Chi Square Statistics - Hobart and William Smith Colleges

Chi-Square Test of Independence

Chi-Square Test of Independence - Statistics Solutions

The distribution of the test statistic under the null hypothesis is approximately the same as the theoretical chi-square distribution. This means that once you know the chi-square value and the number of degrees of freedom, you can calculate the probability of getting that value of chi-square using the chi-square distribution. The number of degrees of freedom is the number of categories minus one, so for our example there is one degree of freedom. Using the CHIDIST function in a spreadsheet, you enter =CHIDIST(2.13, 1) and calculate that the probability of getting a chi-square value of 2.13 with one degree of freedom is P=0.144.

17/08/2016 · In various articles on the Chi-squared statistical test is referenced a threshold value of 0.05 for the probability density to reject the null hypothesis.

Pearson's chi-squared test - Wikipedia

You calculate the test statistic by taking an observed number (O), subtracting the expected number (E), then squaring this difference. The larger the deviation from the null hypothesis, the larger the difference between observed and expected is. Squaring the differences makes them all positive. You then divide each difference by the expected number, and you add up these standardized differences. The test statistic is approximately equal to the log-likelihood ratio used in the . It is conventionally called a "chi-square" statistic, although this is somewhat confusing because it's just one of many test statistics that follows the theoretical chi-square distribution. The equation is

Chi-Square Independence Testing | Real Statistics Using …

McDonald (1989) examined variation at the Mpi locus in the amphipod crustacean Platorchestia platensis collected from a single location on Long Island, New York. There were two alleles, Mpi90 and Mpi100 and the genotype frequencies in samples from multiple dates pooled together were 1203 Mpi90/90, 2919 Mpi90/100, and 1678 Mpi100/100. The estimate of the Mpi90 allele proportion from the data is 5325/11600=0.459. Using the Hardy-Weinberg formula and this estimated allele proportion, the expected genotype proportions are 0.211 Mpi90/90, 0.497 Mpi90/100, and 0.293 Mpi100/100. There are three categories (the three genotypes) and one parameterestimated from the data (the Mpi90allele proportion), so there is one degree of freedom. The result is chi-square=1.08, 1 d.f., P=0.299, which is not significant. You cannot reject the null hypothesis that the data fit the expected Hardy-Weinberg proportions.

Hypothesis testing - Handbook of Biological Statistics


How to Test Hypotheses - Statistics and Probability

It is important to distinguish between biological null and alternative hypotheses and statistical null and alternative hypotheses. "Sexual selection by females has caused male chickens to evolve bigger feet than females" is a biological alternative hypothesis; it says something about biological processes, in this case sexual selection. "Male chickens have a different average foot size than females" is a statistical alternative hypothesis; it says something about the numbers, but nothing about what caused those numbers to be different. The biological null and alternative hypotheses are the first that you should think of, as they describe something interesting about biology; they are two possible answers to the biological question you are interested in ("What affects foot size in chickens?"). The statistical null and alternative hypotheses are statements about the data that should follow from the biological hypotheses: if sexual selection favors bigger feet in male chickens (a biological hypothesis), then the average foot size in male chickens should be larger than the average in females (a statistical hypothesis). If you reject the statistical null hypothesis, you then have to decide whether that's enough evidence that you can reject your biological null hypothesis. For example, if you don't find a significant difference in foot size between male and female chickens, you could conclude "There is no significant evidence that sexual selection has caused male chickens to have bigger feet." If you do find a statistically significant difference in foot size, that might not be enough for you to conclude that sexual selection caused the bigger feet; it might be that males eat more, or that the bigger feet are a developmental byproduct of the roosters' combs, or that males run around more and the exercise makes their feet bigger. When there are multiple biological interpretations of a statistical result, you need to think of additional experiments to test the different possibilities.

Given below are some of the terms used in hypothesis testing: 1

There are web pages that will perform the chi-square test and . None of these web pages lets you set the degrees of freedom to the appropriate value for testing an intrinsic null hypothesis.

Excel for Business Statistics - Personal Web Space Basics

The probability that was calculated above, 0.030, is the probability of getting 17 or fewer males out of 48. It would be significant, using the conventional PP=0.03 value found by adding the probabilities of getting 17 or fewer males. This is called a one-tailed probability, because you are adding the probabilities in only one tail of the distribution shown in the figure. However, if your null hypothesis is "The proportion of males is 0.5", then your alternative hypothesis is "The proportion of males is different from 0.5." In that case, you should add the probability of getting 17 or fewer females to the probability of getting 17 or fewer males. This is called a two-tailed probability. If you do that with the chicken result, you get P=0.06, which is not quite significant.