Cough-a-Lot children’s cough syrup is supposed to contain 6 ounces of medicine per bottle. However since the filling machine is not always precise, there can be variation from bottle to bottle. The amounts in the bottles are normally distributed with σ = 0.3 ounces. A quality assurance inspector measures 10 bottles and finds the following (in ounces):
5.95 | 6.10 | 5.98 | 6.01 | 6.25 | 5.85 | 5.91 | 6.05 | 5.88 | 5.91
I know that Z=m-u divided by √(SD/n)
I am just not sure what numbers to plug in. I know the (m) mean is 6 but I don't know what the "u" would be.
Your question is unclear, because you ask "what do I plug in for the $Z$-score," but the context of the word problem itself suggests that a hypothesis test is desired. Yet you don't state the entirety of the question. You give us a sample from a normally distributed population with known standard deviation, and you give us background about what the data means, but we don't know what the question requires of us. If we do a hypothesis test, what is the significance level? What is the hypothesis to be tested?
With this information lacking, I will simply make something up. Suppose the hypothesis is that the machine is underfilling the bottles, and that we want to test if the data suggests with 95% confidence that this is the case. That is to say, $$H_0 : \mu = \mu_0 \quad \text{vs.} \quad H_a : \mu < \mu_0$$ is a one-sided test with the hypothesized mean of $\mu_0 = 6$. The test statistic under the null hypothesis is therefore $$Z = \frac{\bar x - \mu_0}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1),$$ where $\bar x$ is the sample mean. The test will reject $H_0$ in favor of $H_a$ if $Z < z_{0.05} \approx -1.645$, the lower 5th percentile of the normal distribution (i.e., the z-score for which $\Pr[Z \le z_{0.05}] = 0.05$). After calculating the test statistic, compare it against $-1.645$. If it is greater, then we fail to reject $H_0$ and conclude the sample contains insufficient evidence to suggest with 95% confidence that the average volume of liquid in each bottle is less than 6 ounces (i.e., there is not enough evidence to suggest the machine is underfilling the bottles). If, however, $Z < -1.645$, then we reject $H_0$ in favor of $H_a$, and state that with 95% confidence, the data suggests the machine is underfilling the bottles.
Note however, that if the test of interest is that the machine is deviating from the mean of 6 ounces per bottle, then we would use a two-sided test, and the above calculation would be different. Also, I don't know the test's significance level. Many tests are done at the $\alpha = 0.05$ level (95% confidence), but it could just as easily be set to 99% confidence or 90% confidence. In each case, the critical value will be different, leading to a different rejection criterion.
Now, if all you wanted to do is calculate the test statistic (which happens to be a z-score), then all you need is the formula above, which is the same under any choice of null and alternative hypotheses for the mean, and for any significance level. But there is no meaning to that test statistic until you compare it against a critical value.