Three "meh" p-values that together clear the bar

Question

Three independent studies of the same hypothesis return p-values of 0.10, 0.15, and 0.08, none significant on its own. You suspect the accumulated evidence is stronger than any single study.

Use Fisher's method to combine the independent p-values, and interpret the result.

Accepted Answer

The construction. Under the null, a p-value is uniform on [0,1], and -2\ln(p) has a chi-square distribution with 2 degrees of freedom. Summing k independent such terms gives a chi-square with 2k degrees of freedom: X = -2\sum{i=1}^{k} \ln(pi) \ \sim\ \chi^2{2k} \quad 	ext{under } H0. Large X (many small p-values) is evidence against the combined null. Plug in the three p-values. \ln(0.10) = -2.303,\quad \ln(0.15) = -1.897,\quad \ln(0.08) = -2.526. X = -2\,(-2.303 - 1.897 - 2.526) = -2 	imes (-6.726) = 13.45, with 2k = 6 degrees of freedom. Get the combined p-value. The 95th percentile of \c…

Three "meh" p-values that together clear the bar

Your answer

More Statistics questions