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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

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Answer:

(18.1042, 35.2292) is a 90% confidence interval for the mean number of journal articles read monthly by professors.

Explanation:

We have a small sample of size n = 12,
\bar{x} =  26.6667 and s = 16.5163. A pivotal quantity for this case is given by
T = \frac{\bar{X}-\mu}{S/√(n)} 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
\bar{x}\pm t_(\alpha/2)((s)/(√(n))) where
t_(\alpha/2) is the
\alpha/2th quantile of the t distribution with n - 1 = 12 - 1 = 11 degrees of freedom. As we want the 90% confidence interval, we have that
\alpha = 0.1 and the confidence interval is
26.6667\pm t_(0.05)((16.5163)/(√(12))) where
t_(0.05) is the 5th quantile of the t distribution with 11 df, i.e.,
t_(0.05) = -1.7959. Then, we have
26.6667\pm (-1.7959)((16.5163)/(√(12))) and the 90% confidence interval is given by (18.1042, 35.2292).

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