I ended up stuck on this review problem and I'm not quite sure where to go from where I got stuck.
The question is: A population distribution is known t have standard deviation 20. Determine the p-value of a test of the hypothesis that the population mean is equal to 50, if the average of 64 observations is 52.5.
std dev = 20
pop. mean = 50
n = 64 observations
avg = 52.5
z = absval(xbar - 50) / (20 / sqrt(64))
I plug in 52.5 for xbar, but I'm not sure where to go from here to obtain the p-value.
Is it p-value = P(z > 2.5/(20/8)), where 52.5 is xbar plugged in? The problem here is that I don't have a level of significance given, so I can't really look-up the z-distribution table. Can someone please give me an explanation for this?
You don't need a level of significance to simply find the $p$-value. All you need to do is compute the test statistic: \begin{equation} z_t = \frac{52.5-50}{\frac{20}{\sqrt{64}}}=1 \end{equation} Since you are doing a two-tailed test, using $z_t=1$ yields a $p$-value of $0.3174$