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Schirmer test = 0,00 in Sjögren's syndrome. The test works by the principle of capillary action, which allows the water in tears to travel along the length of a paper test strip in an identical fashion as a horizontal capillary tube. The rate of travel along the test strip is proportional to the rate of tear production.
During World War II, production of German tanks such as the Panther was accurately estimated by Allied intelligence using statistical methods. In the statistical theory of estimation , the German tank problem consists of estimating the maximum of a discrete uniform distribution from sampling without replacement .
[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 .
The table shown on the right can be used in a two-sample t-test to estimate the sample sizes of an experimental group and a control group that are of equal size, that is, the total number of individuals in the trial is twice that of the number given, and the desired significance level is 0.05. [4]
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A randomness test (or test for randomness), in data evaluation, is a test used to analyze the distribution of a set of data to see whether it can be described as random (patternless). In stochastic modeling , as in some computer simulations , the hoped-for randomness of potential input data can be verified, by a formal test for randomness, to ...
Digital marketing is the secret to scaling your business to the top. It is easier for businesses to stand out in their respective industries if they develop digital marketing skills, but handing ...
The rule can then be derived [2] either from the Poisson approximation to the binomial distribution, or from the formula (1−p) n for the probability of zero events in the binomial distribution. In the latter case, the edge of the confidence interval is given by Pr(X = 0) = 0.05 and hence (1−p) n = .05 so n ln(1–p) = ln .05 ≈ −2