1. Non-Parametric Statistics Part I: Chi-Square

2. x2 Operates on FREQUENCY Data
Suppose we have a plot of land on which we hope to harvest wood.
Maple is more valuable than Oak and Oak more valuable than pine.
We take a sample of the trees (the whole plot is too big) and we
ask whether there are significantly unequal amounts of each type (a=.05).
Pine
Maple
Oak
# of trees
145
301
289
We cannot get a mean from these data but there are clear differences
between the amounts in each category. This is categorical or nominal
data experessed as frequencies. So we use the x2

3. x2 : Homogeneity = (145 + 301 + 289)/3 = 245 245 245 245
What are the null and alternative hypotheses?
H0: The groups do not have different frequencies.
H1: The groups do have different frequencies.
Find the critical value:
x2 table (k-1 df = 3-1= 2) = 5.99
Calculate the obtained statistic:
Pine
Maple
Oak
# of trees observed
145
301
289
# of trees expected
= ( )/3
= 245
= 61.52
Make a decision:
Our obtained value is larger than our critical value.
Reject the null; the groups do have different frequencies.

4. x2 : Homogeneity Example
Is political affiliation distributed equally in our class? (use alpha=.01)
What are the null and alternative hypotheses?
H0: The groups do not have different frequencies.
H1: The groups have different frequencies.
Find the critical value:
x2 table (k-1 df = 3-1= 2) = 9.21
Calculate the obtained statistic:
Democrat
Republican
Other
# of people observed 10 15 5
= ( )/3
= 10
# of people
expected
= 5
Make a decision:
Our obtained value is smaller than our critical value.
Retain the null; the groups do not have different frequencies.

5. x2 : Goodness of Fit
Pine proportion = 255/473 = 0.54
Maple proportion = 115/473 = 0.24
Oak proportion = 103/473 = 0.22
Pine expected = 0.54(735) = 396.9
Maple expected = 0.24(735) = 176.4
Oak expected = 0.22(735) = 161.7
Five years ago the tree-lot was also sampled. Has the composition of the lot changed since then (use alpha=.05)?
We need a different expected value based on the previous sample.
Pine
Maple
Oak
# trees
2014
145
301
289
Total #
735
473
Pine
Maple
Oak
# trees
2009
255
115
103
Notice we’re trying to compare the frequencies from two time points, but the total # of trees categorized in 2014 is different from the 2009 total!
Pine
Maple
Oak
# trees
expected
396.9
176.4
161.7

6. x2 : Goodness of Fit Example
What are the null and alternative hypotheses?
H0: The composition of the lot has not changed.
H1: The composition of the lot has changed.
Find the critical value:
x2 table (k-1 df = 3-1= 2) = 5.99
Calculate the obtained statistic:
Pine
Maple
Oak
# trees
2014
145
301
289
Pine
Maple
Oak
# trees
expected
396.9
176.4
161.7 =
Make a decision:
Our obtained value is larger than our critical value.
Reject the null; the composition of the lot has changed.

7. x2 : Independence Pine Maple Oak
Mirkwood
Old Forest
H0: Tree type and Forest are independent.
H1: Tree type and Forest are not independent.

8. x2 : Independence Example
(assume alpha=.05)
What are the null and alternative hypotheses?
H0: Tree type and forest are independent.
H1: Tree type and forest and not independent.
Find the critical value:
df for this test is (r-1)(c-1)
We have 2 rows and 3 columns, so (2-1)(3-1) = 2
x2 table (df = 2) = 5.99
Calculate the obtained statistic:

9. How to calculate expected values:
x2 : Independence
How to calculate expected values:
Pine Maple Oak
Mirkwood
Old Forest R
702
798
356 C
Grand Total: 1500
Expected value = (R x C)/ grand total
Expected Mirkwood-Pine = (702 x 356)/1500 =
Expected Old Forest-Pine = (798 x 356)/1500 =

10. S x2 : Independence Observed Values Expected Values Pine Maple Oak
Mirkwood
Old Forest
Pine Maple Oak
Mirkwood
Old Forest
( O - E)2
x2 = E S
= 28.18

11. x2 : Independence Example
(assume alpha=.05)
What are the null and alternative hypotheses?
H0: Tree type and forest are independent.
H1: Tree type and forest and not independent.
Find the critical value:
df for this test is (r-1)(c-1)
We have 2 rows and 3 columns, so (2-1)(3-1) = 2
x2 table (df = 2) = 5.99
Calculate the obtained statistic:
28.18
Make a decision:
Our obtained value is larger than our critical value.
Reject the null; tree type and forest are not independent.