STAT 511 HW 4

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16. C(5,2) = 10 p value = 0.2 18. - Treatment group = 3 + 5 + 6 = 14 1. 123 = 6 - 2/20 give a rank sum as larger or larger than the one observed 2. 124 = 7 - p value = 0.1 3. 125 = 8 4. 126 = 9 5. 134 = 8 6. 135 = 9 7. 136 = 10 8. 145 = 10 9. 146 = 11 10. 156 = 12 11. 234 = 9 12. 235 = 10 13. 236 = 11 14. 245 = 11 15. 246 = 12 16. 256 = 13 17. 345 = 12 18. 346 = 13 19. 356 = 14 20. 456 = 15

20. a) Determining the rank transformation means to replace each value with it’s rank, which was done in the previous chapter here b) Calculate the rank sum Trauma Patients(n1): 15+12+7+10+14+13+11 = 82 = T Nontrauma Patients(n2): 3+9+1+4.5+4.5+8+6+2 = 38 c) Find z statistic R bar = 8 Sample standard deviation = 4.468 n1 = 7 n2 = 8 Mean (T) = 7*8 = 56 SD(T) = 4.468*(sqrt((7*8)/(7+8)) = 8.633 (T+0.5) - Mean(T)/SD(T) = (82.5 - 56) / 8.633 = 3.0696 d) Find p value One sided p value = 0.99889 25. - In MiniTab I performed a Welch t-test by doing a 2 sample t-test and not assuming equal variances. This gets the two-sided p value (0.002) and confidence interval (-165.8,-39.6). - I am assuming that additivity means that there is only one independent variable that has a dependent variable. I think with guinea pig lifetimes, there are a lot more variables like genetics and living situation that can change the lifetime so no, I do not think it is a good model.

26. a) Histograms from Minitab. Yes, the differences look right skewed b) I took the natural log of the Unaffected (ln(Unaffected)) and the Affected (In(Affected)) and then the differences (Diff of ln). The histogram of the differences looks less skewed and yes, more symmetric than the one above.

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c) Transformed data p value: 0.006 Untransformed data p value: 0.006 The p values are the same. d) Std dev = 0.155678 SE = 0.155678/(sqrt(15)) = 0.4019 Sample mean = 0.1284 Upper CL = 0.1284 + 1.96(0.4019) = 0.9161 Lower CL = 0.1284 - 1.96(0.4019) = -0.6593 Back Transform: Mean = 0.0594 Upper CL = 0.0594 + 1.96(0.4019) = 0.8471 Lower CL = 0.0594 - 1.96(0.4019) = -0.7283 27. - 15+7+10+8+12+3+6+13+4+11+14+1+2+5 = 111

- There are 34 assignments that are less than the negative rank of 9, so p value = 34/32768 = 0.00104 - The p value is the same as the untransformed data. This p value is so small that it does not have statistical significance. You cannot get this kind of data from the histograms so you need to perform the signed-rank test. 29. a) Rank sum to test for logged and unlogged differences. 95 conf interval for diff of medians - j: nq - z sqrt(nq(1-q)) 16*(0.5) - 1.96 * sqrt(16*(0.5) * (1-0.5)) 8 - 3.92 = 4.08 = 4 - k: nq + z sqrt(nq(1-q)) 8 + 3.92 = 11.92 = 12 The confidence interval is between the 4th and 12th observation = (18.2, 53.1) b) The confidence intervals are similar, the lower intervals are different by 3.8 and the upper intervals are different by 1.2 so they aren’t that off. By using the rank sum test, you add up the ranks and the one with the lower ranks will have a lower T and that is better for this survey because you want the least percentage lost. Therefore, the Unlogged parts have a lower percentage loss. The mean is 38.16.

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