This page provides 2 calculations for large contingency tables of counts, the chi squares test and the non-parametric Spearman's correlation coefficient ρ

Power(parametric)	Power(non-parametric)
0.80	0.818
0.90	0.909
0.95	0.955
0.99	0.991

There is no specific calculations to estimate sample size requirements for Spearman's vorrelation coefficient. According to Siegel S and Castellan Jr. N J (2000) however, the efficientcy of non-parametric correlation coefficient such as Spearman or Kendall is 91% that of Pearson's (see reference). The appropriare modification in the power value in the calculations need to be made, where

β(non-parametric) = β(paramtric) x 0.91

The table to the tight shows the required modifications of commonly used powers. Calculations to estimate sample size for correlation coefficients are in SSizCorr.php

References

Siegel S and Castellan Jr. N J (2000) Nonparametric Statistics for the Behavioral Sciences (Second Edition). McGraw-Hill International Edition Sydney ISBN 0-07-100326-6

Chi Squares : p.191-200
Spearman's correlation : p.235-244, Table Q in p.360-361.

Data Entry Using Case Values : Data a table of 2 columns   -Each row is data from an individual subject   -Column 1 is the x value, row variable   -Column 2 is the y value, column variable    Data Entry Using Table of Counts : uses a table of counts   -The rows represent x values in rank order from top to bottom   -The columns represent y values in rank order from left to right   -Each cell the number of subjects for that row / column

R codes for Chi Square (large contingency table) and Spearman's Correlation Coefficient

Two programs, for entry by raw data or by table of counts

Program 1: raw data data table of 2 columns x and y

# data entry
dat = ("
X       Y      
1	1
2	2
3	3
2	3
5	6
2	3
4	3
5	5
       ")
df <- read.table(textConnection(dat),header=TRUE)  # conversion to data frame    
result <- chisq.test(df$X,df$Y)
result # result of chi square test
result <- cor.test(df$X,df$Y, method="spearman")  # Correlation
result # result of Spearman Correlation

The results are as follows

> result # result of chi square test

	Pearson's Chi-squared test

data:  df$X and df$Y
X-squared = 17.333, df = 16, p-value = 0.3644

> result # result of Spearman Correlation

	Spearman's rank correlation rho

data:  df$X and df$Y
S = 9.2934, p-value = 0.003111
alternative hypothesis: true rho is not equal to 0
sample estimates:
      rho 
0.8893639 

Program 2: data is a table of counts

# data entry
dat = ("
6	 0	5
6	 0	10
3	 5	8
0	15	7
       ")
mx <- read.table(textConnection(dat),header=FALSE)  # conversion to data matrix 
# create X and Y from matrix of counts mx
# vectors of X and Y
X <- vector()
Y <- vector()
for(i in 1:nrow(mx))
{
  for(j in 1:ncol(mx))
  {
    if(mx[i,j]>0)
    {
      for(k in 1:mx[i,j])
      {
        X <- append(X,i)
        Y <- append(Y,j)
      }
    }
  }
}
# Calculations
result <- chisq.test(X, Y)
result # result of chi square test
result <- cor.test(X, Y, method="spearman")  # Correlation
result # result of Spearman Correlation

The results are as follows

> result # result of chi square test

	Pearson's Chi-squared test

data:  X and Y
X-squared = 31.644, df = 6, p-value = 1.909e-05

> result # result of Spearman Correlation

	Spearman's rank correlation rho

data:  X and Y
S = 42101, p-value = 0.5266
alternative hypothesis: true rho is not equal to 0
sample estimates:
       rho 
0.07996065