Testing hypothesis using significance levels A random sample of n=26 observations on escape time (sec) for oil workers in a test resulted in sample mean 370.69 and sample standard deviation 24.36. Suppose the investigators had believed a priori that true average escape time would be at most 6 minutes. Does the data contradict the prior belief? Assuming normality of escape times, test the appropriate hypotheses using a significance level of a=0.05
I've tried using the follow formula to get my results.
z_0.05 = 1.645
$z=\dfrac{\overline{x}-\mu_0}{s/\sqrt{n}}$
$= \frac{370.69 - \mu}{24.36/\sqrt{26}} $
My problem is I'm not sure what $\mu$ is supposed to be in this problem and also I don't if I'm supposed to be checking that z > z_0.05  to prove that the data contradicts the belief or vice versa.
EDIT:
t_.05 = 1.708
t = 2.23762
t_.05 < t so the data contradicts the prior belief.
 A: $\mu_0$ should be the value in the null hypothesis (which you failed to state). The null hypothesis is essentially given in words in the question, you need simply turn that null in text into a symbolic form and then it's obvious what $\mu_0$ should be in the formula.
I believe you're correct in thinking it's after a one-tailed alternative.
Note, however, that you have been given a sample standard deviation, not a population standard deviation, so this statistic won't have a normal distribution (it's not '$Z$', then, but '$t$').
Can you do it now?

Working out degrees of freedom in a t-test
The degrees of freedom in a t-test come down to how many independent pieces of information there are in your estimate of $s$. In a one sample t-test, you have $n$ observations, but the degrees of freedom will be reduced by 1 because of the estimation of the sample mean. That is, the degrees of freedom will be the same as the $n-1$ in your denominator when you worked out the variance before taking the square root to get $s$. Similarly when computing an equal-variance two-sample $t$, or a one-way ANOVA, or a regression, your degrees of freedom will be the same as the  denominator for the variance of the numerator (that is, the thing you divide by immediately before taking the square root to get the $s$ for the t-statistic in question).
[See the discussion here for some extensive discussion of degrees of freedom, or see here for some briefer, more basic information.]
