# The Chi-square Statistic. Goodness of fit 0 This test is used to decide whether there is any difference between the observed (experimental) value and.

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The Chi-square Statistic

Goodness of fit 0 This test is used to decide whether there is any difference between the observed (experimental) value and the expected (theoretical) value.

Goodness of Fit

Free from Assumptions 0 Chi square goodness of fit test depends only on the set of observed and expected frequencies and degrees of freedom. This test does not need any assumption regarding distribution of the parent population from which the samples are taken. 0 Since this test does not involve any population parameters or characteristics, it is termed as non- parametric or distribution free tests. This test is also sample size independent and can be used for any sample size.

It is all about expectations O i = an observed frequency (i.e. count) for measurement i E i = an expected (theoretical) frequency for measurement i, asserted by the null hypothesis.

Expected Value 0 F = the cumulative Distribution function for the distribution being tested. 0 Y u = the upper limit for class I 0 (maximum possible observations for any category) 0 Y l = the lower limit for class I 0 (minumum possible observations for any category) 0 N = the sample size

Hypothesis testing Choose a level of alpha – usually 0.05 This implies a 95% level of comfort that the observation is correct.

Example The number of cubs delivered to a population of bears in the wild is tested to see if there is no difference in probability of twins. (N = 50 females) Number of cubs 0123 Observed15359 Expected12.5 Degrees of Freedom = Number of groups – 1 df = 4 – 1 = 3

CHI-SQUARE DISTRIBUTION TABLE

Decision Rule 0 Based on the alpha and the degrees of freedom, look up the value in the table. 0 For our example of alpha=.05 and df=3 0 If chi square is greater than 7.82 then reject the null hypothesis that bears normally birth twins.

Calculate the value Chi-square = (1-12.5) 2 /12.5 + (5-12.5) 2 /12.5 + (35-12.5) 2 /12.5 + (9- 12.5) 2 /12.5 = 10.58 + 4.5 + 40.5 + 0.98 = 56.56 Number of cubs 0123 Observed15359 Expected12.5 Since 56.56 > 7.82 we reject the null hypothesis that the number of bear cubs is equally possible for 0-3 cubs

Interpret the result 0 Since we rejected the null hypothesis, what conclusions (inferences) can we come to?

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