Testing whether more than 50% of a population is aware of something I have to prove  a hypothesis that more than 50% of a population is  aware of invasive species. I did a survey and found that 83% of the population is aware of invasive species. Now, I have to statistically analyse the results and I am stuck. I have no idea which test to use for this. 
 A: If you know the frequency counts, you could use a one-sample chi-squared test!
(Note: this assumes that you've collected a representative sample)
Imagine for a moment that we collected a sample of 200 and 160 were aware, while 80 were not. Your data could be:
Wariness   Observed    Expected
Aware       160         100
Unaware     40          100
total       200         200
The chi-square statistic is:
$$\chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i}$$
Where O = observed and E = expected. 
Plugging in the numbers, we get a chi-squared value of 72 (in our example). The degrees of freedom will be 1, because we have 2 levels to your categorical variable and in this case df = levels-1. 
Use the degrees of freedom and the chi-squared value to calculate a p-value for significance. 
Using the numbers from our example, you'd say:
In this case we found evidence to reject the null hypothesis that all members were equally aware and unaware. This supports the alternative hypothesis that the population was more aware than unaware. 
$$χ^2(1) = 72.0, p < 0.00001 $$
EDIT
As suggested in comments (thanks Glen_B)!, a binomial test would also yield comparable results. Not Sure what stats packages you're using, but here's the general format in python. 
from scipy import stats
stats.binom_test(x, n, p, alternative)

where: 
x = number of 'successes'
n = number of trials
p = null probability
alternative = specify 'greater' for one-sided or 'two-sided' for two sided.
So in your example, you'd specify:
stats.binom_test(160, 200, 0.5, alternative='greater') 

