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