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Chi-square Test Test of independence A vaccine trail produced the following results: infected uninfected placebo 81 1427 vaccine 179 2824The null hypothesis: there is no relationship between row and column frequencies, i.e., vaccine/placebo will not have any effect in the infected/uninfected frequencies. Using test of independence, we find p = 0.42 and conclude that the null hypothesis cannot be rejected, i.e., the vaccine is not effective. Goodness of fit The cross of A a B b X a a b b is used to determine whether the two loci are linked. The results are shown below: A a B b 310 a a b b 315 A a b b 287 a a B b 288The null hypothesis: the two loci are not linked, thus the ratio should be 1:1:1:1. The expected frequencies are: A a B b 300 a a b b 300 A a b b 300 a a B b 300Using the Goodness of fit test, we find p = 0.55. So the null hypothesis cannot be rejected: the two loci may not be linked.  What is a USB Flash Disk? | |