To answer the first part of your question, i.e. what you are doing wrong. In your example, you are testing the following hypotheses:
$H_0: \mu_s \le \mu$ and $H_1: \mu_s > \mu$ where $\mu = 2.366$
In other words, you are trying to accept the alternative hypothesis that states that the true mean is greater than 2.366... but you failed to reject the null hypothesis (which states that the true mean is less or equal to 2.366)!
EDIT FOR FUTURE READERS INTERESTED BY THIS QUESTION
Comments on (accepted) answer proposed by @user334895: testing variance does not lead to equivalence. The collected information could have a different mean but a similar variance and, therefore, lead to an incorrect conclusion.
What is implied in this answer is that if the mean of the obtained sample is very close to the population mean, then we only need to test the variance of the sample to conclude to equivalence. The problem is that it is actually a mix of descriptive statistics (use of the sample mean and assume it is equal the true population mean) and inferential statistics (testing variance to draw conclusion about the population variance).
However, the obtained sample is one possible out of many others, so there is no certainty about its mean. Based on the sample we have, we can empirically illustrate/simulate this sampling distribution of the sample means with bootstrap:
A mean of 2.6 is also possible for example, so the variance test is no longer valid in this scenario because even if variance matches the target, the data collected will differ from the population mean.
An approach to solve this problem would be to run a TOST (Two-One-Sided T-test) procedure (equivalence testing). It is already well documented on this site https://stats.stackexchange.com/a/500121/321901.