The single-classification Chi-square (2) test

    The critical value for a single-classification 2 test ("either / or": observed and expected values are either one way or the other), therefore with degrees of freedom = df = 1 is 3.84 at p = 0.05. That is, for a test of a hypothesis with two equally-likely alternative outcomes (heads or tails, round or wrinkled, left or right, etc), 2 = 3.84 or greater indicates that the chances are 5% or less that so large a deviation could have been obtained simply by chance. [Alternatively, in a large number of similar experiments, such a result would occur about one time in 20]. Given a 2 = 6.635 or greater, the chance is less than 1%. A statistical outcome that occurs less than 5% of the time is said to be significant marked and  *; a results obtained less than 1% of the time is highly significant and marked **. Non-significant outcomes are marked ns.

    The formula for the calculation for a single-classification test is as follows, where w + z = x + y, the deviation (d) is the difference between the observed (o) and expected (e) results for each class, and the chi-square value* is the squared deviation divided by the expected value, summed for both classes:


o
e
d
d2/e
I
x
w
(x-w)
(x-w)2 / w
II
y
z
(y-z)
(y-z)2 / z
2 = [d2/e] = [(x-w)2 / w] + [(y-z)2 / z]

Consider two experiments in which coins are tossed, and the two possibilities are heads (H) or tails (T). The first involves n = 10 tosses, the second n = 100 tosses


o
e
d
d2/e
H
6
5
1
1 / 5
T
4
5
-1
1 / 5
2 = 1/5 + 1/5 = 0.40, ns


o
e
d
d2/e
H
60
50
10
100 / 50
T
40
50
-10
100 / 50
2 = 2.0 + 2.0 = 4.0*p < 0.05   [or 0.05 > p > 0.01]

    Note that the same 60% / 40% proportion is not significantly different from random in the smaller experiment, but is significant in the larger. This emphasizes that actual numbers, not proportions, must be tested, and that statistical power (ability to detect a difference of a particular magnitude) is greater in larger experiments.

    *For the advanced student: show that, for a single-classification test with a total sample size of n where X is the value of the first observed class and the two expected values are equal or "even" (x = y = 0.5), the value of chi-square simplifies to 
2 = (n - 2X)2 / n. Then, calculate the sample size n necessary to show that a deviation of 1% from even is significant at p = 0.05.

All text material 2013 by Steven M. Carr