I was solving this problem from a textbook
In a random sample of three pupils, $x_i$ is the mark of the $i$th pupil in a test on volcanoes and $y_i$ is the mark of the $i$th pupil in a test on glaciers. All three pupils sit both tests.
It is believed that the difference between the results in these two tests follows a normal distribution with variance $16$ marks. If the mean mark on the volcano test was $23$ and the man mark for the glacier test was $30$, find a $95\%$ confidence interval for the improvement in marks from the volcano test to the glacier test.
My attempt was, letting the desired distribution be $\overline{Y-X}$. We know that the mean is $\overline{y-x} = \bar{y} - \bar{x} = 30-23= 7$. Standard deviation is given as $16$, $\Phi^{-1}(0.975) = 1.959$, and number of samples is $n=3$. Hence average improvement, $\mu$, would lie in the interval, $$ \bar{x} - 1.959 \frac{4}{\sqrt{3}} < \mu < \bar{x} + 1.959 \frac{4}{\sqrt{3}}$$ giving our confidence interval $[ 2.478, 11.524 ]$.
However, the actual interval from the textbook is $ [ -11.1, 25.1 ] $ May someone explain why? Thank you so much!
In your confidence interval, you are using $\bar x = 7$, which is inconsistent with your earlier notation. It should be $\bar d = \bar y - \bar x= 7.$ Also, to be clear, 16 is the variance for the difference in scores for one individual, so the variance for $\bar d$ is $16/3$ and the standard deviation used in the confidence interval is $4/\sqrt{3},$ as you have it.
With those assumptions, your confidence interval is correct. I cannot see any path (correct or making 'slight' errors) that leads to the interval given by the textbook. (As a textbook author, I am perhaps somewhat more willing than most students to believe that wrong answers can get into print.)