Go back to Exercise 21.2 investigating the genetic model of heat resistance in r

Question

Go back to Exercise 21.2 investigating the genetic model of heat resistance in rice plants. What are the expected counts for each of the two phenotypes? Use these expected counts to compute the chi-square statistic.

Exercise 21.2:

Example 21.2 described the color of 429 second-generation A. caudatus seeds after crossing black-seeded and pale-seeded populations. The null hypothesis, based on a dominant epistatic model of genetic inheritance, assumed that

If H0 was true, we would expect the distribution of seed color to be

We can now calculate the chi-square statistic to test H0.

Think of the chi-square statistic X2 as a measure of the distance of the observed counts from the expected counts. Like any distance, it is always zero or positive, and it is zero only when the observed counts are exactly equal to the expected counts. Small values of X2represent small deviations from H0 that do not provide sufficient evidence to reject H0.Inversely, large values of X2 are evidence against H0, because they say that the observed counts are far from what we would expect if H0 was true. The alternative hypothesis Hafor the chi-square test is nonspecific, or nondirectional, because any violation of H0 tends to produce a large value of X2.

Explanation / Answer

The computation process and value of Chi-square test statistic is correct. The p value is 0.669449 (df=2, alpha=0.05). Therefore, fail to reject null hypothesis. There is not sufficient sample evidence to conclude that frequncy of caudatus seeds do not follow the given distribution.

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