# 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.

• You appear to have calculated a two-tail critical value for the t, when you calculated a one-tail critical value for the Z. Why did you change? The diagrams at the calculator you linked to aren't ambiguous. – Glen_b Apr 29 '14 at 3:52
• Oh that was my mistake, accidentally doing for a confidence of 95% rather than 5%. I think I understand things a bit clearer now, obviously I need to read up on this some more, but at least I have a better understanding of what to do – Valrok Apr 29 '14 at 4:06

## 1 Answer

$\mu_0$ 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.]

• So wait, would it be the 6 minutes which when converted to seconds would be 360 seconds for $\mu$? And as for t, the formula would stay the same, just that the letter would change? – Valrok Apr 29 '14 at 3:01
• Six minutes is indeed 360 seconds, and it makes sense to work in seconds since all your other information is in seconds. The formula you have is correct, but its distribution under the null hypothesis isn't normal, so you don't use z-tables; you use the appropriate t-table. So you don't compare with 1.645. – Glen_b Apr 29 '14 at 3:04
• I've always had trouble getting what the value of t should be. I went off some notes where I noticed that given a 5% confidence it was always 1.645, is there a way I could confirm that since I really don't know how to get t? – Valrok Apr 29 '14 at 3:06
• You have a 5% significance level ('confidence' doesn't enter into it). I'm not sure what you want me to confirm, but what I said was correct. As for degrees of freedom, I'll include some discussion in my answer. – Glen_b Apr 29 '14 at 3:09
• Some discussion of degrees of freedom in my answer now – Glen_b Apr 29 '14 at 3:19