# Size of sample for mean

I am trying to calculate the sample size, in order to calculate the mean of my experiment. For that I read a book and here say that it is can be calculated using the next formula:

$$n=\dfrac{\sigma ^2z_{y}^2}{\epsilon^2},$$

where $\sigma^2$ is the standard deviation, $\epsilon$ is sample error and $z_{y}$ is calculated from $y$ using the normal distribution table. Also, $\sigma^2$ can be calculated using a pilot sample.

It is expected that my experiment has a log-normal distribution.

Thus, if the random variable $X$ is log-normally distributed, then $Y = \ln ⁡ ( X )$ has a normal distribution

Then my question is: For my experiment, It is necessary to use a log-normal (to calculate the mean) distribution table or it can be calculated using the normal distribution table?

EDIT

The data of my experiments are running times of a algorithm A, and a book say that that a exp of the sample, always, has log-normal distribution. I need to known How many times I need to run A (with certain input) in order to get a aproximation for the mean. I do not if in this case it is important to get the normal mean or the log normal mean

Often when data $X$ are lognormal so that $Y = \ln(X) \sim Norm(\mu, \sigma)$ it is the normal mean $\mu$ that is to be estimated. If that is the case, then take natural logs of your data, and use $\bar Y \pm t^*S_Y/\sqrt{n}$ to get a confidence interval (CI) for $\mu,$ where $S_Y$ is the sample standard deviation of the Y's. Or if you know the normal $\sigma$ from a pilot experiment then use $\bar Y \pm z^*\sigma/\sqrt{n}.$
You could get $t^*$ from a table of Student's t distribution. For example if $n = 20$ and you want a 95% CI use $df = n - 1 = 19$ to obtain $t^* = 2.093.$ In the case where $\sigma$ is known from a pilot experiment and you want a 95% CI, then $z^* = 1.96$ (from a standard normal CDF table).
The quantity $z_y$ in your equation is my $z^*.$ For a 95% CI, if $n > 30$ then $t^* \approx z^*.$ The width of the CI is $2\epsilon$ (twice the Margin of Error).
If you need to estimate the mean $E(X) = \exp(\mu + \sigma^2/2)$ of the lognormal distribution, then the problem is a little more complicated. See Wikipedia on 'lognormal distribution'.