I just finished this unit in stats. The premise of a 95% confidence interval is that we are 95% confident that the true proportion lies between two values. In this case, one must find what the proportion 120/600 in the given sample can tell us about the true proportion of the population who is in favor of raising taxes.
First, calculate the decimal that 120/600 represents. Then you have to find margin of error, given by the formula:
z(a/2) is the critical value that separates an area of a/2 in the right tail of the standard normal distribution. a=1-confidence level (.95)
n is the sample size, in this case 600
p̂ is the sample proportion (expressed as a decimal), in this case 120/600
q̂ is simply 1-p̂, in this case 480/600
In this case, it is simply known from the critical value [z(a/2)] is 1.96. This can be proven with a graphing calculator or some other technology.
So your confidence interval is shown by p̂-E<p<p̂+E. In words, the true population proportion should lie between 1 margin of error in either direction of your sample proportion. Calculate the confidence interval. If it does capture the true value p=.35, then the sample is a good representation of the population. If not, then it isn't.
Remember to convert your percentages to decimals, and good luck!