# Standard error of mean $\mu_{\bar x}$ vs standard deviation of means $s_x$

$$70$$ bags of sugar selected at random from a large batch are weighed. The mean weight of the sample is: $$\bar x=227\ g$$ and the mean of the sample is: $$s=7.5\ g$$

Calculate a $$95\%$$ confidence interval for the mean weight of all packs in the batch.

Do I use: ($$\mu_{95}=\bar x \pm 1.96s$$) or ($$\mu_{95}=\bar x \pm 1.96\frac{s}{\sqrt{n}}$$) ?

My textbook gives the formula: $$\mu_{95}=\bar x \pm 1.96\frac{\sigma}{\sqrt{n}}$$

where $$\sigma$$ is the known standard deviation of sugar bags.

My textbook says: If $$\sigma$$ the standard deviation of the population is not known, use $$s$$ the standard deviation of the sample as an approximation.

I take this as meaning: if $$\sigma$$ is not known then: $$\mu_{95} \approx \bar x \pm 1.96\frac{s}{\sqrt{n}}$$

But my textbook also says that: the standard deviation of the sampling distribution is given by $$\mu_{\bar x}=\frac{\sigma}{\sqrt{n}}$$

It's not clear to mean the difference between this $$\mu_{\bar x}$$ and $$s$$.

$$s$$ is the standard deviation of the sample means. Is $$\mu_{\bar x}$$ not the same thing ?

Does this mean that the standard deviation of the sample means $$s=\frac{\sigma}{\sqrt{n}}$$ ?

So if $$\sigma$$ is not known but $$s$$ is known do I use: ($$\mu_{95}=\bar x \pm 1.96s$$) or ($$\mu_{95}=\bar x \pm 1.96\frac{s}{\sqrt{n}}$$) ?

I assume that the standard deviation of the mean of the sample is $$s=7.5\,\mathrm{g}$$.

A 95% confidence interval for the mean weight of the population of all packs in the batch is

$$\mu_{95}=\bar x \pm 1.96\frac{s}{\sqrt{n}}$$

To be more correct, you should use a t-distributrion in stead of a z-distribution to calculate the confidence interval. So use $$t_{n-1,0.05}$$ in stead of 1.96, but if $$n$$ is large the difference will be minimal.

The symbol for the standard deviation of the sampling distribution should be something like $$\sigma_{\bar X}$$ and not $$\mu_{\bar x}$$.

We have

$$\sigma_{\bar X}=\frac{\sigma}{\sqrt{n}}.$$

$$s$$ can be used as an estimation for $$\sigma$$.