Answer:
(9.6, 25.7) is a 80% confidence interval for the average net change in a student's score after completing the course.
Step-by-step explanation:
We have n = 6, [tex]\bar{x} = 17.6667[/tex] and s = 13.3367. The confidence interval is given by
[tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the t distribution with n-1=5 degrees of freedom. As we want the 80% confidence interval, we have that [tex]\alpha = 0.2[/tex] and the confidence interval is [tex]17.6667\pm t_{0.1}(\frac{13.3367}{\sqrt{6}})[/tex] where [tex]t_{0.1}[/tex] is the 10th quantile of the t distribution with 5 df, i.e., [tex]t_{0.1} = -1.4759[/tex]. Then, we have [tex]17.6667\pm (1.4759)(\frac{13.3367}{\sqrt{6}})[/tex] and the 80% confidence interval is given by (9.6, 25.7)
