practical 1

Practical 1 1. Produce a figure showing the distribution of each species as a function of location along the transect. 2. Examine the distribution of all species. Was there evidence for aggregated occurrence in any species? (Provide a table with the results of the run tests and the conclusions). Species P Value Z Value Run s Occurrence s Conclusion Cyperus.p. 1 0 1 Positive = 50 Negative = 0 As the P – Value is greater than 0.05, we accept the null hypotheses. The Z – Value of 0 is evident as the species was present throughout the entire transect. Ipomea.b. 2.159e-07 -5.1851 4 Positive = 43 Negative = 7 The P – Value is extremely small, so we reject the null hypothesis the at the distribution of the range is random. The Z – Value is negative indicating that there are fewer runs than expected suggesting aggregated distribution. Canavaliar.r. 7.401e-08 -5.3811 4 Positive = 42 Negative = 8 The P – Value is extremely small, so we reject the null hypothesis for the alternative hypothesis that the true number of runs is not equal to the expected number of runs. The negative Z – Value demonstrates that there are fewer runs than what is expected showing aggregation. Pandanus.s. 1 0 3 Positive = 1 Negative = 49 The P – Value is 1 which is greater than 0.05 so we accept the null hypothesis. However, this species was only present 1 time across the transect with 3 runs suggesting a possible outlier. Panicum.m. 1.734e-05 -4.2966 3 Positive = 3 Negative = 47 The P – Value is extremely small, so we reject the null hypothesis that the distribution across the range is random for the alternative. This is further confirmed by the negative Z – Value which indicates fewer runs than what is expected indicating aggregation. Cenchrus.e. 0.0001227 -3.8407 8 Positive = 10 Negative = 40 The P – Value is extremely small, so we reject the null hypothesis for the alternative hypothesis where the negative Z – Value indicates that there are fewer runs than expected indicating an aggregated distribution. Heteropogon.c . 2.267e-09 -5.9774 4 Positive = 15 Negative = 35 The P – Value is extremely small, so we reject the null hypothesis for the alternative that the true number of runs is not equal to the expected. This is further confirmed by the negative Z – Value indicating fewer runs indicating aggregation. Spinifex.s. 2.422e-05 -4.2219 5 Positive = 6 Negative = 44 The P – Value is extremely small, so we reject the null hypothesis for the alternative hypothesis that the true number of runs is not equal to the expected number of runs. The negative Z – Value shows there are fewer runs than what is expected representing aggregation. Unknown 6.613e-07 -4.9724 3 Positive = 4 Negative = 46 The P – Value is extremely small, so we reject the null hypothesis for the alternative hypothesis where the negative Z – Value indicates that there are fewer runs than expected indicating an aggregated distribution. Desmodium.r. 1.56e-10 -6.3994 2 Positive = 7 Negative = 43 The P – Value is extremely small, so we reject the null hypothesis the at the distribution of the range is random. The Z – Value is negative which indicates that there are fewer runs than expected which suggests an aggregated distribution.