SECTION ONE Module 1-2: Chi-Square Analysis Please answer the following question
ID: 281700 • Letter: S
Question
SECTION ONE Module 1-2: Chi-Square Analysis Please answer the following question. Show your work as completely as possible. I suggest that you solve the problem on a separate piece of paper and scan your work/answer for submission In maize, the recessive genes su and sh (when homozygous) produce sugary and shrunken endosperm, respectively. Their dominant alleles Su and Sh produce normal starchy and full endosperm. A true breeding sugary variety is crossed to a true breeding shrunken variety and the F1 progeny are all starchy and full. The F1 generation is crossed to a sugary & shrunken variety (su su and sh sh) and produce 1000 progeny in the following numbers: 160 starchy/full, 350 sugary/full, 340 starchy/shrunken and 150 sugary/shrunken plants. The NULL Hypothesis "The progeny classes are in equivalent numbers SHOW ALL WORK i) Use the chi-square analysis to evaluate your hypothesis? (8 pts) ii) Do you reject or fail to reject the NULL hypothesis (at the 0.05 level)? What does this mean? (2 pt) Chi Square valuesExplanation / Answer
chi square calculated value = 32.4+40+32.4+40 = 144.8
at degree of freedom = 4-1=3 and 0.05 significance level, chi square tabulated value is = 7.81
As, chi square calculated value > chi square tabulated value
Null hypothesis is rejected, which means progeny are not in equivalent numbers.
Category observed expected (O-E) (O-E)2 (O-E)2/E Starchy/full 160 250 -90 8100 32.4 sugary/full 350 250 100 10000 40 Starchy/shrunken 340 250 90 8100 32.4 sugary/shrunken 150 250 -100 10000 40Related Questions
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