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This reduces the chi-squared value obtained and thus increases its p-value. The effect of Yates's correction is to prevent overestimation of statistical significance for small data. This formula is chiefly used when at least one cell of the table has an expected count smaller than 5. = =
The sign test is a statistical test for consistent differences between pairs of observations, such as the weight of subjects before and after treatment. Given pairs of observations (such as weight pre- and post-treatment) for each subject, the sign test determines if one member of the pair (such as pre-treatment) tends to be greater than (or less than) the other member of the pair (such as ...
The p-value for the permutation test is the proportion of the r values generated in step (2) that are larger than the Pearson correlation coefficient that was calculated from the original data. Here "larger" can mean either that the value is larger in magnitude, or larger in signed value, depending on whether a two-sided or one-sided test is ...
MSWD < 1 if the observed scatter is less than that predicted by the analytical uncertainties. In this case, the data are said to be "underdispersed", indicating that the analytical uncertainties were overestimated. MSWD > 1 if the observed scatter exceeds that predicted by the analytical uncertainties.
For the test of independence, also known as the test of homogeneity, a chi-squared probability of less than or equal to 0.05 (or the chi-squared statistic being at or larger than the 0.05 critical point) is commonly interpreted by applied workers as justification for rejecting the null hypothesis that the row variable is independent of the ...
These values can be calculated evaluating the quantile function (also known as "inverse CDF" or "ICDF") of the chi-squared distribution; [24] e. g., the χ 2 ICDF for p = 0.05 and df = 7 yields 2.1673 ≈ 2.17 as in the table above, noticing that 1 – p is the p-value from the table.
From this representation, the noncentral chi-squared distribution is seen to be a Poisson-weighted mixture of central chi-squared distributions. Suppose that a random variable J has a Poisson distribution with mean λ / 2 {\displaystyle \lambda /2} , and the conditional distribution of Z given J = i is chi-squared with k + 2 i degrees of freedom.
For that value of λ, the chi-square statistic is about 3.062764. There are 10 cells. There are 10 cells. If the null hypothesis had specified a single distribution, rather than requiring λ to be estimated, then the null distribution of the test statistic would be a chi-square distribution with 10 − 1 = 9 degrees of freedom.