Answer:
(18.1042, 35.2292) is a 90% confidence interval for the mean number of journal articles read monthly by professors.
Step-by-step explanation:
We have a small sample of size n = 12, [tex]\bar{x} = 26.6667[/tex] and s = 16.5163. A pivotal quantity for this case is given by [tex]T = \frac{\bar{X}-\mu}{S/\sqrt{n}}[/tex] which has a t distribution with n-1 degrees of freedom if we suppose that we take the sample from the normal population. 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 = 12 - 1 = 11 degrees of freedom. As we want the 90% confidence interval, we have that [tex]\alpha = 0.1[/tex] and the confidence interval is [tex]26.6667\pm t_{0.05}(\frac{16.5163}{\sqrt{12}})[/tex] where [tex]t_{0.05}[/tex] is the 5th quantile of the t distribution with 11 df, i.e., [tex]t_{0.05} = -1.7959[/tex]. Then, we have [tex]26.6667\pm (-1.7959)(\frac{16.5163}{\sqrt{12}})[/tex] and the 90% confidence interval is given by (18.1042, 35.2292).