advertisement

BIOL 300 Assignment 6 Fall 1013 Chapter 12 15. The difference in white blood cell count is 1.87, with the more promiscuous species having the higher count. (b) 0.10 < µd < 3.62. (c) The null hypothesis that there is no difference means that µ0 = 0, so t = 1.87 / 0.52 = 3.56 > α(2)0.01, 8 df, so P < 0.01. We reject the null hypothesis: promiscuous primates have higher white blood cell counts. 17. (a) Two sample t-test: The difference in means is 19.9 - 17.5 = 2.4. The null difference in means is zero. The pooled sample variance is 8.46, so the standard error of the difference between two means is 1.05. t = 2.4 / 1.05 = 2.29, with 29 df. The critical value for α(2) = 0.05 = 2.05, so P < 0 .05. We reject the null hypothesis that the copulation times are equal for the two circadian rhythm mutations. Chapter 13 23. The distribution is left skewed, and has both positive and negative values. Therefore, we're back to a sign test. 21 of the 36 showed an increase in biomass. ! 36 ! 36 $ $ i 36−i P = 2 ## ∑ # & ( 0.5) ( 0.5) && = 0.405 .We do not see a change in biomass using the " i=21" i % % data in this way.