Some researchers have used the following method to perform tests of association in two-way tables. Instead of
Question:
Where n is the number of observation units in the sample. The sum of the new weights
wk , then, is n. The observed count for cell (i, j) is
xij = sum of the w k s for observations in cell (i, j)
and the expected count for cell (i, j) is mË ij = (xi+x+j)/n. Then compare the test statistic
to a Ï2 (r1) (c1) distribution.
Does this test give correct p-values for data from a complex survey? Why, or why not?
Fantastic news! We've Found the answer you've been seeking!
Step by Step Answer:
Related Book For
Question Posted: