2 years ago

ALGEBRA

Mathematics

1 answer:

storchak [24]2 years ago

8 0

Answer:

Check images below.

Step-by-step explanation:

Check images below.

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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 a

goblinko [34]

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, $\bar{x} = 26.6667$ and s = 16.5163. A pivotal quantity for this case is given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ 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 $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ 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 $\alpha = 0.1$ and the confidence interval is $26.6667\pm t_{0.05}(\frac{16.5163}{\sqrt{12}})$ 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 $26.6667\pm (-1.7959)(\frac{16.5163}{\sqrt{12}})$ and the 90% confidence interval is given by (18.1042, 35.2292).