November 1st, 2009, 04:27 AM  #1 
Newbie Joined: Nov 2009 Posts: 6 Thanks: 0  Chi Square
This is a question I am having trouble with I have done some of the answer. Can someone please check the ones that I have done and explain the others. Thanks Two traits which have been studied extensively in tomato plants are height (tall vs. dwarf) and leaf type (cut vs. potato). In a particular breeding experiment, 1611 tomato plants were grown and classified by leaf type and height. The results are given below Leaf Type Cut Potato Total Tall 926 288 1214 Height Dwarf 293 104 397 Total 1219 392 1611 (a) A genetic theory predicts that leaf type and height are independent. Perform a chisquare test of independence for the hypothesis implied by this genetic theory by completing the following work: i. List H0 and Ha. H0 : p1 =p2 and Ha: p1 not equal to p2 ii. Find the expected value under H0 for each observation in the table. E(11) = (1214 x 1219)/1611 = 918.6 E(12) = (1214 x 392)/1611 = 295.4 E(21) = (397 x 1219)/1611 = 300.4 E(22) = (397 x 392)/1611 = 96.6 iii. Calculate the chisquare test statistic and specify its degrees of freedom. ChiSq = (((926918.6)^2)/918.6) + (((288295.4)^2)/295.4) + (((293300.4)^2)/300.4) + (((10496.6)^2)/96.6) = 0.06 + 0.185 + 0.182 + 0.567 = 0.994 d.f. = either 3 or 1? iv. Find the range of the pvalue of the test. 0.319 v. Draw a conclusion of the test based on the significance level 0.05. 0.319>0.05 Therefore we accept the null hypothesis (b) A more detailed theory predicts that the 4 types of tomato plant given above appear in a 9:3:3:1 (tall cut : tall potato : dwarf cut : dwarf potato) ratio. Perform a chisquare goodness of fit test to check whether the data support this theory or not. i. List Ho and Ha. Ho: p1= 0.5625, p2= 0.1875, p3= 0.1875 and p4= 0.0625 H1: not all p1 are as specified in Ho ii. Find the expected value under Ho for each observation in the table. E(11) = 0.5625 x 1611 = 906.19 E(12) = 0.1875 x 1611 = 302.06 E(21) = 0.1875 x 1611 = 302.06 E(22) = 0.0625 x 1611 = 100.69 iii. Calculate the chisquare test statistic and specify its degrees of freedom. ChiSq = (((926906.19)^2)/906.19) + (((288302.06)^2)/302.06) + (((293302.06)^2)/302.06) + (((104100.69)^2)/100.69) = 0.433 + 0.654 + 0.272 + 0.11 = 1.469 d.f. = 41 = 3 iv. Find the rejection region of the test at the significance level 0.05. v. Draw a conclusion of the test based on the significance level 0.05. 

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