CHAPTER 4: GOODNESS OF FIT TESTING.
PURPOSE: Up until now, we have made a decision about whether a distribution is clumped,
uniform or stochastic in two ways: 1) we visually compared the observed frequencies to the expected
frequencies and 2) we computed the CD and compared it to some value. In both of these cases, if
differences are large we have no trouble believing the result, but, if the differences are small, we
really aren’t sure. Now we will learn how to use a statistical test to help make a decision with some
degree of confidence.
We will use a statistical test, the Log-likelihood Ratio Goodness of Fit test (a.k.a. the G-test), to
decide if differences between observed and expected are real. If we decide that the differences are not
real, then we will conclude that the distribution in question is stochastic. If we do decide that the
differences are real, we will compute the CD to tell us whether the distribution is either clumped or
uniform.
Hypotheses:
The Null Hypothesis or Ho is: The observed frequency distribution FITS the expected frequency
distribution with the given parameters.
The Alternative Hypothesis or Ha is: : The observed frequency distribution DOES NOT FIT the
expected frequency distribution with the given parameters.
Interpretation of Hypotheses
If you accept Ho, you conclude that the observed frequency distribution is stochastic. If you reject Ho
and accept Ha, you conclude that the observed frequency distribution is either clumped or uniform;
then you need to inspect the graph or the Coefficient of Dispersion to determine which.
Computing the Test - Basic Steps.
1) First determine what your results would mean with respect to your original
question.
For example, what would it mean it the pattern was uniform or clumped or stochastic?
2) Determine which statistical error to avoid.
There are two types of statistical errors that you can make with this test. You can accept Ho and
conclude that the expected distribution fits the observed but it really doesn’t or you can reject Ho
and conclude that the expected distribution does not fit the observed but it really does. Accepting
Ho incorrectly is a Type II error. Rejecting Ho and accepting Ha incorrectly is a Type I error.
The way to decide which error to avoid is to determine which error will cause the most harm. To
determine which error to avoid, complete the following table.
4 -1
Table 4- 1: Selecting which statistical error to avoid.
Error Avoidance Decision Matrix
Statistical
Conclusion
Decision
Accept Ho:
Observed fits
the expected
Reject Ho:
Observed does
not fit the
expected
Action
What if I’m
wrong?
Type of
error
II
I
If you decide to avoid Type I error, set alpha in tests of hypotheses to 0.025
If you decide to avoid Type II error, set alpha in tests of hypotheses to 0.050
3) Determine if the parameters are Extrinsic or Intrinsic.
They are EXTRINSIC if the values are KNOWN (e.g. p and q for a coin flip are 0.5). They are
INTRINSIC if they had to be estimated (ie. computed from the data).
4) Compute the appropriate Expected Frequency distribution.
5) For Poisson Only – Lump classes when an Expected Frequency <5.
If the expected frequency is less than 5 for classes with values greater than 2.0, then adjacent
classes must be grouped together before testing such that each class with a value greater than 2
should have an expected frequency of 5 or greater.
6) Compute degrees of freedom for the test.
df=ν
Table 4- 2: Determining degrees of freedom for the Goodness of Fit test.
Parameters
Lumping
Yes
No
Extrinsic
a*-2
a-1
*where a = number of classes after lumping
4-2
Intrinsic
a-3
a-2
7) Compute the G-statistic
G  2 * Observed Freq * ln(Observe d Freq/Expec ted Freq)
8) Compute Gadj


 a2 1 
Gadj= G/ where   1  
 , n=total frequency (∑f ) and v=degrees of freedom
 6n 
9) Test the statistic. There are two ways to do this. Method a) is the oldest method but
method b) is more useful for reporting results. Both give the same conclusion
a) Look up the critical value in the Chi-Square table or in Excel and compare to
Gadj
i) If Gadj ≤ Х2 then ACCEPT Ho and conclude that the observed distribution is stochastic.
ii) If Gadj > Х2 then REJECT Ho and compute the CD to determine whether the observed
distribution is uniform or clumped.
b) Use Excel to look up the probability and compare to Alpha
i) If α ≤ P then ACCEPT Ho and conclude that the observed distribution is stochastic.
ii) If α > P then REJECT Ho and compute the CD to determine whether the observed
distribution is uniform or clumped.
10) If you reject Ho in Step 9, compute the CD and compare to the Expected CD to
determine if the distribution is clumped or uniform.
The expected CD for the Binomial is 1-p (or q) and the expected CD for the Poisson is 1.
If the Computed CD is greater than the Expected CD, the distribution is clumped. Otherwise,
the distribution is uniform
11) Draw conclusion
EXAMPLE 1
Someone has published an article claiming that the environment can influence gender of offspring
such that families would end up with either more males or more females than expected by chance.
You suspect that the finding may be inaccurate so you set up a sampling scheme to find out. You
randomly select 30 families that have four children.
What is the measured variable? Female (yes or no)
Is the sample size fixed? Yes, k=4
What does “p” represent in this experiment? The probability of having a female child by chance
alone.
4 -3
Do you have to compute it from the data or do you already know the expected result?
know that p=0.5.
We already
What does “q” represent in this experiment? The probability of the next child not being female.
What is the value of q? 1-p=1-0.5= 0.5
How would you conduct the experiment? You would randomly select 30 families from families of
four. Then you would record the gender of each sibling.
1) Determine what your results would mean with respect to your original question.
What would it mean if the results showed a clumped distribution?
It would mean that there was a higher probability of getting mostly males or mostly females than
would be expected by chance alone; something was controlling the distribution of males and
females.
What would it mean if the results showed a uniform distribution?
If the genders were distributed uniformly, it would mean that families tended to have the same
distribution of genders and that families were too alike to be a function of chance.
What would it mean if the observed frequencies were essentially the same as the expected
frequencies?
It would mean that there wasn’t any pattern to the distribution of females or males.
2) Determine Statistical Error to Avoid (alpha)
Table 4- 3: Determining which statistical error to avoid - Example 1 - Binomial
Conclusion
Action
What if I’m wrong?
Accept Ho:
Observed fits
the expected
There is no
pattern to
gender
Incorrectly critized a
fellow scientist’s
work
Reject Ho:
Observed does
not fit the
expected
There is a
pattern to
gender
Publish a
rebuttal to the
previous author’s
findings.
Do nothing
Lost out on a chance
for recognition
Type of
error
II
I
Conclude that the worse error is Type II so alpha (α) will equal 0.05.
3) Determine if the parameters are Extrinsic or Intrinsic
If the number of females is stochastic, then p should be 0.5 because there should be an equal
chance of obtaining a male or female. Since this is not computed from the data, the parameters
are Extrinsic! Otherwise, you would have to estimate p.
4-4
4) Compute the appropriate Expected Frequency distribution
Table 4- 4: Computing the expected frequencies for Example 1- Binomial.
# of Females in Observed
Family (Y)
Frequency (f)
0
7
1
4
2
2
3
7
4
10
TOTAL
30
Probability
equations
q4
4pq3
6p2q2
4p3q
p4
Probabilities
0.0625
0.2500
0.3750
0.2500
0.0625
1.0000
Expected
Frequencies
1.9
7.5
11.3
7.5
1.9
30
5) Do any necessary lumping (for Poisson Only) where an Expected Frequency <5.
Not necessary here as we are not using the Poisson
6) Compute the degrees of freedom for the test.
This is Extrinsic with no lumping and a=5 so the degrees of freedom(ν) = a-1 = 4
7) Compute the G-statistic
Table 4- 5: Computing the G-statistic for Example 1: Binomial.
# of Females in Observed
Family (Y)
Frequency (f)
0
1
2
3
4
TOTAL
7
4
2
7
10
30
Expected
Frequencies
1.9
7.5
11.3
7.5
1.9
30.1
2 *Observed
Freq*ln(Observed
Freq/Expected Freq).
18.257
-5.029
-6.927
-0.966
33.215
38.550
G = 39.550
8) Compute Gadj
=1+ ( (a2-1)/6nν)) = 1+((52-1)/(6*30*4))=1.033 where ν = degrees of freedom
Gadj = 39.550/1.033 =38.287
4 -5
9) Test the statistic. There are two ways to do this. Method a) is the oldest method
but method b) is more useful for reporting results. Both give the same conclusion).
a) Look up the critical value in the Chi-Square table or in Excel and compare to
Gadj
The critical value of the Chi-square distribution can be computed using the CHIINV(α, ν )
function in Excel. Enter =CHIINV(0.05,4) into a cell and you will get 9.488 (rounded to 3
decimal places). So Х2 (0.05,4)=9.488
Because Gadj (38.287) > Х2 (9.488) we REJECT Ho.
b) Use Excel to look up the probability and compare to Alpha
P (actual Type I error) can also be computed directly using the CHIDIST (X, ν ) function
where X is the computed value (in this case Gadj). Enter =ROUND(CHIDIST(37.77,4),3)
into a cell and you will get 0. P is never zero but it is a very small number in this case so you
would report it as p<0.001.
Because α (0.05) > p (<0.001) we REJECT Ho
10) If you reject Ho in Step 9, compute the CD and compare to the Expected CD to
determine if the distribution is clumped or uniform.
Table 4- 6: Computing the CD.
# of Females in Observed
Family (Y)
Frequency (f)
0
7
1
4
2
2
3
7
4
10
TOTAL
30
Y  69
30
 2.3
s2 
f*Y2
f*Y
0
4
4
21
40
69
235  69
0
4
8
63
160
235
2
30  1
30  2.63
CD  8.10
2.3
 1.144
Since this is a Binomial Distribution, we compare 1.144 to the Binomial Expected CD =1-p
=1-0.5=0.5. Because 1.144>0.5, the distribution is clumped.
1) Draw conclusion.
The distribution of female births in families of 4 is clumped (Gadj=38.287, ν = 4. p<0.001); there
are more families with all females or all males than expected by chance. Therefore the published
idea is correct.
EXAMPLE 2:
Suppose you are working for Fish and Game and you have been assigned to investigate road kills
on Highway 26 south of Hollister, about 100 miles from San Jose. In this case, your biggest
4-6
priority is to try to see if there are specific areas where road kills are excessively high. If you
identify problem areas, you will try to develop safe corridors (e.g. culverts) for the animals.
However, you do have a tight budget so you must be sure that there really are problem areas
before you authorize the construction of corridors. You have divided the 100 mile distance into 1mile segments (units of space) and count the number of roadkills on the route in each segment
(you are measuring the abundance of road kills).
What is the measured variable? Road Kills
Is the sample size fixed? No
1) Determine what your results would mean with respect to your original question.
What would it mean if the results showed a clumped distribution?
If the results showed a clumped distribution, you would conclude that there are “hot spots” where
animals are more likely to be killed and “safe spots” where animals are less likely to be killed than by
chance alone.
What would it mean if the results showed a uniform distribution?
If the distribution is uniform, it would mean that virtually every segment has exactly the same
mortality rate. This would be very strange and you would suspect some outside influence
What would it mean if the observed frequencies were essentially the same as the expected
frequencies?
There is no pattern to the number of road kills per segment.
2) Determine Statistical Error to Avoid (alpha)
Table 4- 7: Determining which statistical error to avoid - Example 2 - Poisson
Statistical
Decision
Accept Ho:
Observed fits
the expected
Reject Ho:
Observed does
not fit the
expected
Conclusion
Action
What if I’m wrong?
There is no
pattern
Do not need to
do any special
control measures
Need to develop
safe corridors
Animals are dying
needlessly
II
Extra work and
expense for nothing
I
There is a
pattern
Type of
error
Want to avoid a Type I error so alpha(α) = 0.025
3) Determine if the parameters are Extrinsic or Intrinsic
Because the mean had to be determined from the data, the parameters are Intrinsic.
4 -7
4) Determine Expected Frequencies
Table 4- 8: Determine expected frequencies for Example 2 - Poisson.
# of kills per
segment
0
1
2
3
4
5
≥6
TOTAL
Expected
probabilities
0.124
0.259
0.270
0.188
0.098
0.041
0.020
1.000
Observed
frequencies
13
26
26
18
10
5
2
100
Expected
frequencies
12.4
25.9
27.0
18.8
9.8
4.1
2.0
100.0
5) Do any necessary lumping (for Poisson Only) where an Expected Frequency <5.
Notice that the expected frequency for class ≥6 = 2. Because this is lower than 5, we must
combine the observed and expected frequencies for class ≥6 with class 5 to make a new class ≥5
(Table 4-8). The new class, ≥5, now has an expected frequency of 6.1 which is fine. However, if
the expected frequency had been less than 5, we would have to lump again.
Table 4- 9: Lumping - combining observed and expected frequencies for classes that have an expected
frequency less than 5.
5
≥6
≥5
0.041
0.020
becomes
0.061
5
2
4.1
2.0
7
6.1
6) Compute the degrees of freedom for the test.
This is Intrinsic with lumping and a=6 so the degrees of freedom(ν) =a-3= 3
7) Compute the G-statistic
Table 1: Computing the G-statistic for Example 2 - Poisson
# of kills per
segment
0
1
2
3
4
≥5
TOTAL
G = 0.23
4-8
Observed
Frequency
(f)
13
26
26
18
10
7
100
Expected
Frequencies
12.4
25.9
27.0
18.8
9.8
6.1
100.0
2 *Observed
Freq*ln(Observed
Freq/Expected Freq).
1.294
0.300
-1.989
-1.603
0.336
1.895
0.233
8) Compute Gadj
=1+ ( (a2-1)/6nν) = 1+((62-1)/(6*100*3))=1.019 where ν= degrees of freedom
Gadj = 0.233/1.019 =0.228
9) Test the statistic. There are two ways to do this. Method a) is the oldest method
but method b) is more useful for reporting results. Both give the same conclusion).
a) Look up the critical value in the Chi-Square table or in Excel and compare to
Gadj
The critical value of the Chi-square distribution can be computed using the CHIINV(α, ν )
function in Excel. Enter =CHIINV(0.025,3) into a cell and you will get 9.348 (rounded to
3 decimal places). So Х2 (0.025,3)=9.348
Because Gadj (0.228) < Х2 (9.348) we ACCEPT Ho.
b) Use Excel to look up the probability and compare to Alpha
P (actual Type I error) can also be computed directly using the CHIDIST (X, ν ) function
where X is the computed value (in this case Gadj). Enter =ROUND(CHIDIST(0.228,3),3)
into a cell and you will get 0.973. Because α (0.05) < p (0.973) we ACCEPT Ho
10) If you reject Ho in Step 9, compute the CD and compare to the Expected CD to
determine if the distribution is clumped or uniform.
Because we did not reject Ho, we do not need to compute the CD.
11) Draw conclusion.
There is no pattern (Gadj =0.228, ν =3, p=0.973) to the distribution of road Kills along a 100 mile
stretch of Hwy 26 south of Hollister. Therefore it is not necessary to construct safe corridors.
4 -9
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?
What is Ha?
.
2) What is the measured variable?.
3) Is the sample size fixed? If so, what is the value?
4) What is the appropriate theoretical probability distribution for this problem?
5) What would it mean if the results showed a clumped distribution?
6) What would it mean if the results showed a uniform distribution?
7) What would it mean if the observed frequencies were essentially the same as the expected
frequencies?
4-1 0
8) 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
Conclusion
Action
What if I’m
wrong?
Accept Ho:
Observed fits
the expected
Reject Ho:
Observed
does not fit
the expected
Type
of
error
II
I
9) Alpha (α) =
10) Are the parameters Intrinsic or Extrinsic?
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
fY
fY2
11) How many seastars did you examine?
12) How many of those were black?
13) What is the value for p?
14) What is the value for q?
4 -1 1
15) 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)
Observed
Frequency
(f)
0
13
1
7
2
8
3
22
Probability
equations
Probabilities
Expected
Frequencies
TOTAL
16) For Poisson Only – Lump classes when an Expected Frequency <5
17) Compute the degrees of freedom for the test.
18) 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
TOTAL
Observed
Frequency
(f)
13
7
8
22
G= __________________
19) Compute q and Gadj
=
Gadj=
20) Test the statistic using method b).
4-1 2
Expected
Frequencies
2 *Observed
Freq*ln(Observed
Freq/Expected Freq).
21) If you reject Ho, compute the CD and compare to the Expected CD to determine if the
distribution is clumped or uniform.
20) Draw conclusion
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.
22) What is Ho?
23) What is the measured variable?
24) Is the sample size fixed? If so, what is the value?
25) What is the appropriate theoretical probability distribution for this problem?
26) What would it mean if the results showed a clumped distribution?
27) What would it mean if the results showed a uniform distribution?
28) What would it mean if the observed frequencies were essentially the same as the expected
frequencies?
4 -1 3
29) Determine Statistical Error to Avoid (alpha). Complete Table 5
Table 10: Select the statistical error to avoid for Problem 1 – Pisaster ochraceus color morph frequencies.
Statistical
Decision
Conclusion
Action
What if I’m
wrong?
Accept Ho:
Observed fits
the expected
Reject Ho:
Observed
does not fit
the expected
I
30) Alpha (α) =
31) Determine if the parameters are Extrinsic or Intrinsic
Data
Table 11: Data for Problem 2- Frequency distribution for Dipodomys deserti.
# of Rats per
Quadrat (Y)
0
1
2
3
4
5
6
TOTAL
32) Mean =
33) Variance =
4-1 4
Type
of
error
II
Observed
Frequency (f)
25
118
97
54
32
7
2
fY
fY2
34) Compute the appropriate Expected Frequencies. Complete Table 7
Table 12: 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
2
Probability
equations
Probabilities
Expected
Frequencies
TOTAL
35) For Poisson Only – Lump classes when an Expected Frequency <5.SHOW WORK in Table 8.
Table 4- 13: Lump small frequencies for Problem 2 – distribution of Dipodomys deserti.
# of Rats per
Quadrat (Y)
Observed
Frequency
(f)
Probability
equations
Probabilities
Expected
Frequencies
TOTAL
36) Compute the degrees of freedom for the test.
4 -1 5
37) Compute the G-statistic. Complete Table 9
Table 4- 14: 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
6
25
118
97
54
32
7
2
TOTAL
Expected
Frequencies
2 *Observed
Freq*ln(Observed
Freq/Expected Freq).
48
94
91
58
28
11
5
-32.616
53.664
12.387
-7.718
8.546
-6.328
-3.665
24.270
G = ______________
38) Compute Gadj
=
Gadj =
39) Test the statistic using method b).
40) If you reject Ho, compute the CD and compare to the Expected CD to determine if the
distribution is clumped or uniform.
41) Draw conclusion
4-1 6