# Hypothesis Testing Question: Average Given, no Level of Significance

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: $$z_t = \frac{52.5-50}{\frac{20}{\sqrt{64}}}=1$$ Since you are doing a two-tailed test, using $z_t=1$ yields a $p$-value of $0.3174$
• You are correct in that $0.1587$ would be the $p$-value for a one-tailed test. Here, you are asked to test equality; the alternative hypothesis would be "not equal to", which is a two-tailed test. To compute the $p$-value for a two-tailed test, simply double what you see in your table. – Sloan Jun 1 '15 at 21:59
• It's $0.1587$ if your alternative hypothesis is a one-tailed test. It's twice that if your alternative hypothesis is a two-tailed test -- I read the question as having a two-tailed alternative hypothesis. – Sloan Jun 2 '15 at 18:23