Statistics A Level Errors in Hypothesis Testing In a large company the time taken for an employee to carry out a certain task is a normally distributed
Random variable with mean $78.0s$ and unknown variance. A new training scheme is introduced and after its introduction the times taken by a random sample of $120$ employees are recorded. The mean time for the sample is $76.4s$ and an unbiased estimate of the population variance is $68.9s$.
(i) Test, at the $1$% significance level, whether the mean time taken for the task has changed.
(ii) It is required to redesign the test so that the probability of making a Type I error is less than $0.01$ when the sample mean is $77.0s$. Calculate an estimate of the smallest sample size needed, and
explain why your answer is only an estimate. 
This question is taken from a past A level stats $2$ paper. I'm having trouble understanding part ii) of this (part i included in case it's relevant). The mark scheme says to standardise $1$ with $n$ and $2.576$, which I understand is the critical level of the test, but I really don't understand what's going on here so any explanation would be greatly appreciated!
 A: For part (ii), you are being asked to provide the minimum sample size where a difference of only one in the mean is significant at the $99\%$ level. The probability of making a type I error is $1\%$ at the $99\%$ level so this is just an indirect way of saying "the $99\%$ level". In other regards it's kind of backwards because you don't know the mean you are dealing with until you've calculated it with a known $n$.
For both $t$ and $Z$ equations, the only difference is in the standard deviation, sample versus population. $$t(Z) = \frac{\bar x - \mu_0}{s(\sigma)/\sqrt{n}}=\frac{1\sqrt n}{\sqrt{68.9}} = \frac{\sqrt n}{8.3}$$
Looking at a $t$ table, the lowest value in the column at the $99\%$ significance level is $2.576$ to which they reference a $Z$ for the $df$, meaning $Z = 2.576$ at the $99\%$ level. $$2.576 = \frac{\sqrt n}{8.3}$$
$$n = (2.576\cdot 8.3)^2 = 457.14$$
The reason this is only an estimate is because we used the sample standard deviation instead of the population standard deviation for a $Z$ score calculation of n.
