usernamed Posted May 4, 2016 Report Share Posted May 4, 2016 Since paper 3 is tomorrow I have a question that is buzzing in my head I'm doing option G: Ecology and conservation And a chi-squared question might pop up And my question is on how to calculate the expected value for chi-squared questions????? Reply Link to post Share on other sites More sharing options...
ZhenXS Posted May 4, 2016 Report Share Posted May 4, 2016 10 hours ago, Nadsxo2 said: Since paper 3 is tomorrow I have a question that is buzzing in my head I'm doing option G: Ecology and conservation And a chi-squared question might pop up And my question is on how to calculate the expected value for chi-squared questions????? I'm not quite sure but... Isn't the options this year only until Option D? Never heard of Option G before... Reply Link to post Share on other sites More sharing options...
TheNintendoChip Posted May 5, 2016 Report Share Posted May 5, 2016 (edited) 12 hours ago, Nadsxo2 said: Since paper 3 is tomorrow I have a question that is buzzing in my head I'm doing option G: Ecology and conservation And a chi-squared question might pop up And my question is on how to calculate the expected value for chi-squared questions????? Heads up, Ecology is now Option C! (I'm doing it too) I can't recall seeing Chi-Squared in the actual option part of the syllabus, but I guess it could appear in Section A. From what I understand, you pretty much have to figure out the expected values yourself. Let's say you were doing a Chi-Squared test relating to Mendelian Genetic, and you did a cross between tall (TT) and short (tt) pea plants, resulting in the F1 generation (Tt), and then crossed those to get the F2 generation (i.e., crossing two heterozygous). In the F2 generation, 787 plants were tall and 277 were short. What you're testing is if this follows Mendel's experiments. Since the cross was between two heterozygous plants (I'll say Tt and Tt), the resulting offspring would be (according to Mendel) : TT, Tt, Tt, and tt. 3/4 of those will express the dominant phenotype, while 1/4 will express the recessive phenotype. So the expected value of tall plants should be 3/4 of the total bred (total = 787+277=1064), and the expected value for the short plants should be 1/4 of the total bred (again 1064). 3/4*1064=798, and 1/4*1064=266. Those are your expected values, and the observed are 787 and 277. I hope that made sense and helped!!! Edited May 5, 2016 by TheNintendoChip Reply Link to post Share on other sites More sharing options...
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