Considering the given test, the p-value is of 0.1586, and the null hypothesis is not rejected at the 5% significance level.
It depends it is less or greater than the critical value.
In this problem, we use a t-distribution calculator, with t = 1.45, 27 df and a two-tailed test, as we are testing if the mean is different of a value. Doing it, the p-value is of 0.1586.
Since it is greater than 0.05, the null hypothesis is not rejected at the 5% significance level.
More can be learned about p-values at https://brainly.com/question/26454209