So the question I'm trying to answer looks like this:
We are interested in p, the population proportion of all people who are currently happy with their cell phone plans. In a small study done in 2012, it was found that in a sample of 150 people, there were 90 who were happy with their cell phone plans.
Find the upper confidence limit of an 82% confidence interval for p.
I know the formula is
$$ estimate \pm (critical value)(standard error) $$
But I'm not sure how to calculate for standard deviation in standard error ($\sigma$) or estimate ($\overline{x}$)
The mean of a distribution of sample proportions is equal to the population proportion (p). Thus, the standard deviation of this sample proportion can be found by using the standard error. This formula is Sqrt[(p*(1-p)/n)].. You can solve for the standard error now. You have the population proportion estimate which is just 90/150 or .6, can you handle it from there?? The critical value can be found by using a table.
In addition, you might need to use a "continuity correction" because we need to correct for the fact that we are approximating a discrete distribution (the sampling distribution of p) with a continuous distribution (the normal distribution)... This continuity correction is given by .5/N, thus your final equation for the upper limit will look something like:
.60 + (critical value)(.04) + .5/150
Alternatively, The lower limit would simply be:
.60 - (critical value)(.04) - .5/150