I believe that your problem is simply the testing of two proportions.

You can first notice that there are two populations in your problem. The silk and non-silk populations. In this setting the question you are asking is if the proportion of green items in the silk population is significantly smaller than the proportion of green items in the non-silk population.

There is a simple answer to this question if you can assume that the observations are independent of one-another. If you are not able to ad-hoc impose the independence condition there are tests for it: https://en.wikipedia.org/wiki/Pearson%27s_chi-squared_test

Let's assume that the silk population is $\{X_1,...,X_n\}$ and that the non-silk population is $\{Y_1,...,Y_m\}$. In this notion I mean that $X_i = 1$ if the i-th observation in the silk population is green and $X_i = 0$ otherwise. The same rule applies to the $Y$'s. In this formulation the $X$'s are independent Bernoulli random variables with mean $p_X$ and the $Y$'s are also independent Bernoulli random variables with mean $p_Y$. The statistical test you want to perform is:

$$H_0: p_Y \leq p_X$$
$$H_1: p_Y > p_X$$

A test for those statistical hypothesis can be performed as follows. The proportion of green and silk items is $\bar X =\frac{\sum_{i = 1}^{n}X_i}{n}$ and the proportion of green and non-silk items is $\bar Y =\frac{\sum_{i = 1}^{m}Y_i}{m}$. Than under fairly simple conditions,

$$T = \frac{\bar Y - \bar X}{\sqrt{p^*(1-p^*)(\frac{1}{n}+\frac{1}{m})}}$$

where $p^* = \frac{\sum_{i-1}^nX_i + \sum_{i-1}^mY_i}{n+m}$, has approximately a normal distribution with mean 0 and variance 1.

You can reject $H_0$ if $T$ is greater than the critical value related to the level of significance of your choice.

This http://www.dummies.com/how-to/content/how-to-compare-two-population-proportions.html might be useful.