St. John Fisher College

For a two-tailed test, P = 2(area in tail). From the normal table the area corresponding to z = 2.14, the area in the right tail is 1 – 0.9838 = 0.0162.So the P-value for the two-tail test with z =2.14 is P = 2(0.0162) = 0.0324. Since 0.0324 < 0.05, then reject H0. Examples: Find the P-value for a left-tailed test with a test statistic of z ... ................
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