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Suppose we were given a sample with mean of $\overline{x} = 5$ and a standard deviation of $s = 2$, and asked to estimate the mean $\mu$.
Certainly, the best point estimate of $\mu$ is $\overline{x}$. However, if we want a confidence interval for $\mu$, we will need to consider the spread of the distribution of sample means.
Recall the Central Limit Theorem tells us that $$\mu_{\overline{x}} = \mu \quad \textrm{ and } \quad \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$$
As such, under an assumption that the distribution of sampling means is approximately normal, we can construct a confidence interval for $\mu$ with confidence level of $(1 - \alpha)$ between $$\overline{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$ This begs two questions:
Supposing that $\sigma$ is known for the moment, recall the Central Limit Theorem tells us that the distribution of sample means gets more and more normal as the sample size increases.
Thus, if $n$ is sufficiently large (let's say $n \ge 30$), then we can rely on the fact that the distribution of sample means should be approximately normal.
If $n \lt 30$, then knowing that the original distribution was approximately normal should be enough to assure us that the distribution of sample means is normal.
If $\sigma$ is unknown, however, we are forced to approximate its value with the sample standard deviation, $s$. This in turn, causes the nature of the associated distribution to change from a standard normal distribution to a $t$-distribution.
Thus, if $\sigma$ is unknown, and $n \ge 30$ or the original population is approximately normal, the confidence interval for the mean with confidence level of $1-\alpha$ falls between $$\overline{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$$ where $t_{\alpha/2}$ is taken from a $t$-distribution with $n-1$ degrees of freedom.
Just like with proportions, one can assure a certain margin of error by choosing a large enough sample size. To determine what that sample size might be, simply solve for $n$ in the formula for $E$ used -- always rounding up to ensure $n$ is large enough.
So, in the case of $\displaystyle{E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}}$, we would have $$n = \left( \frac{z_{\alpha/2} \sigma}{E} \right)^2$$ Of course, calculating this $n$ requires knowledge of $\sigma$. We can use the sample standard deviation from a previous study if it is available. Also, as a rule of thumb, $$\sigma \approx \frac{\textrm{range}}{4}$$ Last, but not least, one should also consider the return rate when designing a study -- it can be very difficult to get a 100% return rate, so without a little "padding" one might not get the margin of error one desires, even when everything else was done perfectly.
Many of the same considerations made in the construction of confidence intervals for means apply in hypothesis tests for means:
If $\sigma$ is known and $n \ge 30$ or the distribution is approximately normal, then we can use the standard normal distribution and a test statistic of $$z = \frac{\overline{x} - \mu}{\displaystyle{\frac{\sigma}{\sqrt{n}}}}$$
If $\sigma$ is unknown and $n \ge 30$ or the distribution is approximately normal, then we can use a $t$-distribution with $n-1$ degrees of freedom and a test statistic of $$t = \frac{\overline{x} - \mu}{\displaystyle{\frac{s}{\sqrt{n}}}}$$
Recall when checking a data set to see if the underlying distribution is normal, one should remove any outliers (even when $n \ge 30$), check for skewness, and visually inspect the histogram to ensure it looks approximately normal.