Grouped monthly sample sizes This will be really simple, but I'm having a hard time figuring out how to Google this.
Let's say we have 12 months of trailing data on something like "Calls per month," and that data is on the low side. We'll say it looks like this:
3, 4, 4, 6, 14, 11, 3, 5, 8, 15, 15, 18

Now to me off the top of my head, this is a sample size of 12. But there are underlying days/hours that make this sample size much larger than it seems, as we could have a calls by hour of year and we'd have a huge sample of mostly 0s. Is the fact that it would be mostly 0s make it go back to conforming around a very small number of points?
Anyways, for this kind of data, what's the best way to try to calculate a mean of the data(+/- from actual mean for something like 95% confidence), and then the standard deviation around the current mean and the 95% confidence actual mean? 
 A: Calculating the mean and the standard deviation will ultimately be a question of what you want to examine. If your goal is to estimate the mean number of calls each month, then your sample should be counts of calls each month.  If you split your data up into calls per day but you aggregate it to estimate the mean number of calls, then you're not actually doing anything.  You can't try to estimate the mean number of calls in a month but divide your data up into calls in a day in order to get a larger sample size.  If your goal is to estimate the mean number of calls in a year, then it sounds like you only have one year of data so you couldn't calculate a confidence interval.
Let's assume that you want to estimate the mean number of calls each month.  The formula for a confidence interval of a mean is given by $\bar{x}\pm t^*_{df} \frac{\sigma_X}{n}$, where $\bar{x}$ is the sample mean, $t^*_{df}$ is the critical $t$-value for your desired level of confidence with your number of degrees of freedom $df$, $s_X$ is the sample standard deviation of your data set, and $n$ is the sample size of your data set. (If $n>30$, people generally use $z$ instead of $t$, but your sample size is too small.)
\begin{eqnarray*}
\bar{x}\pm t^*_{df}\cdot\frac{s_X}{n} &=& \bar{x}\pm t^*_{11}\cdot\frac{s_X}{n} \\
&=& 8.8333\pm 2.20\cdot\frac{5.4745}{12} \\
&=& (7.8296,9.8370)
\end{eqnarray*}
Your 95% confidence interval for the mean number of calls in a month is $(7.8296,9.8370)$, with an estimated mean of $8.8\bar{3}$.
