New answers tagged

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There are a couple of things to note here. In the case of the tests you've done the p-value of the 2-sided test will be twice that of the ">" test. In hypothesis testing we are generally looking at values as or more extreme than the observed test statistic. For 1-sided ">" t-test we calculate the (tail) probability of getting a ...


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I will prefer you install jtools library and use summ function jtool package to run the output model which will give you the p_value of the fixed effects. Library (lme4) Library (jtools) model=lmer(value~status+1|experiment); summ(model);


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Its probably due to increasing sample size, which by the manual formulas we can see that the larger the sample size, the "easier" it gets to get a low p value


3

We can simplify the power calculation by approximating the sample distributions as normal distributions with $$\sigma \approx \sqrt{pq/n}$$ approximation $pq \approx 0.5^2$ such that $$\sigma\approx \frac{0.5}{\sqrt{n}}$$ approximating the power by considering the entire left tail as non-rejection of the hypothesis (which is not entirely true because a tiny ...


0

Why do we take as p-value the proportion/probability of permutation with test statistic equal or larger than the observed one? This is contextual, and not all permutation tests work this way. In any hypothesis test there is a test statistic that is used to establish an ordering over all possible data outcomes, describing whether outcomes are more or less ...


3

Why do we take as p-value the proportion/probability of permutation with test statistic equal or larger than the observed one? A p-value is defined to be the probability, according to the statistical model, of getting a test statistic value at least as large as that observed when the null hypothesis is true. The “getting a test statistic value” is often ...


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There are various ways you could examine this problem analytically, but a typical way is to frame the problem as a hypothesis test for a stipulated probability for the coin. Suppose we let $X_1,X_2,X_3, ... \sim \text{IID Bern}(\theta)$ denote the outcomes of the coin-flips where $\theta$ is the probability of flipping a head (here denoted by a one). We ...


1

We use the likelihood method. Suppose that you have that $$\mathcal{P}(X_n = 1 ) = p$$ for some $p \in [0,1]$. Then the likelihood that you will see the sequence of coinflips $X=X_1...X_n = s_1...s_n=s$ $$\mathcal{P}^{(n)}(X = s) = p^{\mathcal{N}_1(s)}(1-p)^{\mathcal{N}_0(s)} $$ where $\mathcal{N}_i(s)$ is the number of $i$'s in $s$. Thus the likelihood ...


4

Your "upper/lower bounds" form 95% confidence intervals. Their interpretation is: if you were to repeat your experiment many times, calculating 95% CIs each time, then 95% of these CIs would contain the true parameter (assuming your model specification is correct). Yes, that is a very cumbersome definition, but it is the best that frequentist ...


0

For complicated settings it is important to validate your modelling assumptions. If your distributional assumptions appear correct, then you have confidence that although you may have not selected the true distribution, you have adequately approximated the true distribution, and hence your p-values are still relevant.


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In case you are still interested, I found this paper that seems to address the problem of integrating different analysis algorithms on the same set of RNA-seq data: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4344485/. It's basically proposing a weighted sum of all p-values and the weights are based on the ROC of each algorithm. The name of the paper is: ...


1

Next to the mean figures in brackets is standard deviation ANOVA is an omnibus test, that means when you have a significant model F test, this tells you that overall there are differences between groups. You then need to follow this up with a post-hoc procedure (if not doing planned contrasts, i.e., if you have no apriori hypotheses you want to test). The ...


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This comment is for the future, not for your immediate frequentist statistics-laden question. To obtain direct evidence for fairness of a coin you need to define the state of prior knowledge and to define fairness, e.g, the probability of heads $\theta$ being between 0.49 and 0.51. Then acquire a lot of data and compute the Bayesian posterior probability ...


0

Because the problem does not provide enough information, I will show four relevant exact binomial tests in R. For such small numbers of tosses, I would not use normal approximations. Right-sided alternative. Four tosses. right-sided $(H_0: p=.5$ against: $H_a: p > .5).$ Getting four Heads in four tosses does not quite lead to rejection at the 5% level. ...


0

This is called a chunk test. You would make some full_model that includes all of the types of anesthesia, plus whatever other variables you want to include (could be none, but it looks like you have some). Then you would make some reduced_model that includes those other variables but not the any of the anesthesia variables. Finally, you compare the nested ...


2

Let's look at the P-value of a one-sided t test. Suppose we have data from $n=10$ observations from a population distributed $\mathsf{Norm}(\mu,\sigma),$ with $\mu$ and $\sigma$ both unknown. We want to test $H_0: \mu =20$ against $\mu < 20.$ Null hypothesis true. We might have data from a population for which $H_0$ is true, sampled using R, as follows: ...


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1 They don’t mean what people think they mean Am I right that this is not a p-value (which is the probability to see this or more extreme value of a test statistic)? Is it a correct procedure for a statistical testing? I have a gut feeling that it is a wrong situation to apply hypothesis testing, but I can not formally answer why. One could argue that ...


4

1. Citizenship example This seems to be a poor but valid test. It makes sense with an extreme p-value cut off $\alpha = 0$. This way we would only reject those citizens whose profession is not found anywhere in the US. So maybe we would say "Robbert cannot be a US citizen because he is a suicide bomber and there are no suicide bombers in the united ...


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"Am I right that this is not a p-value (which is the probability to see this or more extreme value of a test statistic)?" Good question! Yes, you're right, it's not a p-value. What's more the example is not a hypothesis test and it's not a significance test. Anyone who uses it as an argument to discard p-values or hypothesis tests is either ...


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I agree that confidence intervals provide a lot more for performing inference than a single p-value for a single hypothesis, but there is no reason to ditch the p-value and no reason to rely solely on bootstrap confidence intervals. The confidence interval is the set of all hypotheses that are not significant (one would fail to reject) at a specific alpha ...


13

The author of the article suffers from not understanding that hypothesis tests and confidence intervals serve different inferential purposes: The confidence interval (bootstrap or otherwise) serves to provide a plausible range of estimates for a target parameter. The hypothesis test serves to make a decision as to whether there is evidence or lack of ...


0

There are already some very good answers here, but I figured I would offer my own take. You're correct in identifying the interpretation of the p value of the test: The probability of seeing 9 or more heads from a fair coin assuming all other assumptions are correct. Note that the binomial test is in terms of the coin's bias, not in terms of the number of ...


0

Thanks to @BruceET and @Glen_b some confusion was clarified, which I figured I'd write here to be helpful for anyone who also had my issue. The main issue that I had was that I previously did a simulation to look at the frequency at which fair coins produce p-values that would be false-positives. If I have a p-value of 3.4% and test it for p-value<3.5% (...


1

There are two kinds of tests for binomial proportions. One uses a normal approximation to binomial distributions. It works fine if $n$ is large enough and $p$ is sufficiently near $1/2$ (roughly speaking, so that $np$ and $n(1-p)$ both exceed $5).$ Example: Suppose you have $x = 45$ Successes out of $n = 50$ Bernoulli trials. You wish to test $H_0: p= 0.5$ ...


2

shouldn't my binomial test just be the binomial distribution Beware speaking so loosely that concepts become muddled. A test is not just a statistic (e.g. it also needs data, hypotheses and a rejection rule for example) and a statistic is not a distribution (though it has a distribution). Given the assumptions, the test statistic will have the null ...


1

IMHO, confidence intervals are a better method of expressing results. This is especially true when comparing results to be included in meta analysis and for "not significant" answers. This avoids the all too common misrepresentation of not significant results as significantly insignificant. I don't know in which "camp" that puts me, ...


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