How does one find sample size without a given variance?

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Eclair_de_XII
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Homework Statement


"The mean pH value of a certain chemical is to be controlled at ##\mu = 5##. Deviation from this target value in either direction is to be detected with high probability. For this purpose it is proposed to measure a ceratin number of samples from each batch and decide that the mean pH is different from 5 if the sample mean differs significantly from 5 at the 10% level of significance."
(b) "What sample size is needed if the probability of not detecting a change of one standard deviation is to be no more than 1%?"

Homework Equations


##H_0: \mu = 5##
##H_1: \mu ≠ 5##

The Attempt at a Solution


Basically, I'm interpreting this probability as the probability of a type-II error. So ##\beta = 0.01##. I know that the formula for the required sample size in this case would be: ##n=(\frac{\sigma}{E})^2##, but I don't know what ##\sigma## is. For that matter, I don't know what ##E## is supposed to be, either. So I'm kind of stuck, here.
 
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mjc123 said:
What is the relationship between the variance of a sample and the variance of the mean?

Let's see... The variance of a sample is: ##s^2=\frac{1}{n-1}∑(\bar X - X_i)^2## and the variance of the mean, was ##Var(\bar X)=\frac{Var(X)}{\sqrt{n}}##? I'll get back to you on the matter of their relationship.

Stephen Tashi said:
Is there a part (a)?

(a) "State the null hypothesis and the alternate hypothesis. What can you infer about the statistic ##\mu##?"
I didn't think it was related to the problem, so I decided to exclude it.
 
So what I'm gathering is to find an ##n## such that ##T=\frac{\mu - \bar x}{s/\sqrt{n}}=t_{n-1,\alpha /2}##.

In any case, though, my instructor told me to use ##-z_{\alpha /2}+\frac{\delta \sqrt{n}}{\sigma}=z_{\beta}##, and by setting ##\delta=\sigma##, I get ##n=(z_{\alpha /2}+z_{\beta})^2##.