The degrees of freedom for a chi ^ 2 test of independence is calculated as (number of rows - 1) * (number of columns - 1) = (5 - 1) * (4 - 1) = 4 * 3 = 12.
Given that the chi ^ 2 value is 22.5 with 12 degrees of freedom, we can determine the P-value for this statistic.
Looking at a chi ^ 2 distribution table or using statistical software, we find that the P-value for a chi ^ 2 value of 22.5 with 12 degrees of freedom is less than 0.01.
Therefore, the correct answer is:
E. 0.02 < P < 0.01
For a chi ^ 2 test of independence, a researcher computes a chi ^ 2 value of 22.5. The two-way table from which this value was computed had 5 rows and 4 columns. What's the P-value for this statistic?
A. 0.05 < P < 0.1
B. 10 < P < 0.15
C. 0.025 < P < 0.05
OD. P > 0.25
E. 0.02 < P < 0.01
1 answer