suggestions about what I might do in your shoes. The output of margins in an average marginal effect of mino, ...

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