When df = 1, the P-value from the chi-squared test of independence is the same as the P-value for the two-sided test comparing two proportions with the z test statistic. This is because of a direct connection between the standard normal distribution and the chi-squared distribution with df = 1: Squaring a z-score yields a chi-squared value with df = 1 having chi-squared right-tail probability equal to the two-tail normal probability for the z -score.
a. Illustrate this with z = 1.96, the z -score with a two-tail probability of 0.05. Using the chi-squared table or software, show that the square of 1.96 is the chi-squared score for df = 1 with a P-value of 0.05.
b. Show the connection between the normal and chi-squared values with P-value = 0.01.

  • CreatedSeptember 11, 2015
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