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I know that if 

H0: μ = 1.35

then if H1: μ != 1.35 means p-value is 2P(Z ≥ |z|)

else if

H1: μ > 1.35 then P(Z ≥ z) and H1: μ < 1.35 means p-value is P(Z ≤ z)

However, I am stumped about how to do it with following sets of hypothesis. 

H0: μ ≥ 1.35 
H1: μ < 1.35

p-value=??

H0: μ ≤ 1.35 
H1: μ > 1.35

p-value=??

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First, the p-value is not z or 2*z. P values range from 0 to 1, z scores can be any real number. But perhaps you mean how to look up p in a z table? I am not sure.

Second, if the variable that $\mu$ is the average of is continuous, then $p(\mu \le 1.35) = p(\mu \lt 1.35)$ because the probability of any exact value of $\mu$ is infinitesimal.

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  • $\begingroup$ Yes, I agree p-value is not z or 2*z. Just a typo. Please see my edit above. Now, how to compute p-value when H0 is not a fixed constant but a range of values? (inequality sign) $\endgroup$
    – t t
    Commented Nov 20, 2013 at 23:19
  • $\begingroup$ You already wrote the solution to that in your problem, in the second box $\endgroup$
    – Peter Flom
    Commented Nov 20, 2013 at 23:55
  • $\begingroup$ Are you talking about H1: μ > 1.35 then P(Z ≥ z) and H1: μ < 1.35 means p-value is P(Z ≤ z)? but this assumes H0 = 1.35. The question I'm asking has H0 with inequality too! Are you saying it doesn't matter? (Both H0 and H1 have inequalities in my question!) $\endgroup$
    – t t
    Commented Nov 21, 2013 at 0:04
  • $\begingroup$ No it doesn't assume $\mu = 1.35$, it assumes it is either <= 1.35 or >= 1.35. An H0 that $\mu = 1.35$ could be rejected 100% of the time with no error. $PR(\mu = 1.35) = 0$ $\endgroup$
    – Peter Flom
    Commented Nov 21, 2013 at 0:35

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