Gamewizard Posted January 3, 2011 Posted January 3, 2011 If the mean mercury concentration in fish caught in a particular lake is below 0.5ppm then the state of florida would class that lake as acceptable. On the basis of the data provided in the case study how should the state classify each lake ? so there are basically two lakes, East Lake and Lake Iamonia. East lake mean= 1.1765 sd=0.2802 Lake Iamonia mean= 0.598 sd= 0.2836 formula used = t= mean-value(0.5)/s/square root of n The t-value for the first lake is 15.27, and the total population is 40(fishes) so the answer given by my teacher is t39(0.025) = 3.307< 15.27. So hes looked at the 99.9 % column in the t-table. For the next lake, hes looked at the 97.5% column, and so his answer is t29(0.025)=2.045 What I do not understand is why is he looking at diffrent percentiles? and how would I know which percentile column to look for in diff questions? please help
alpha2cen Posted January 4, 2011 Posted January 4, 2011 my teacher is t39(0.025) = 3.307< 15.27. So hes looked at the 99.9 % column in the t-table. We don't know total number of the fish in each lake. So we assume East lake Total =1,000,000, sampling=40 Lake lamonia Total= 1,500,000, sampling=30 t39(0.025) =3.307 ------>97.5% column value I did not see the exact value, but it might be trivial mistake, because every human is not a machine.
Gamewizard Posted January 4, 2011 Author Posted January 4, 2011 We don't know total number of the fish in each lake. So we assume East lake Total =1,000,000, sampling=40 Lake lamonia Total= 1,500,000, sampling=30 t39(0.025) =3.307 ------>97.5% column value I did not see the exact value, but it might be trivial mistake, because every human is not a machine. No, if you look at the t-percentiles table, (df 40) 3.307 is in the 99.9% column. 2.021 is the value in 97.5% column. Just want to know why hes choosing to look in the 97.5% column and 99.9% column is it because of the diffrence in samples then?
alpha2cen Posted January 5, 2011 Posted January 5, 2011 No, if you look at the t-percentiles table, (df 40) 3.307 is in the 99.9% column. 2.021 is the value in 97.5% column. Just want to know why hes choosing to look in the 97.5% column and 99.9% column is it because of the diffrence in samples then? 2.021 is correct. I think he knew it well. It is only trivial mistake. In general 97.5% confidence interval is used. Because it's good for reducing type 2 error .
Gamewizard Posted January 5, 2011 Author Posted January 5, 2011 2.021 is correct. I think he knew it well. It is only trivial mistake. In general 97.5% confidence interval is used. Because it's good for reducing type 2 error . oh ok thanks
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