How to compare correlations In R I have a dataset data for each day a week, ie 7 datasets. For the Monday I have data_monday which gives
promotions   views
41           231
7            11
120          236
...

I get similar output for the other days a week. In R I have calculated the correlation between promotion and views. For example for data_monday we get
cor.test(data_monday[,1], data_monday[,2], alternative="greater",method="pearson")

and this gives me a low p-value, ie we have significant positive correlation and we also get an estimate of the correlation - in this case it's 0.494.
For all the days we have positive correlation but we have different values of the correlation. For example Sunday we have an estimated correlation at only 0.291. 
My question is this: How can I compare the correlation-values and use it to determine that some days a week have a higher correlation than others? For example one can say that Monday we have a significant higher correlation than Sunday and therefore we should use more money on making the promotion higher than Sunday.
Update
Using the advise from the second answer the Peason correlation I get and confident interval but I can't see how to use this to compare the correlations. For example Friday gives me this
cor.test(data_friday[,1],data_friday[,2],alternative="greater",methid="pearson")

and this gives me this output
        Pearson's product-moment correlation

 t = 7.1374, df = 60, p-value = 7.29e-10
alternative hypothesis: true correlation is greater than 0
95 percent confidence interval:
 0.5445291 1.0000000
sample estimates:
      cor 
0.6776269 

I want to use Fisher r to z transform to calculate this in R using the packages
library(psych)

I'm not sure how to calculate this in R. 
My Answer
This is how I would compare the correlation and draw my conlsusion:
In R I have this function
diff.corr <- function( r1, n1, r2, n2 ){ 

Z1 <- 0.5 * log( (1+r1)/(1-r1) ) 
Z2 <- 0.5 * log( (1+r2)/(1-r2) ) 

diff   <- Z1 - Z2 
SEdiff <- sqrt( 1/(n1 - 3) + 1/(n2 - 3) ) 
diff.Z  <- diff/SEdiff 

p <- 2*pnorm( abs(diff.Z), lower=F) 
cat( "Two-tailed p-value", p , "\n" ) 
  } 

For Monday I have r1=0.494 and n1=60, and for Sunday I have r2=0.291 and n2=60. Then diff.corr(r1=0.494,n1=60,r2=0.291,n2=60) which give a two-tailed p-value at 0.19, ie there is no significant difference in the correlations. If the p-value was very low there had been a significant difference between the two correlations. 
Is my answer and argument valid?
 A: In regards to your last sentence, drawing predictive conclusions this way requires more information about the population and the sampling.  Have you established that the results on each day of the week are independent of every other day?  Lets say you have a promotion on Sunday that gives you an incentive to look on Monday instead.  In this case, any increased slope, correlation, etc. observed on Monday would be dependent on the promotion on Sunday.  This implies that we shouldn't assume there is anything more than a correlation here, as the Sunday event could be a confounding variable.  If I'm understanding your question correctly and your day-of-the-week samples are truly independent, you could apply the Fisher r-to-z transformation as shown here to determine the statistical significance of the difference between the two correlations (i.e. what is the probability that they would occur when there is no relationship between the phenomena):  http://vassarstats.net/rdiff.html.
A: If you can live with linear regression instead of Kendall tau, than try Pearson correlation or linear regression. cor.test used for pearson will give you a 95% confidence intervall for the regression coefficient and linear regression will give you confidence intervalls for the coefficients and you could test the significants of dummy variables for the days.
