Margin of Error question from a stats amateur Our organisation is currently putting together a report based on a salary survey of employees within a specific industry (a huge array of questions with varying types of answer). We surveyed 11,000 people, and there are 400,000 people nationwide within this industry.
We're now putting the report together and want to include margin of error in our statistics. I've been charged with doing this, and while I'm mostly confident in super-basic statistical analysis, I just want to check that I'm doing things right. 
As someone who isn't particularly well versed in this side of stats, I defaulted to some online calculators. The first one that came up was from SurveyMonkey, and while I wasn't expecting that to be particularly scientific (just scientific enough), but I was a bit confused by the method. The user-fillable boxes asked me for sample size + population size and the confidence interval I wished to calculate at (95%), and returned a 1% margin of error.
However, below it explains what the formula to calculate margin of error is, and this is where I got lost:
$$z\times\frac{\sigma}{\sqrt{n}}$$
where $n$ = sample size, $σ$ = population standard deviation, and $z$ = z-score.
I hadn't entered standard deviation in the initial box (which is difficult to nail down in our survey due to the varied nature of it, with lots of plain text answers). When I used the standard deviation of salaries (I chose this as it's the key variable in the report, came out as 7) in the above equation, I ended up with a very different margin of error. I tried to do some reading around this and it seems there are couple of different methods to get to margin of error (something that often seems to be completely misinterpreted as a statistic, which must be endlessly frustrating!) but I'll admit that I struggled to get my head around them.
Apologies for the long winded explanation of something that is probably very elementary, but I guess my questions are:


*

*Is there a way to calculate margin of error if you only have sample
size/population size?

*How did SurveyMonkey arrive at the 1% figure?

*Is there a simple way to calculate standard deviation here that I've
completely overlooked?


If it helps, tools-wise I'm using Excel and general internet advice for everything.
Thanks in advance – I feel like there is probably a very simple answer to this problem that is eluding me.
 A: Here are some comments. Other than mentioning 'salaries', you have not
given much information about what kind of information you collected in
your survey. And you said nothing about the purpose of stating margin of 
error.
In statistics a common use of the terminology 'margin of error' has to
to with confidence intervals. A common type of 95% confidence interval
is of the form $\bar X \pm 1.96 S/\sqrt{n},$ where $\bar X$ is the mean
of the sample, $S$ is the standard deviation of the sample, and $n$ is the sample size. The
number 1.96 is used because the interval $(-1.96, 1.96)$ contains 95%
of the probability in a standard normal distribution (or, for very
large sample sizes, also in the relevant t distribution). 
This kind of
confidence interval is made in such a way that it should contain the 
unknown population mean $\mu$ in 95% of the cases where it is used.
If your data are not extremely skewed toward high values (with extremely
high values much more often than extremely low values), then this is
a useful and often used kind of confidence interval. The quantity
$1.96S/\sqrt{n}$ is called the 95% margin of error of such a confidence
interval. I doubt that the readers of your report will be trying to
estimate population means, but this is a very common use of the
term margin of error and it might be good enough for your purposes
to give the sample mean and the 95% margin of error.
If your data are highly skewed (and salaries are often skewed),
of if you are reporting opinions on job satisfaction (perhaps on a scale
from 'very unsatisfied' to 'very satisfied"), then this kind of
margin of of error might not be so useful. Then you could give an
idea of the variability of the data by giving the population median
and the lower and upper quartiles ($Q_1$ and $Q_3,$ respectively). That way readers of the report will
know that 50% of the people surveyed  gave answers below (and above) the
median and that the interval from $Q_1$ to $Q_3$ contains the middle
half of the responses (sorted in order). [You might also use lower or upper deciles to
give an interval that contains 80% of the responses.] 
Note: You should not
give the maximum or minimum values, because you have probably promised
not to reveal individual information. And if your survey contained
a famous industry leader, giving the maximum salary could reveal the
reported salary of such a person.
There are other possibilities, but I have seen both the 95% margin of error
and  median-quartiles methods used to report variability in industry surveys. If you can tell us
more about your data and objectives, maybe one of us can give you
more specifically targeted suggestions. 

Example: Below is a histogram of 1000 right-skewed salaries in thousands of dollars (simulated, but roughly modeled after real data) with $\bar X \approx 75$ and
$S/\sqrt{n} \approx 50.$ (For a larger sample from a similar population, the margin of error would
typically be smaller.) Vertical red lines are at $Q_1,$ Median, and $Q_3,$
respectively.

