# Confidence interval for difference in means

I need to obtain a $95$-$\%$ confidence interval for the indifference in the mean score overall.

I have the following data which states subject ($A$-$L$) and then Test and Retest $$\begin{matrix} A & B & C & D & E & F & G & H & I & J & K & L \\ 93 & 89 & 84 & 85 & 90 & 99 & 97 & 96 & 91 & 88 & 79 & 85 \\ 91 & 89 & 78 & 82 & 97 & 98 & 95 & 96 & 88 & 90 & 81 & 85 \end{matrix}$$ I understand the formula for a confidence interval is $$\bar{x} \pm Z \left( 1-\dfrac{a}{2} \right) \cdot \dfrac{s}{\sqrt{n}}$$ where $s$ is the standard deviation and $Z \left( 1-\dfrac{a}{2} \right) = 1.96$ due to $100(1-a) = 95\%$, so $a = 0.05$ and $Z(0.975)$.

I might be being really silly, but I dont get what mean they are asking for? Are they asking for the mean of Test and Retest? And then, how do I work out a confidence interval of the difference? Thanks.

Since the sample size is only 12, I would not use $1.96$, but rather I would use a corresponding number for the t-distribution with $12-1=11$ degrees of freedom.