Answer:
Explained below.
Step-by-step explanation:
The information provided is as follows:

(1)
A single mean test is to be performed in this case.
As the population standard deviation is not provided, a one-sample <em>t</em>-test will be used.
The correct option is b.
(2)
The null hypothesis is:
<em>H</em>₀: The average temperature in the population is 98.6°F, i.e. <em>μ </em>= 98.6°F.
The correct option is b.
(3)
The alternative hypothesis is:
<em>Hₐ</em>: The average temperature in the population is less than 98.6°F, i.e. <em>μ </em>< 98.6°F.
The correct option is c.
(4)
The standard deviation of the sample mean is as follows:

Thus, the value of SD is 0.3°F.
(5)
Compute the value of test statistic as follows:


Thus, the value of test statistic is -1.67.
(6)
The degrees of freedom of the test are:
df = n - 1
= 9 - 1
= 8
Thus, the degrees of freedom of the test is 8.