Construct a Confidence Interval of $95\%$ Based on a random sample of 20 values from a normal distribution with mean $\mu$ and variance $\sigma^{2}$, it was calculated that $\bar{X}=8$ and $s=4$. Provide a $95\%$ confidence interval for the population mean. 
I know that I have to find $\bigl[\bar{X}-E,\bar{X}+E\bigr]$ since the sample is normally distributed. I also know that $E=Z_{c}\cdot\dfrac{\sigma}{\sqrt{n}}$. But after this, I'm not really sure what to do. 
 A: It looks as if the variance is not known: only the sample variance is known. 
Since $20$ is a fairly small number, it is more appropriate to use Student's t-distribution.
A: If the random variable $X$ is normally distributed, then its sample mean $\overline{X}$ is normally distributed. The requirement that $n$ be sufficiently large is only needed for cases where the distribution of $X$ is arbitrary.
Then $T=\frac{\overline{X}-\mu}{s/\sqrt{n}}$ follows the Student distribution with $n-1$ degrees of freedom (http://mathworld.wolfram.com/Studentst-Distribution.html). Because you don't know $\sigma$, you cannot use the normal distribution for setting a random variable $Z =\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}$ which would be normally distributed (and even if $n$ is small, the base random variable $X$'s distribution is normal, so if $\sigma$ was known, it would still work).
So instead of the z-table, you need to look up the t-table. One is given here: http://math.usask.ca/~oshaughn/Student.pdf.
Correspondingly, the formula for $E$ is slightly different from the one where a $Z$ follows a normal distribution. $E=t_{n-1,\alpha/2} \frac{s}{\sqrt{n}}$.
