Question

A random sample of 550 physicians was selected and classified according to the type of practice they operate and whether or not their practice was full (meaning, not accepting new patients). The counts are given in the table below.

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The test statistic for a test of independence is ?² = 34.893. What is the p-value for this test?

a. p < .0001

b. p = .004

c. p = .0025

d. p = .05

Answer 1

As given in the question: The calculated value of chi-square statistic x? = 34.893 The number of rows in the given contingency table = 3 The number of columns in the given contingency table = 2 The degrees of freedom are = (The number of rows - 1)*(The number of columns - 1) = (3-1)(2-1) = 2x1 = 2 The p-value of the test at x² = 34.893 and 2 degrees of freedom is less than 0.0001. (By using the chi-square table.) Therefore, the option (a) is correct.