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Since the procedure is step-down, we first test = (), which has the smallest p-value = =. The p-value is compared to α / 4 = 0.0125 {\displaystyle \alpha /4=0.0125} , the null hypothesis is rejected and we continue to the next one.
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 ...
The following table defines the possible outcomes when testing multiple null hypotheses. Suppose we have a number m of null hypotheses, denoted by: H 1, H 2, ..., H m. Using a statistical test, we reject the null hypothesis if the test is declared significant. We do not reject the null hypothesis if the test is non-significant.
The following table defines the possible outcomes when testing multiple null hypotheses. Suppose we have a number m of null hypotheses, denoted by: H 1, H 2, ..., H m. Using a statistical test, we reject the null hypothesis if the test is declared significant. We do not reject the null hypothesis if the test is non-significant.
The sign test is a special case of the binomial test where the probability of success under the null hypothesis is p=0.5. Thus, the sign test can be performed using the binomial test, which is provided in most statistical software programs. On-line calculators for the sign test can be founded by searching for "sign test calculator".
Statistical hypothesis testing is based on rejecting the null hypothesis when the likelihood of the observed data would be low if the null hypothesis were true. If multiple hypotheses are tested, the probability of observing a rare event increases, and therefore, the likelihood of incorrectly rejecting a null hypothesis (i.e., making a Type I ...
Step 5: The F-ratio is = / The critical value is the number that the test statistic must exceed to reject the test. In this case, F crit (2,15) = 3.68 at α = 0.05. Since F=9.3 > 3.68, the results are significant at the 5% significance level. One would not accept the null hypothesis, concluding that there is strong evidence that the expected ...
The new multiple range test proposed by Duncan makes use of special protection levels based upon degrees of freedom.Let , = be the protection level for testing the significance of a difference between two means; that is, the probability that a significant difference between two means will not be found if the population means are equal.