A random sample of $20$ items is drawn from a population that is known to be normally distributed. The sample has a mean of $60$ and a standard deviation of $16$. Test the hypothesis that the true mean of the population from which the sample is taken is $68$.
My calculations:
Find std error of mean = $\frac{16}{\sqrt{20}}= 3.58$
Therefore, the population mean is very likely to be between 7.16-10.73 of $60$. Is my conclusion correct? Did I set this up correctly?
It isn't clear from what you have said what your conclusion is. You have also not stated the significance level you are working to.
A model answer will look something like this:
$H_0: $ population mean $\mu = 68$
$H_1: $ population mean $\mu \ne 68$
Assume that the standard deviation of the sample is a good estimate of the population standard deviation.
Conduct a two-tailed test at 5% significance level (which is customary).
z-statistic $z=\frac {\bar x-\mu}{\sigma / \sqrt{n}} = \frac {60-68}{16 / \sqrt{20}}=-2.236 $
This is greater in magnitude than the critical value for a 5% two-tailed test (which is 1.960), so there is evidence to suggest that the true mean is not equal to 68. Conclude that $H_1$ is true.
Alternatively you could calculate the $p$ value for the test statistic, which is 0.0254 or 2.54%