Answer:
0.0263
Step-by-step explanation:
We have the null hypothesis [tex]H_{0}: \mu = 98.6[/tex] vs the alternative hypothesis [tex]\mu > 98.6[/tex] (upper-tail alternative). Besides, we have a small sample size n = 6, therefore, if we suppose that [tex]H_{0}[/tex] is true, then, the observed value t = 2.528 comes from a Student's t-distribution with n-1 = 6 - 1 = 5 degrees of freedom. Therefore, the p-value is computed as P(T > 2.528) = 0.0263