In a random sample of three pupils, $x_{i}$ is the mark of the ith pupil in a test on volcanoes and $y_{i}$ is the mark of the ith pupil in a test on glaciers. All three pupils sit both tests.
It's believed that the difference between the results in the two tests follows a normal distribution with variance 16 marks. If the mean mark of the volcano test was 23 and the mean mark of the glacier test was 30, find a 95% confidence interval for the improvement in marks from the volcano test to the glacier test.
I got that:
$$(Y-X)^- \pm t \biggl(\frac{s_{n-1}}{\sqrt{n}}\biggr) = (30-23) \pm 4.303 \biggl(\frac{\sqrt{\frac{3}{2}(16)}}{\sqrt{3}}\biggr)= 7 \pm 12.171.$$
where t is the t-score associated with a 95% confidence interval for v=2.
Thus, I got the confidence interval [-5.171, 19.171]
But the answer key says the correct answer is [-11.1, 25.1].
Any ideas where I went wrong?
Because you are told that the population variance of differences in scores is $\sigma^2 = 16,$ this should be a paired z-interval, not a t-interval.
You have $\bar D = \bar V - \bar G = 7,\, \sigma = 4,\, n = 3,$ so that the interval is $7 \pm 1.96(4/\sqrt{3}).$
If an answer key is wrong, that would hardly be the first time ever, but it does warrant a second look. First, please check that you have posted the exact wording of the Question; second, please make sure you have looked up the answer key correctly. [Technically, if the variance of the difference in test scores is $\sigma^2=16$ marks$^2$; then the standard deviation would be $\sigma = 4$ marks. I mention this inconsistency because $7 \pm 1.96(16/\sqrt{3})$ computes to $( -11.1057, 25.1057).$ Is it possible you typed 'variance' instead of 'standard deviation'?]