Name Pts Lab Section On your own PROBLEM #1 You are studying the distribution of a seastar, Pisaster ochraceus, which has two color morphs, Black and ochre. You want to know if color is adaptive and may provide some type of advantage in certain environments. You randomly sample 50 locations and record the color morph, black or not-black, of the first three seastars you encounter. Assume that, if a pattern exists, you will spend time and effort in trying to find out why. If not, you will just move on to a different problem: 1) What is Ho? The observed distribution of black seastars fits a binomial distribution with p=0.593 and K=3 2) What is Ha? The observed distribution of black seastars fits a binomial distribution with p=0.593 and K=3 . 3) What is the measured variable?. Presence or absence of the black seastar for each of 3 seastars 4) Is the sample size fixed? If so, what is the value? The sample size is fixed at 3 (k=3) 5) What is the appropriate theoretical probability distribution for this problem? Binomial 6) What would it mean if the results showed a clumped distribution? That the black seastars are found in clumps and most likely responding to some aspect of that microhabitat. 7) What would it mean if the results showed a uniform distribution? That the black seastars are over dispersed which most likely means that they are defending territories. 8) What would it mean if the observed frequencies were essentially the same as the expected frequencies? The distribution of black seastars is unpredictable (stochastic). 4 -1 9) Determine statistical error to avoid. Complete Table 1 Table 1: Select the statistical error to avoid for Problem 1 – Pisaster ochraceus color morph frequencies. Statistical Decision Accept Ho: Observed fits the expected Reject Ho: Observed does not fit the expected Conclusion Action What if I’m wrong? No pattern Find another problem Lost out on finding something interesting. Type of error II I Pattern Try to find out why Wasted time and energy 10) Alpha (α) = 0.025 11) Are the parameters Intrinsic or Extrinsic? Intrinsic Data Table 2: Data for Problem 1- Color morph frequencies for Pisaster ochraceus. # of black morphs per quadrat (Y) 0 1 2 3 TOTAL Observed Frequency (f) 13 7 8 22 50 fY2 fY 0 7 16 66 89 0 7 32 198 237 12) How many seastars did you examine? 3*50=150 13) How many of those were black? 89 14) What is the value for p? 89/150 = 0.593 15) What is the value for q? 1-0.593 = 0.407 4 -2 16) Compute the appropriate Expected Frequency distribution. Complete Table 3 Table 3: Compute expected frequencies for color morphs of Pisaster ochraceus. # of black morphs per quadrat (Y) 0 1 2 3 TOTAL Observed Frequency (f) 13 7 8 22 50 Probability equations 1p0q3 3p1q2 3p2q1 1p3q0 Probabilities Expected Frequencies 0.067254 0.294372 0.429494 0.208880 3.4 14.7 21.5 10.4 17) For Poisson Only – Lump classes when an Expected Frequency <5 N/A 18) Compute the degrees of freedom for the test. Intrinsic no lumping so df=4-2=2 19) Compute the G-statistic. Complete Table 4. Table 4: Compute the G statistic for Problem 1 - Pisaster ochraceus color morph data. # of black morphs per quadrat (Y) 0 1 2 3 Observed Frequency (f) 13 7 8 22 TOTAL 50 Expected Frequencies 3.4 14.7 21.5 2 *Observed Freq*ln(Observed Freq/Expected Freq). 34.87052 -10.3871 -15.8178 10.4 50 32.96641 41.63203 G= __41.632________________ 20) Compute q and Gadj = 1+ ((42-1)/(6*50*2)) = 1.025 Gadj= 40.617 21) Test the statistic using method b). p<0.001 so pvalue<alpha, REJECT Ho 22) If you reject Ho, compute the CD and compare to the Expected CD to determine if the distribution is clumped or uniform. Sample Mean= Sample var= Sample CD= 1.78 1.603673469 0.900940151 4 -3 20) Draw conclusion Because you rejected Ho, the distribution is not stochastic but either clumped or uniform. Since the Sample CD (0.901) was greater than the Expected CD (q=0.407), then the distribution is clumped. Therefore, the black seastars are seen more often in clumps so it is likely that they are responding to some aspect of the microhabitat. PROBLEM #2: Seeds are the main food source for Dipodomys deserti, the desert kangaroo rat. Because the seed distribution appears to have a clumped distribution, you expect that the rodents will also have a clumped distribution. If you find a pattern, you will try to find out why. You have counted rats in 335 quadrats selected randomly from a desert region where the rodents are found. 23) What is Ho? The observed distribution of desert kangaroo rats fits a Poisson distribution with μ=1.94 24) What is the measured variable? Number of rats per quadrat 25) Is the sample size fixed? If so, what is the value? Sample size is NOT fixed 26) What is the appropriate theoretical probability distribution for this problem? Poisson 27) What would it mean if the results showed a clumped distribution? Desert kangaroo rats are found in clumps together which may mean that they are responding to clumped resources. 28) What would it mean if the results showed a uniform distribution? Desert kangaroo rats are evenly dispersed which most likely means that they are competing for resources (i.e. maintaining territories). 29) What would it mean if the observed frequencies were essentially the same as the expected frequencies? There is no pattern to the distribution of desert kangaroo rats. 4 -4 30) Determine Statistical Error to Avoid (alpha). Complete Table 5 Table 1: Select the statistical error to avoid for Problem 1 – Pisaster ochraceus color morph frequencies. Statistical Decision Accept Ho: Observed fits the expected Reject Ho: Observed does not fit the expected Action What if I’m wrong? No pattern Find another problem Lost out on finding something interesting. Pattern Try to find out why Wasted time and energy Conclusion Type of error II I 31) Alpha (α) = 0.025 32) Determine if the parameters are Extrinsic or Intrinsic Data Table 2: Data for Problem 2- Frequency distribution for Dipodomys deserti. # of Rats per Quadrat (Y) 0 1 2 3 4 5 6 TOTAL Observed Frequency (f) 25 118 97 54 32 7 2 335 fY2 fY 0 118 194 162 128 35 12 649 0 118 388 486 512 175 72 1751 33) Mean = 649/335 = 1.94 34) Variance = (1751-(6492/335))/(335-1) = 1.478 4 -5 35) Compute the appropriate Expected Frequencies. Complete Table 7 Table 3: Compute expected frequencies for Problem 2 – distribution of Dipodomys deserti. # of Rats per Quadrat (Y) Observed Frequency (f) 0 25 1 118 2 97 3 54 4 32 5 7 6 TOTAL 2 Probability equations e-1.94 P0* (1.94/1) P1* (1.94/2) P2* (1.94/3) P3* (1.94/4) P4* (1.94/5) 1(P0+P1+P2+ P3+P4+P5) 335 Probabilities Expected Frequencies 0.144091 48.3 0.279149 93.5 0.270399 90.6 0.174616 58.5 0.084571 28.3 0.032768 11 0.014406 4.8 1 335 36) For Poisson Only – Lump classes when an Expected Frequency <5.SHOW WORK in Table 8. Table 4- 4: Lump small frequencies for Problem 2 – distribution of Dipodomys deserti. # of Rats per Quadrat (Y) 0 1 2 3 4 ≥5 TOTAL Observed Frequency (f) 25 118 97 54 32 9 335 Expected Frequencies 48.3 93.5 90.6 58.5 28.3 15.8 335 37) Compute the degrees of freedom for the test. Since parameters are intrinsic and there was lumping (6 classes after lumping), Df=6-1-1-1 = 3 4 -6 38) Compute the G-statistic. Complete Table 9 Table 4- 5: Compute the G statistic for Problem 2 – distribution of Dipodomys deserti. # of Rats per Quadrat (Y) Observed Frequency (f) 0 1 2 3 4 ≥5 TOTAL Expected Frequencies 25 118 97 54 32 9 48.3 93.5 90.6 58.5 28.3 15.8 2 *Observed Freq*ln(Observed Freq/Expected Freq). -32.928 54.923 13.242 -8.645 7.864 -10.130 335 335 24.326 G = __24.326_____ 39) Compute Gadj = 1+ ((62-1)/(6*335*3) = 1.006 Gadj = 24.186 40) Test the statistic using method b). Pvalue <0.001, Since Pvalue<alpha (0.025), REJECT Ho 41) If you reject Ho, compute the CD and compare to the Expected CD to determine if the distribution is clumped or uniform. Sample Mean= Sample var= Sample CD= 1.937313 1.478095 0.762961 42) Draw conclusion: Since Ho was rejected, the distribution of desert kangaroo rats is not stochastic. Since the Sample CD < Expected CD (1.00), the distribution is Uniform. Therefore the rats are most likely competing with each other and maintaining territories rather than following the distribution of seeds. 4 -7