Exam 1 Answers number/m

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1 Exam 1 Answers (15 pts) a) Construct a two-sided 95% confidence interval for the mean abundance of pipevine swallowtail butterflies (Battus philenor) per m 2 of foliage of their host plant Aristolochia californica. The measured values are given below and are the result of random sampling. Note your assumptions. observation # number/m Confidence interval on the mean abundance: Assumes underlying population is normally distributed P [ x ( tα 2, n 1 )( s n) µ x + ( tα 2, n 1 )( s n)] = 1 α Calculations in R # enter data abund=c(17,52,48,22,31,41,53,19) # calculate mean, sd, and sample size mabund=mean(abund) mabund [1] sdabund=sd(abund) sdabund [1] n=length(abund) n [1] 8

2 # determine critical t for n-1 df's critt=qt(0.975,n-1) # calculate upper confidence interval limit ciup=mabund+(critt*(sdabund/sqrt(n))) ciup [1] #calculate lower confidence interval limit cilow=mabund-(critt*(sdabund/sqrt(n))) cilow [1] The estimated 95% confidence interval on mean abundance of the pipevine swallowtail on foliage of the host plant is: P ( µ 47.94) = (10pts) What four factors affect the power of a statistical hypothesis test? Which of these factors are most appropriate to attempt to change in order to achieve a more powerful test? Answer The power of a statistical test depends on the Type I error rate, α, the inherent observational variability, σ, the magnitude of the difference between treatments that one wants to detect, the effect size or δ, and on the sample size n. It is most appropriate to increase sample size, or to alter the design of the experiment in order to reduce σ to improve the power of a test. 3. (20 pts) A geologist was interested in the effectiveness of a hydrologic restoration on soil moisture, and ultimately the biota of a montane meadow. Before restoration the meadow appeared to have a shortened hydroperiod - moisture was not retained in the meadow late into the growing season, due to the presence of a severely head cut stream. Runoff from snowmelt exited the meadow quickly via a deep stream channel that developed due to erosion caused by logging. The restoration involved excavating ten ponds along the stream and using the soil from these excavations to create a series of dikes to plug a deeply incised steam channel. Spillways were constructed at each pond so that excess water would flow onto the meadow and presumably increase soil moisture and enhance plant growth rather than flow rapidly downstream. The geologist reasoned that any increase in soil moisture would be detected initially just downstream from of the spillways at each pond. Therefore, she removed soil cores from both the upstream and downstream sides of three of the spillways and estimated the percent moisture content by measuring the initial mass of the core, drying the core at 60 o C for 48 hrs and then reweighing the

3 core. Percent moisture content was then calculated as [(initial mass-final mass)/initial mass]. Use these data to determine the sample size necessary to detect a 25% increase in soil moisture downstream from the spillways. Would such a full study be feasible? Spillway Upstream Downstream % 15.1% % 15.5% % 16.4% # this is a paired t-test like design # enter data up=c(11.9,10.3,11.4) down=c(15.1,15.5,16.4) # calculate mean soil moisture up mup=mean(up) mup [1] 11.2 # calculate the standard deviation of the differences in soil mositure above and below the spillways diff=up-down diff [1] sddiff=sd(diff) sddiff [1] # conjecture effect size effect=0.25*mup effect [1] 2.8

4 # calculate sample size required power.t.test(delta=effect, sd=sddiff, sig.level=0.05, power=0.8, type="paired", alternative="one.sided") Paired t test power calculation n = delta = 2.8 sd = sig.level = 0.05 power = 0.8 alternative = one.sided NOTE: n is number of *pairs*, sd is std.dev. of *differences* within pairs # calculate sample size with power = 0.95 Paired t test power calculation n = delta = 2.8 sd = sig.level = 0.05 power = 0.95 alternative = one.sided NOTE: n is number of *pairs*, sd is std.dev. of *differences* within pairs 1. Calculate the standard deviation of the differences in soil moisture above and below each spillway. (sd of differences = Calculate the effect size. Given that the preliminary data indicate that percent soil moisture above the spillways is 11.2%, a 25% increase would be to 14%, for a difference between the observed and conjecture effect means of Using power.t.test with (delta=2.8, paired=true, sig.level=0.05, Power = 0.8, and sd= , I calculate the necessary sample size to be Given that there are only 10 ponds in the meadow, the fact that only 3 are needed to have adequate power suggests that such a study is feasible. Even with power = 0.95, the sample size necessary to detect a 25% increase in soil moisture is only (15 pts) A geographer studying hammock islands (small patches of tropical hardwood forest) in the Florida Everglades was interested in determining the effect of invasion by the shrub Brazilian Pepper (Schinus sp.) on the species richness of native plants in hammocks. She sampled 13 un-invaded and 9 invaded hammock islands and determined that the average species richness (mean±standard error) was 28.2±3.32 and 18.7±4.55 on un-invaded and invaded islands, respectively.

5 Would she be safe in concluding that the invaded islands have lower species richness than un-invaded islands? Perform a statistical test and report the results of that test to support your interpretation of the data. Answer: This problem calls for applying a test that is based on having two independent treatment groups. If I had given you the data it would have been possible to apply the Wilcoxon Rank- Sum test. However, since I only gave you the treatment means and standard errors, the only option was to apply the independent groups t-test. In this instance you could have assumed that variances were equal, or assumed they were unequal. Either could have been argued. H o : μ d 0 H a : μ d > 0 # input values n1=13 n2=9 x1=28.2 x2=18.7 se1=3.32 se2=4.55 #calculate sd's and vars from se's sd1=se1*sqrt(n1) sd1 [1] sd2=se2*sqrt(n2) sd2 [1] var1=sd1^2 var2=sd2^2 # perform t-test assuming equal variances # calculate pooled estimate of sd sp=sqrt((((n1-1)*(var1))+((n2-1)*(var2)))/((n1+n2)-2)) sp [1]

6 # calculate t value tev=(x1-x2)/(sp*sqrt((1/n1)+(1/n2))) tev [1] # calculate dfs dfev=(n1+n2)-2 siglevel=1-pt(tev,dfev) siglevel [1] When assuming equal variances, given a t-value of with 20 dfs and p= reject null hypothesis. # perform t-test assuming unequal variances t=(x1-x2)/sqrt(((sd1^2)/n1)+((sd2^2)/n2)) t [1] # calculate df's var1=sd1^2 var2=sd2^2 df=(((var1/n1)+(var2/n2))^2)/((((var1/n1)^2)/n1+1)+(((var2/n2)^2)/n2+1))-2 df [1] # determine significance level of test 1-pt(t,df) [1] When assuming unequal variances, t = with dfs = and p= Therefore do not reject the null hypothesis. So in this instance the assumption one makes concerning the variances determine whether or not the null hypothesis is rejected. 5. (10 pts) Design an experiment to determine the impact of introduced Argentine ants on soil insects and arthropods. State your null and alternative hypothesis explicitly, and explain what preliminary data you would collect to help you decide how to design your experiment.

7 Answer: I would place pit fall traps in areas invaded by Argentine ants and in nearby areas not invaded by Argentine ants. These traps could be used to estimate the abundance of soil arthropods and insects. I would probably view traps in adjacent invaded/un-invaded areas as samples of the same subject (site), so multiple sites (subjects) would be used to test the null hypothesis that Argentine Ants have no effect on the abundance of soil arthropods and insects versus the alternative that they reduce the abundance of soil arthropods and insects. This experiment would be a within subjects design. I would need preliminary data on the average abundance of soil arthropods and insects in the invaded areas, and preliminary data on variability in the difference between invaded and un-invaded areas to perform a sample size analysis. 6. (20 pts) The following data represent the abundance of seeds of a native species of thistle in the seed bank on serpentine soils and on adjacent non-serpentine soils. Test the hypotheses that thistle abundance is unrelated to soil type. Analyze this data assuming both that they are either markedly non-normal or are normally distributed. Abundance (seeds/ cm 3 ) Site Serpentine Non-Serpentine Soil Soil To examine these data assuming normality apply the paired t-test on the data. Note that the serpentine and non-serpentine data some for adjacent sites suggestion a paired design.

8 # input data serp=c(16.4,19.8,18.5,17.4,21.2,21.3) serp [1] nonserp=c(9.6,0.6,4.8,3.5,2.5,8.2) nonserp [1] # assume normality t.test(serp,nonserp,paired=true,alternative="two.sided") Paired t-test data: serp and nonserp t = , df = 5, p-value = alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of the differences To examine these data assuming non-normality apply the Wilcoxon signed-rank test of these data. Note that the serpentine and non-serpentine data some for adjacent sites suggestion a paired design. # assume non-normality wilcox.test(serp,nonserp,paired=true) Wilcoxon signed rank test data: serp and nonserp V = 21, p-value = alternative hypothesis: true location shift is not equal to 0

9 7. (10pts) The graphs depicted below illustrate instances where hypothesis tests for differences in means were not significant and the α = 0.05 level. Do either of these plots suggest that a Type II error is likely to have been made? A. B Density Control Treatment Control Treatment Answer: The pattern of the means and standard errors in Plot B suggest that it is possible that the failure to reject the null hypothesis is likely to be due to a Type II error. This is because there are substantial differences in the treatment means, but the standard errors are so large that this difference could not be detected statistically. In Plot A the means are so similar that repeating the experiment with larger sample sizes is not likely to lead to finding the treatment means are different, since the sample mean is an unbiased estimate of the population mean. Extra Credit 8. (10 pts) What are the assumptions of an independent groups t-test? a paired t- test? What are the consequences of violating these assumptions (e.g., how will it affect the Type I and Type II error rates)? Answer: A. Assumption of the independent groups t-test 1. independent random sampling 2. normality of the underlying population distributions 3. variances equal among the treatment groups B. Assumption of the paired t-test 1. Independent random sampling 2. normality of the distribution of differences Independence - The assumption of independence can never be violated. If the data are not obtained by independent random sampling then the statistics estimated from the

10 data are likely to be biased estimates. All conclusions drawn from such data are unreliable. Normality and Equality of Variances Both tests are fairly robust with respect to violation of the assumption of normality when both variances and sample sizes are equal. However, in the presence of variance heterogeneity and inequality of sample sizes the Independent groups t-test may result in too few or too many Type I errors depending on whether or not the treatment with the largest variance also received the largest sample size. In the paired t-test, since there is only one variance and one sample of differences, the test is always robust with respect to violation of the assumption of normality. For the Independent groups t-test, one can also relax the assumption of equal variances and perform the test with separate variance estimates and for degrees of freedom adjusted using Satterthwaite s or Welch s correction. In the independent groups t-test, power is reduced when variances are unequal unless a larger sample size is allocated to the group with the largest variance. 9. (20 pts) The following data represent the maximum distances that birds and monkeys in rain forests in Cameroon dispersed the seeds of several species of trees. A. Construct a 95 % confidence interval on the underlying population mean of the maximum dispersal distances. B. Tests the hypothesis that the maximum dispersal distance for bird dispersed tree species equals that for monkey dispersed tree species. State the assumptions of the test you apply. Bird Monkey Species 1 Species 2 Species 3 Species 4 Species 5 Species m 210 m 300 m 124 m 90 m 200 m First place 95% confidence intervals on either the overall mean maximum dispersal distance, or the means for birds or monkeys separately. # input data bird=c(473,210,300) bird [1] monkey=c(124,90,200) monkey [1]

11 # combined data to get overall values combined=c(bird,monkey) combined [1] # calculate overall mean maximum dispersal distance mcomb=mean(combined) mcomb [1] # calculate sd of overall maximum dispersal distance sdcomb=sd(combined) sdcomb [1] # get combined sample size ncomb=length(combined) ncomb [1] 6 # calculate confidence interval limits for overall data cc=qt(0.975,ncomb-1) clow=mcomb-cc*(sdcomb/sqrt(ncomb)) clow [1] chigh=mcomb+cc*(sdcomb/sqrt(ncomb)) chigh [1] # calculate confidence limits for bird or monkeys alone mbird=mean(bird) mbird [1] nbird=length(bird) mmonkey=mean(monkey) mmonkey [1] 138 nmonkey=length(monkey)

12 sdbird=sd(bird) sdmonkey=sd(monkey) calone=qt(0.975,nbird-1) # for birds birdlow=mbird-calone*(sdbird/sqrt(nbird)) birdlow [1] birdhigh=mbird+calone*(sdbird/sqrt(nbird)) birdhigh [1] # for monkeys monkeylow=mmonkey-calone*(sdmonkey/sqrt(nmonkey)) monkeylow [1] monkeyhigh=mmonkey+calone*(sdmonkey/sqrt(nmonkey)) monkeyhigh [1] Confidence interval for overall mean maximum dispersal distance P[87.37 μ ] = 0.95 Confidence interval for the mean maximum dispersal distance for birds. Note covert impossible vales (a dispersal distance of to 0). P[0 μ ] = 0.95 Confidence interval for the mean maximum dispersal distance for monkeys. Note covert impossible vales (a dispersal distance of to 0). P[0 μ ] = 0.95 #test hypothesis of no difference in maximum dispersal distance between birds and monkeys

13 # check normality assumption library(quantpsyc) Loading required package: boot Loading required package: MASS Attaching package: 'QuantPsyc' The following object is masked from 'package:base': norm norm(combined) Statistic SE t-val p Skewness Kurtosis # perform 2 sample t-test with unequal variances t.test(bird,monkey) Welch Two Sample t-test data: bird and monkey t = , df = , p-value = alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of x mean of y Given a t =2.2649, df=2.6885, and p=0.1187, we fail to reject the null hypothesis of no differences between birds and monkeys in their mean maximum dispersal distances for fruits.