# Finding the sample size given the 95% confidence interval bounds, standard error, and sample mean?

I am not given the standard deviation in this problem. I understand that the following holds when the data is normally distributed:

Upper 95% bound $= (\bar{X} + (SE) \cdot 1.96)$

I have all of these values, but the only way I see to relate them to $n$ is through:

$$SE = \frac{\sigma}{\sqrt{n}}$$

but because I dont have the stdev, I don't understand how all of these pieces connect so that I can find $n$.

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As a two sided bound (X¯-(SE)⋅1.96, X¯+(SE)⋅1.96) is a 955 bound. But as a one-sided bound (X¯+(SE)⋅1.96) is a 97.5% bound. –  Michael Chernick Aug 27 '12 at 1:50
Since you don't know the true variance ($\sigma^2$), you can use the sample variance instead:
You now must use the t-distribution confidence interval, which is used when the samples are IID normal with unknown mean and unknown variance. You will have to replace $1.96$ with the corresponding value from a t-distribution table, with $n-1$ degrees of freedom. Here's such a table.
However, I have seen people make a further simplification, and use $1.96$ in place of the proper value from the t-distribution table. This simplification is actually pretty good for large values of $n$.