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An example of Neyman–Pearson hypothesis testing (or null hypothesis statistical significance testing) can be made by a change to the radioactive suitcase example. If the "suitcase" is actually a shielded container for the transportation of radioactive material, then a test might be used to select among three hypotheses: no radioactive source ...
Test statistic is a quantity derived from the sample for statistical hypothesis testing. [1] A hypothesis test is typically specified in terms of a test statistic, considered as a numerical summary of a data-set that reduces the data to one value that can be used to perform the hypothesis test.
Statistical tests are used to test the fit between a hypothesis and the data. [1] [2] Choosing the right statistical test is not a trivial task. [1] The choice of the test depends on many properties of the research question. The vast majority of studies can be addressed by 30 of the 100 or so statistical tests in use. [3] [4] [5]
A one-sample Student's t-test is a location test of whether the mean of a population has a value specified in a null hypothesis. In testing the null hypothesis that the population mean is equal to a specified value μ 0, one uses the statistic = ¯ /,
The likelihood-ratio test, also known as Wilks test, [2] is the oldest of the three classical approaches to hypothesis testing, together with the Lagrange multiplier test and the Wald test. [3] In fact, the latter two can be conceptualized as approximations to the likelihood-ratio test, and are asymptotically equivalent.
Neyman–Pearson lemma [5] — Existence:. If a hypothesis test satisfies condition, then it is a uniformly most powerful (UMP) test in the set of level tests.. Uniqueness: If there exists a hypothesis test that satisfies condition, with >, then every UMP test in the set of level tests satisfies condition with the same .
In statistical hypothesis testing, a two-sample test is a test performed on the data of two random samples, each independently obtained from a different given population. The purpose of the test is to determine whether the difference between these two populations is statistically significant .
Naaman [3] proposed an adaption of the significance level to the sample size in order to control false positives: α n, such that α n = n − r with r > 1/2. At least in the numerical example, taking r = 1/2, results in a significance level of 0.00318, so the frequentist would not reject the null hypothesis, which is in agreement with the ...