MATH SOLVE

3 months ago

Q:
# Part of a university's professor's job is to publish his or her research. This task often entails reading a variety of journal articles to keep up to date. To help determine faculty standards, a dean of a business school surveyed a random sample of 12 professors across the country and asked them to count the number of journal articles they read in a typical month. These data are listed here. Estimate with 90% confidence the mean number of journal articles read monthly by professors.9 17 4 23 56 30 41 45 21 10 44 20

Accepted Solution

A:

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} = Β 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).