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
\\( h_{a}: \mu>11.0 \\) \\( h_{a}: \mu \geq 11.0 \\) \\( h_{a}: \mu \leq 11.0 \\)
d. \\( h_{0}: \mu=11.0 \\) oe \\( h_{0}: \mu \geq 11.0 \\) of. \\( h_{0}: \mu \
eq 11.0 \\)
\\( h_{a}: \mu \
eq 11.0 \\) \\( h_{a}: \mu<11.0 \\) \\( h_{a}: \mu=11.0 \\)
(b) use technology to find the p - value.
\\( p = 0.186 \\) (round to three decimal places as needed.)
(c) decide whether to reject or fail to reject the null hypothesis.
which of the following is correct?
a. reject \\( h_{0} \\) because the p - value is less than the significance level.
b. reject \\( h_{0} \\) because the p - value is greater than the significance level.
c. fail to reject \\( h_{0} \\) because the p - value is less than the significance level.
d. fail to reject \\( h_{0} \\) because the p - value is greater than the significance level.

Explanation:

Step1: Recall the decision rule for hypothesis testing

If \(P - value<\alpha\), reject \(H_0\). If \(P - value\geq\alpha\), fail to reject \(H_0\). Here, \(\alpha = 0.05\) and \(P=0.186\).

Step2: Compare \(P\) - value and \(\alpha\)

Since \(0.186>0.05\) (i.e., \(P - value>\alpha\)).

Answer:

D. Fail to reject \(H_0\) because the \(P\) - value is greater than the significance level.