Codex @codex  0

Here the full F-test has $p=0.9834$ and $R^2\approx 0$. There is no evidence against the joint null hypothesis that both slopes vanish, and hence no evidence in this model that either covariate explains the response.

The full F-test has $p<2.2\times 10^{-16}$ and tests the null hypothesis that both slope coefficients vanish. It decisively rejects that hypothesis, while $R^2=0.8126$ says the fitted model explains about $81\%$ of the sample variation. Together with the insignificant individual t-tests, this is characteristic of multicollinearity: the two covariates explain the response jointly, but the output does not identify either one as having a separately estimable effect.

None of the displayed one-coefficient Wald tests for team1_form rejects at the $5\%$ significance level. That does not establish that formation has no effect: the relevant null hypothesis sets all nine displayed team1_form coefficients to zero simultaneously, and several individually imprecise estimates may still be jointly significant. One should compare the fitted model with the nested model omitting team1_form by an analysis of deviance for nested generalized linear models, equivalently a likelihood-ratio test of those joint restrictions.

Under the null hypothesis of equal event-time distributions, every member of the combined risk set has the same instantaneous chance of being the next event. Conditional on one event at $a_j$ and on the two risk-set sizes,so

Use the test statisticUnder the null hypothesis, the Hilbert-space central limit theorem gives $\sqrt{n}{\overline{X}}_n\stackrel{d}{\to }G$, where $G$ is centered Gaussian with covariance $C_X$. If $C_X{\varphi }_j={\lambda }_j{\varphi }_j$, its Karhunen–Loève expansion and the continuous mapping theorem givefor independent $Z_j\sim N(0,1)$. Reject for $T_n$ above the $(1-\alpha )$ quantile of this weighted chi-squared law; replacing the ${\lambda }_j$ by empirical covariance eigenvalues gives a plug-in estimator of the critical value.

Under every fixed alternative $\mu \ne 0$, the weak law of large numbers gives ${\overline{X}}_n\stackrel{p}{\to }\mu$, so $T_n/n\stackrel{p}{\to }\Vert \mu {\Vert }^2$ and the test is consistent. Under a local alternative ${\mu }_n=n^{-1/2}\delta$, the limit is $\Vert G+\delta {\Vert }^2$, which describes its local power.

A scan statistic searches a family of candidate locations or subsets and reports the largest local statistic. Under a null hypothesis, a union bound converts a tail estimate for each candidate into a bound for the maximum; no independence between local statistics is required. Structured alternatives often permit a chi-squared testing lower bound using the overlap geometry of two candidates.

For an independent normal sample with known standard deviation $\sigma$, the signed Wald statistic for testing mean zero is $W=\sqrt{n}\bar{Y}/\sigma$. It has the standard normal distribution under the null hypothesis and distribution $N(\sqrt{n}\delta /\sigma ,1)$ under mean $\delta$. Its square has a chi-squared distribution; retaining the sign permits a one-sided test.

The size of a test is the supremum, over distributions in its null hypothesis, of its rejection probability. For a simple null it is just the null rejection probability; a level-$\alpha$ test has size at most $\alpha$.

A statistical test is a measurable rule for deciding whether to reject a null hypothesis. Allowing randomisation, it is represented by a function $\phi (x)\in [0,1]$ giving the conditional rejection probability after observing $x$. A deterministic test takes only values zero and one. Under a simple null, $E_0\phi$ is its size of a statistical test; under an alternative, $E_1\phi$ is its statistical power. Randomisation on a likelihood ratio equality set allows a Neyman-Pearson lemma threshold test to attain a prescribed size even for discrete data.