hw2soln - Biostatistics 100B Solutions To Homework...

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Biostatistics 100B Homework Solutions 2 January 16th, 2007 Solutions To Homework Assignment 2 Warmup Problems (1) Statistics and Agriculture: (a) By looking at the table it seems as if fertilizer 2 is doing a better job. We applied each of the fertilizers to the same number of seeds, and a greater number of the fertilizer 2 seeds germinated. The difference (150 germinated for fertilizer 2 vs 120 for fertilizer 1 out of 200 seeds) seems fairly large to me intuitively, but we will need to verify that with a statistical analysis. The point of doing part (a) first is to realize that you can usually figure out the answer to a contingency table problem using common sense before you calculate a chi-square statistic or a p-value. It is always a good idea to figure out the answer ahead of time! (b) Let p 1 be the percentage of seeds on fertilizer 1 that germinate, and let p 2 be the percentage of seeds on fertilizer 2 that germinate. If there is no relationship between which fertilizer is used and whether the seeds germinate then the two fertilizers must be equally good and we should have p 1 = p 2 . That is probably our default position and is certainly the less interesting or important case. If the fertilizers are equally good it doesn’t matter which we use. On the other hand, if one of them is better, we would want to find that out so we could use the superior fertilizer. Therefore, our alternative is that there is a difference between the fertilizers, p 1 negationslash = p 2 , or alternatively that fertilizer type and germination are related. Of course, for a two sided test, the null is always equality and the alternative lack of equality. Here we have a two sided test because we don’t care which fertilizer is better, if either. To summarize, we have H 0 : p 1 = p 2 (The fertilizers are equally good, or equivalently, there is no relationship between which fertilizer you use and whether or not the seed germinates. Fertilizer type and germination are independent.) H A : p 1 negationslash = p 2 (The fertilizers are not equally good. How likely a seed is to germinate is dependent on which fertilizer you use.) (c) The formula for the expected count in any cell can be obtained from the row and column totals, R i and C j according to the formula E ij = R i C j N Here we have E 11 = (270)(200) 400 = 135 E 12 = (270)(200) 400 = 135 E 21 = (130)(200) 400 = 65 E 11 = (130)(200) 400 = 65 Note that we only got E 11 = E 12 and E 21 = E 22 because the two column totals were the same. To compute the chi-squared statistic we must take the difference between each observed count and expected count, square it, divide by the expected count, and add up the results for all the cells. We get 1
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χ 2 obs = 2 summationdisplay i =1 2 summationdisplay j =1 ( O ij E ij ) 2 E ij = (120 135) 2 135 + (150 135) 2 135 + (80 65) 2 65 + (50 65) 2 65 = 10 . 256 (d) We reject the null hypothesis if χ 2 obs > chi 2 α, ( r 1)( c 1) . Here we have two rows and two columns so the number of degrees of freedom is (2-1)(2-1) = 1. Looking on the chi-squared table we see that χ 2 . 05 , 1 = 3 . 84 .
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