Statistical insignificance I am attempting to show the correlation between two features of a dataset is weak. I am under the impression I should be using a t test for this to measure statistical power but would like to know how many samples I would need to show that the correlation is less than x%. Am I to understand that I would need many samples in order to show a weak correlation between two features or is there a different method which I could use in order to show that a strong correlation does not exist? If there is another method, how might I determine how many samples would be needed to be statistically certain that say, only a 30% or less correlation exists?
I am most familiar with R and MATLAB.
 A: Links to relevant background and a caveat
Your question combines three fairly distinct subjects: estimating a correlation coefficent, equivalence testing, and sample size determination. Good info on estimating a correlation coefficient can be found in Wikipedia's article on the subject. Equivalence testing is relevant because you'd like to positively demonstrate that the correlation is weak (that is, equivalent to zero in practical terms), and not merely demonstrate an absence of statistical significance since that doesn't warrant any strong conclusion. Good info on equivalence testing can found found in this article. The rest of this answer takes the information in these links as given.
But first, a caveat: the following calculation assumes that your data follows a bivariate normal distribution. If it doesn't, I simply have no idea whether or not the confidence intervals I discuss below are even approximately valid.
Sample size determination
To discuss the sample size determination problem, we need to fix some notation: let $\rho$ be an assumed true value for the correlation; let $r$ be the sample correlation; let the confidence level be $1-\alpha$; and let $\rho_w$ be a positive number such that $\pm \rho_w$ defines a zone of weak correlation in practical terms. For brevity, Fisher-transformed quantities will be indicated by tildes, e.g., $\tilde{\rho}=\text{artanh}(\rho)$. I will write equations as if the Fisher transformation yields exact normality of $\tilde{r}$, which it does not. 
In my view, there are two issues that need consideration when deciding on sample size: first, the length of the confidence interval, and second, the probability of observing an interval within the zone of weak correlation. (Per the article on equivalence testing linked above, I'll leave the issue of statistical significance aside.)
Confidence interval length
In Fisher-transformed parameter space, the confidence interval is $\tilde{r} \pm \frac{z_{\alpha/2}}{\sqrt{n-3}}$ where $z_p = \Phi^{-1}(p)$, $\Phi^{-1}(\cdot)$ is the quantile function of the standard normal distribution, and $n$ is sample size. Hence the length of the confidence interval is $\frac{2z_{1-(\alpha/2)}}{\sqrt{n-3}}.$ The inverse of the Fisher transformation shrinks intervals, so this is an upper bound on the length of the interval in the space you actually care about. But the Fisher transformation is essentially equal to identity in the range $\pm 0.5$, which is where you expect your correlations to be, so it should be a tight upper bound. The following R code shows how to plot the interval length (in Fisher-transformed parameter space) as a function of confidence level and sample size:
interval_length <- function(conf_level, n, alpha = 1 - conf_level)
  (2*qnorm(1-(alpha/2)))/sqrt(n-3)
curve(interval_length(0.95, x), from = 4, to = 150, ylim = c(0,2))

Probability of confidence interval inclusion in the zone of weak correlation
The confidence interval will lie within the endpoints of the zone of weak correlation if $\tilde{r}$ is in the interval $\pm(\tilde{\rho}_{w}+\frac{z_{\alpha/2}}{\sqrt{n-3}})$, provided $\tilde{\rho}_w + \frac{z_{\alpha/2}}{\sqrt{n-3}} > 0$. If that inequality is not satisfied, the confidence interval is wider than the zone of weak correlation. Assuming that it's possible for the confidence interval to fit into the zone of weak correlation, the probability that it does so is,
$\Pr\left(-\tilde{\rho}_{w}-\frac{z_{\alpha/2}}{\sqrt{n-3}}<\tilde{r}\le\tilde{\rho}_{w}+\frac{z_{\alpha/2}}{\sqrt{n-3}}\right)=\Pr\left(\left(-\tilde{\rho}_{w}-\tilde{\rho}\right)\sqrt{n-3}-z_{\alpha/2}<\left(\tilde{r}-\tilde{\rho}\right)\sqrt{n-3}\le\left(\tilde{\rho}_{w}-\tilde{\rho}\right)\sqrt{n-3}+z_{\alpha/2}\right),$
$\Pr\left(-\tilde{\rho}_{w}-\frac{z_{\alpha/2}}{\sqrt{n-3}}<\tilde{r}\le\tilde{\rho}_{w}+\frac{z_{\alpha/2}}{\sqrt{n-3}}\right)=\Phi\left(\left(\tilde{\rho}_{w}-\tilde{\rho}\right)\sqrt{n-3}+z_{\alpha/2}\right)-\Phi\left(\left(-\tilde{\rho}_{w}-\tilde{\rho}\right)\sqrt{n-3}-z_{\alpha/2}\right),$
in which $\Phi(\cdot)$ is the cdf of the standard normal distribution. R code:
prob_inclusion <- function(conf_level, n, rho, rho_w,  alpha = 1 - conf_level) 
  max(0, diff(pnorm(c(-1,1)*(atanh(rho_w) + qnorm(alpha/2)/sqrt(n-3)), mean = atanh(rho), sd = 1/sqrt(n-3))))
prob_inclusion_vectorized <- Vectorize(prob_inclusion, "n")
curve(prob_inclusion_vectorized(.95,x,0.1,0.3), from = 4, to = 150)

Concluding remarks
As you will see from the plots, large sample sizes are indeed required. Convincingly demonstrating weak correlation is inherently a difficult problem in the sense that estimates of weak correlations are the ones that are most subject to random variation. Consequently, confidence intervals for the correlation parameters are widest when the true correlations are weak. 
