z-score use of $\sqrt{n}$ I am going through a simple statistics exercise in class that has me thinking a little too much about the outcome. The professor has posed the following data set and told us to "compare it" to $X=4$.
{2.5, 2.8, 1.9, 3.7, 4.2, 3.5, 2.9}
The formula given is:
$z = \frac{X-\mu}{\sigma/\sqrt{n}}=\frac{4-3.0714}{0.7804/\sqrt{7}}=3.1482$
With no assumption of whether $X=4$ should be higher or lower than the set, a two-tailed test results in a p-value of 0.0016, which is quite low. I am struggling with the interpretation.
Usually* the $\sqrt{n}$ is 1 and the interpretation is that "4 is a statistically unlikely draw from the population". This is clearly not the case here, especially considering that there is a 4.2 in the set and there is no indication that the set is a population.
The best I have so far is that the interpretation is that "the mean of the set is statistically unlikely to be 4 or higher". Am I missing something here?
* Usually, meaning when I typically have calculated and interpreted a z-score.

First, if you are testing whether the true mean of the population (in which you have collected a sample of size 7) is equal to μ0 = 4 or not, it should be "3.07 - 4" [sample mean - μ0] in the formula.
Secondly, since the population variance is unknown, you need to use t-statistic instead of z-statistic.
Finally, the small p-value tells you that if the true mean of the population is really 4, the probability that you get that kind of sample {2.5, 2.8, 1.9, 3.7, 4.2, 3.5, 2.9} with an average of 3.07 (or more extreme) is low. In other words, the true mean of the population is unlikely to be 4.
