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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.

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If these are paired, i.e. the subject scoring 93 on the test is the SAME subject that scored 91 on the retest, etc., then you need the 12 differences between the test score and the retest score, and you base the confidence interval on that set of 12 numbers. (On the other hand, if the second set of 12 scores were independent of the first, rather than being paired, one would do this quite differently.)

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.

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