II . We will use a chi-square test for homogeneity of proportions. The samples of tenured and
nontenured instructors are given as random. We determine that the expected values are given by the
matrix Because all expected values are greater than 5, the conditions for the test arepresent.III . X 2 = 1.78, df = (2 – 1)(2 – 1) = 1 0.15 < P -value < 0.20 (from Table C; on the TI-83/84: P =
χ^2 cdf(1.78,1000,1) = 0.182 ).IV . The P -value is not small enough to reject the null hypothesis. These data do not provide strong
statistical evidence that tenured and nontenured faculty differ in their attitudes toward the proposed
salary plan.X 2 = z 2
The example just completed could have been done as a two-proportion z -test where p 1 and p 2 are
defined the same way as in the example (that is, the proportions of tenured and nontenured staff that favor
the new plan). Then
and
Computation of the z -test statistics for the two-proportion z -test yields z = 1.333. Now, z 2 = 1.78.
Because the chi-square test and the two-proportion z -test are testing the same thing, it should come as
little surprise that z 2 equals the obtained value of X 2 in the example. For a 2 × 2 table, the X 2 test
statistic and the value of z 2 obtained for the same data in a two-proportion z -test are the same. Note that
the z -test is somewhat more flexible in the case that a one-sided test allows us to consider a particular
direction of difference, whereas the chi-square test has only one tail and does not allow us to test for a
particular direction of the difference. However, this advantage only holds for a 2 × 2 table since there is
no way to use a simple z -test for situations with more than two rows and two columns.
Digression: You aren’t required to know these, but you might be interested in the following
(unexpected) facts about a χ^2 distribution with k degrees of freedom:• The mean of the χ^2 distribution = k .
• The median of the χ^2 distribution (k > 2) ≈ k – ⅔.
• The mode of the χ^2 distribution = k – 2.
• The variance of the χ^2 distribution = 2k .Rapid Review
A study yields a chi-square statistic value of 20 (X 2 = 20). What is the P -value of the test if