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Normally, the denominator of the chi-square statistic is the variance of the sample. Only in the Poisson distrinbution is the expected value = to the variance. So, what is the justification for this in a non-Poisson distributed sample?

2007-05-15 12:49:32 · 2 answers · asked by stevmg 3 in Science & Mathematics Mathematics

2 answers

"Chi square is a non-parametric test of statistical significance for bivariate tabular analysis (also known as crossbreaks). Any appropriately performed test of statistical significance lets you know the degree of confidence you can have in accepting or rejecting an hypothesis. Typically, the hypothesis tested with chi square is whether or not two different samples (of people, texts, whatever) are different enough in some characteristic or aspect of their behavior that we can generalize from our samples that the populations from which our samples are drawn are also different in the behavior or characteristic.

A non-parametric test, like chi square, is a rough estimate of confidence; it accepts weaker, less accurate data as input than parametric tests (like t-tests and analysis of variance, for example) and therefore has less status in the pantheon of statistical tests. Nonetheless, its limitations are also its strengths; because chi square is more 'forgiving' in the data it will accept, it can be used in a wide variety of research contexts.

Chi square is used most frequently to test the statistical significance of results reported in bivariate tables, and interpreting bivariate tables is integral to interpreting the results of a chi square test, so we'll take a look at bivariate tabular (crossbreak) analysis. "

2007-05-15 12:53:06 · answer #1 · answered by gutuku 2 · 0 0

As stated, the question doesnt make sense. If you really had an expected of 0, there is nothing to analyze - the probability mechanism assigned a value of zero to the likelihood of finding a cure. How was the expected computed? There must have been an underlying probability distribution, and the probability could not have been zero. Are you sure you dont have the Observed and Expected reversed?

2016-05-19 04:08:48 · answer #2 · answered by Anonymous · 0 0

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