The table above shows my Chi-square values and the \"p\" and \"q\" of the fruit
ID: 97930 • Letter: T
Question
The table above shows my Chi-square values and the "p" and "q" of the fruit flies cross. Besides saying thay the chi-square value is higher and that the hypothesis did not was supported what else can we say about the values? F2 Wild Type Sepia observed expected deviation deviation squared (d2) d2/e 173 950 713 237 950 -64 64 4096 4096 5.7 17 7 Chi-square value: 22.7 F2 0.57 0.43 F3 Wild Type Sepia observed expected deviation deviation squared (d2) d2/e 455 101 556 417 139 556 -38 38 3 1444 3.5 1444 10.4 13.9 F3 0.58 0.42 Chi-square value:Explanation / Answer
Answer:
Based on the chi square value of 22.7 with 1 degree of freedom (2 classes, therefore df = n-1 = 2-1 =1) the P-Value is < 0.00001. The result is significant at p < 0.01.
This indicates that there is a statistical difference between the observed genotype frequencies and expected genotype frequencies if the population was in Hardy Weinberg equilibrium. Therefore we will reject the null hypothesis and conclude that the evidence supports that the population is not in Hardy-Weinberg equilibrium.
Similarly for F3, for the chi square value of 13.9 with 1 degree of freedom (2 classes, therefore df = n-1 = 2-1 =1) the P-Value is 0.000193 which is less than the significance level of 0.01. The result is significant at p < 0.01. Therefore we will reject the null hypothesis and conclude that the evidence supports that the population is not in Hardy-Weinberg equilibrium.