Calculating confidence interval - formula I have the following problem that I get the feeling I'm mixing formulas.
A sample of 14 specimens of a particular type gave a sample mean of 8.48 and a sample standard deviation of 0.79.
Calculate and interpret a 95% confidence bounds for the true average proportional limit stress of all such joints. 
What, if any, assumptions did you make about the distribution of proportional limit stress?

I'm using the following formula:
$mean \stackrel{+}{-} z \times \frac{std-dev}{\sqrt{n}}$
Using the formula I'm getting the following to be my answer:
$8.48 \stackrel{+}{-} 1.96 \times \frac{0.79}{\sqrt{14}} $
Now what's confusing me is that I know there exists a similar formula for when comparing two proportions.
$ (p1 - p2) \stackrel{+}{-} z \times \sqrt{\frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2}}$
If I were to use just p1 by itself would the formula work the same as the formula I'm using for the problem above or am I just mixing things together?
 A: Another approach is a t-distribution 
$$\overline{X} + t_{\alpha, n-1} (\frac{S_x}{\sqrt(n)}) $$
The t Critical Value can be retrieved from a Critical Values for t Distributions chart or calculated either with Excel or a TI calculator. The critical t value will be negative. On my TI-89 it is found under the Stats/List Editor > F5 > Inverse > Inverse t. Enter the appropriate area and n-1 Degrees of Freedom. The rest is simple "plug-and-play".
A: The second formula is for comparing two different distributions. This is for example if you were given a problem like:

A sample of 14 specimen had a mean of 1337... etc. Another sample had a mean of 12... etc. Give a 95% confidence interval of the difference between the two means.

So then you'd have:
$$H_0:\mu_a=\mu_b$$
$$H_a:\mu_a \ne \mu_b$$
Or whatever operations other than not equal. But you'd have 2 different distributions and that formula would give you the 95% confidence interval for the mean difference of the means of the two distributions.
For this problem, you only have 1 distribution so you have the formula right:
$$\bar{x}\pm z^*\frac{s}{\sqrt n}$$
Be careful not to mixed matched pairs and 2-samples, though. Make sure you actually have two different distributions of different # of tests or treatments before you use the 2-sample equations.
