One sample T test or Z test? Given is an internet provider stated that the expected speed is at least 90 Mbit/s and the standard deviation is at most 5 Mbit/s. How there is a sample given with a sample mean of 88.36 and sample standard deviation of 6.85.
The question being asked is a CI and a test to prove the claim on the expected speed. I used the One sample T-test for this because I thought the stated standard deviation doesn't have to be the true standard deviation. So I used the sample standard deviation as estimator so it became a T-test. Are my thoughts correct or should I have used the stated standard deviation and have used Z test? 
Note: there is another question where you test the stated standard deviation, so I thought that's why they give the standard deviation.
 A: Yes, what you did is correct, or, I guess that is what the authors of the exercise want you to do. You are supposed to assess the claims of expected speed and SD, and I see no reason to trust any claims internet service providers. So plugging the sample mean and sample SD into the one sample t-test looks correct, because we do not know the true SD nor the true mean.
However, if the exercise really says that you are supposed to „prove“ the claim of „at least“ 90mbit, well... you can‘t really, not like this. The claim is that the mean speed equals or exceeds 90mbit, so you are in part testing for equivalence. With conventional significance testing, you cannot test for equivalence (the absence of a difference). Formally (disregarding the possibility of the speed being higher), the hypotheses are:
H0: Measured speed is the same as claimed speed. 
H1: Measured speed is lower than claimed speed.

If the measured speed is significantly below 90, you can of course reject H0 (the speed claim). If, however, measured speed is not significantly below (or above) 90, you cannot give a definitive answer, because you cannot accept the H0 on the basis of significance tests without adding more assumptions. This reflects the nature of significance testing, and this is why I believe that the exercise is either flawed or more tricky than you thought. Consider the following example with confidence intervals:
If your 95% confidence interval is [91,96], you can say that the speed is significantly higher than 90. If the interval is [82,86] you can say that the speed is significantly lower than 90. However, if the confidence interval is [87, 92], meaning that there is no significant difference, you cannot say that the value is "significantly equal to 90", which means that we still do not know if the speed claim is correct or not. For that you would need to specify a value that counts as the lowest acceptable value and see whether the speed is significantly above it. See the following graph from from https://en.wikipedia.org/wiki/Equivalence_test, which sums it up nicely.

