I'm a bit confused by your question but if I'm understanding it right:
You have two assumptions here: First you have a normal population or large sample and second, the population deviation is known.
State your null and alternative hypothesis.
H0: mu is equal to 20%
HA: mu is not equal to 20%
Because you are dealing with the alternative hypothesis that is not equal to mu naught, you will have a two tailed test with the Reject H0 in the two tails and the Do Not Reject H0 in the middle of the curve.
Now, you compute the value of the test statistic:
z = (sample mean - mu naught)/[(population deviation)(sample size)(1/2)]
I assume you are using a significance level of 5% but since you are dealing with a two tailed test, divide that area by two (in this case, 0.05/2). Essentially, this is the area you will be looking for in the table of z scores in the appendix of any Statistics book.
I don't have a book with me at the moment but look up the z score that corresonds to your significance level. If the test statistic falls in the Do Not Reject H0, obviously, don't reject H0. Likewise, if it falls in the reject zone, reject H0.
Hope this helps.
A bit of an embellishment to my earlier post:
The z score corresponding to your significance level is the numerical "mark" of where the Do not reject/ reject regions are.