|
This shows how pvalue can be calculated from any two arrays, using Welch's 2-sided t-test, which doesn't assume equal variance.
The high "n" value ensures roughly 6 digits convergence with Simpson's integral approximation.
<lang C>double calculate_Pvalue (const double *array1, const size_t array1_size, const double *array2, const size_t array2_size) { ▼
<lang C>#include <stdio.h>
#include <math.h>
▲double calculate_Pvalue (const double *array1, const size_t array1_size, const double *array2, const size_t array2_size) {
if (array1_size <= 1) {
return 1.0;
}
double return_value = ((h / 6.0) * ((pow(x,a-1))/(sqrt(1-x)) + 4.0 * sum1 + 2.0 * sum2))/(expl(lgammal(a)+0.57236494292470009-lgammal(a+0.5)));//0.5723... is lgammal(0.5)
if ((isinfisfinite(return_value) !== 0) || (return_value > 1.0)) {
return 1.0;
} else {
return return_value;
}
}
//-------------------
int main(void) {
const double d1[] = {27.5,21.0,19.0,23.6,17.0,17.9,16.9,20.1,21.9,22.6,23.1,19.6,19.0,21.7,21.4};
const double d2[] = {27.1,22.0,20.8,23.4,23.4,23.5,25.8,22.0,24.8,20.2,21.9,22.1,22.9,20.5,24.4};
const double d3[] = {17.2,20.9,22.6,18.1,21.7,21.4,23.5,24.2,14.7,21.8};
const double d4[] = {21.5,22.8,21.0,23.0,21.6,23.6,22.5,20.7,23.4,21.8,20.7,21.7,21.5,22.5,23.6,21.5,22.5,23.5,21.5,21.8};
const double d5[] = {19.8,20.4,19.6,17.8,18.5,18.9,18.3,18.9,19.5,22.0};
const double d6[] = {28.2,26.6,20.1,23.3,25.2,22.1,17.7,27.6,20.6,13.7,23.2,17.5,20.6,18.0,23.9,21.6,24.3,20.4,24.0,13.2};
const double d7[] = {30.02,29.99,30.11,29.97,30.01,29.99};
const double d8[] = {29.89,29.93,29.72,29.98,30.02,29.98};
const double x[] = {3.0,4.0,1.0,2.1};
const double y[] = {490.2,340.0,433.9};
printf("Test sets 1 p-value = %lf\n",calculate_Pvalue(d1,sizeof(d1)/sizeof(*d1),d2,sizeof(d2)/sizeof(*d2)));
printf("Test sets 2 p-value = %lf\n",calculate_Pvalue(d3,sizeof(d3)/sizeof(*d3),d4,sizeof(d4)/sizeof(*d4)));
printf("Test sets 3 p-value = %lf\n",calculate_Pvalue(d5,sizeof(d5)/sizeof(*d5),d6,sizeof(d6)/sizeof(*d6)));
printf("Test sets 4 p-value = %lf\n",calculate_Pvalue(d7,sizeof(d7)/sizeof(*d7),d8,sizeof(d8)/sizeof(*d8)));
printf("Test sets 5 p-value = %lf\n",calculate_Pvalue(x,sizeof(x)/sizeof(*x),y,sizeof(y)/sizeof(*y)));
return 0;
}
</lang>
Test sets 4 p-value = 0.090773
Test sets 5 p-value = 0.010751</pre>
=={{header|R}}==
<lang R>#!/usr/bin/R
|