Search results
Results from the WOW.Com Content Network
[15] [16] But if the p-value of an observed effect is less than (or equal to) the significance level, an investigator may conclude that the effect reflects the characteristics of the whole population, [1] thereby rejecting the null hypothesis. [17] This technique for testing the statistical significance of results was developed in the early ...
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]
The choice of a significance level may thus be somewhat arbitrary (i.e. setting 10% (0.1), 5% (0.05), 1% (0.01) etc.) As opposed to that, the false positive rate is associated with a post-prior result, which is the expected number of false positives divided by the total number of hypotheses under the real combination of true and non-true null ...
Exact tests that are based on discrete test statistics may be conservative, indicating that the actual rejection rate lies below the nominal significance level . As an example, this is the case for Fisher's exact test and its more powerful alternative, Boschloo's test. If the test statistic is continuous, it will reach the significance level ...
This statistics -related article is a stub. You can help Wikipedia by expanding it.
Z-test tests the mean of a distribution. For each significance level in the confidence interval, the Z-test has a single critical value (for example, 1.96 for 5% two tailed) which makes it more convenient than the Student's t-test whose critical values are defined by the sample size (through the corresponding degrees of freedom). Both the Z ...
This can create a subtle difference, but in this example yields the same probability of 0.0437. In both cases, the two-tailed test reveals significance at the 5% level, indicating that the number of 6s observed was significantly different for this die than the expected number at the 5% level.
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 ...