Table 2A and 3B
3A) we performed A priori chi-square test for these two tables. The main reason for the chi-square is to find out if the expected value is any different from the observed value. This part of the observation tested the null hypothesis, which states that, if when Ear#1(which was test-crossed kernel) is counted and approves of the Mendelian expectation of 1:1:1:1 phenotypic ratio. The chi-square for this data was excepted to accept the null hypothesis. The reason that it was expected is because the hypothesis was established before. The result of our data confirmed this hypothesis, that is our calculated p-value was greater than 0.05, therefore the expected null hypothesis was accepted. Since our collected data proved the mendelian hypothesis, it didn’t matter if the sample was large or small. 3B) Table two of this data was had much smaller sample size and was used to examine the same hull hypothesis as the first table which had much larger sample size. The hypothesis s was the same, which was to prove that the phenotype conforms the mendelian expectation. When we calculated the chi-square for this set of data, we found our p-value to be between .80-.70 which was less than .05, thus we accept the null hypothesis that is the expected ratio of the test cross was indeed 1:1:1:1. Tables 3-4A and 4-4B
3-4A) For this collected set of data, our mission was to test mendilan null hypothesis which stated that the ratio of a dihybrid( Ear#2) would result in a 9:3:3:1 ratio. The result for out p-value was between .70-.50 which was greater than .05. The value of the chi-square was between the approved range for the hypothesis to be accepted, therefore, the null hypothesis was accepted. Since we had a large sample size for this set of data, there wasn’t any conflict. 3-4B) The data collected for this table was same as the previous table. The difference was the mendelian distribution ratio. This data did not meet the expected hypothesis because the hypothesis...
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