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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?

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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 $0.3174$

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  • $\begingroup$ Why is the p-value .3174 instead of .1587? I think I'm using a table that represents something else (one-tailed test, I think it was called?)? Could you please explain that, if you can? EDIT: Oh, never mind, you just multiplied it by 2 to get the two tails. Thank you for the quick answer. $\endgroup$ – Super Rhinocerus Jun 1 '15 at 21:57
  • $\begingroup$ 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. $\endgroup$ – Sloan Jun 1 '15 at 21:59
  • $\begingroup$ Are you sure? I asked a TA and he said the value would be 1 - (z(t) = 1), which is 1 - .8413 and therefore .1587. $\endgroup$ – Super Rhinocerus Jun 2 '15 at 17:42
  • $\begingroup$ 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. $\endgroup$ – Sloan Jun 2 '15 at 18:23
  • $\begingroup$ Understood, I can see it now. There aren't any specifications of making it a one-tailed test. $\endgroup$ – Super Rhinocerus Jun 2 '15 at 18:43

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