Using a z-interval assuming the population variance is known.

Computing two separate confidence intervals for pre-test and post-test scores and then taking their difference.

Ignoring the paired nature of the data and treating the two tests as independent samples.

Using data from 30 students in an SAT preparation class where the improvement (post-test minus pre-test) has a mean of $$42.25$$ and a sample standard deviation of $$27.92$$ to construct a 90% confidence interval of approximately $$ (33.59, 50.91)$$.