Question

the dean of a university estimates that the mean number of classroom hours per week for full - time faculty is 11.0. as a member of the student council, you want to test this claim. a random sample of the number of classroom hours for eight full - time faculty for one week is shown in the table below. at \\( \alpha=0.05 \\), can you reject the deans claim? complete parts (a) through (d) below. assume the population is normally distributed. 10.4 9.4 13.1 7.2 7.7 9.2 13.1 8.6 (a) write the claim mathematically and identify \\( h_{0} \\) and \\( h_{a} \\). which of the following correctly states \\( h_{0} \\) and \\( h_{a} \\)? a. \\( h_{0}: \mu \leq 11.0 \\) \\( h_{a}: \mu>11.0 \\) b. \\( h_{0}: \mu<11.0 \\) \\( h_{a}: \mu \geq 11.0 \\) c. \\( h_{0}: \mu>11.0 \\) \\( h_{a}: \mu \leq 11.0 \\) d. \\( h_{0}: \mu = 11.0 \\) \\( h_{a}: \mu \
eq 11.0 \\) e. \\( h_{0}: \mu \geq 11.0 \\) \\( h_{a}: \mu<11.0 \\) f. \\( h_{0}: \mu \
eq 11.0 \\) \\( h_{a}: \mu = 11.0 \\) (b) use technology to find the p - value. \\( p= \\) 0.279 (round to three decimal places as needed.)

Explanation:

Step1: State the hypotheses

The dean's claim is that the mean number of classroom hours per week for full - time faculty is \(11.0\). So, the null hypothesis \(H_0:\mu = 11.0\) (the claim) and the alternative hypothesis \(H_a:\mu
eq11.0\) (a two - tailed test).

Step2: Analyze the P - value

Given \(P = 0.279\) and the significance level \(\alpha=0.05\).
Since \(P=0.279>0.05\) (the significance level), we fail to reject the null hypothesis.

Answer:

We fail to reject the null hypothesis \(H_0\). So, at the \(\alpha = 0.05\) significance level, there is not enough evidence to reject the dean's claim that the mean number of classroom hours per week for full - time faculty is \(11.0\).