Answer:
The p-value is 0.175
Step-by-step explanation:
We have the null hypothesis [tex]H_{0}: \mu = 18[/tex] and the alternative hypothesis [tex]H_{1}: \mu \neq 18[/tex] (two-tailed alternative). Because the sample size is n = 17 and the test statistic is t=-1.421, we know that this last value comes from a t distribution with n-1=17-1=16 degrees of freedom. Therefore, the p-value is given by 2P(T < -1.421) because the p-value is the probability of getting a value as extreme as the observed value and because of the simmetry of the t distribution. Here, T has a t distribution with 16 df and we are using the t distribution because the sample size is small. So, 2P(T < -1.421) = 0.1745
