# Statistics Hypothesis Testing of Continuous Variables

Longbrite candles are supposed to burn for at least 7 hours. To check, a random sample of 30 are tested. The time (t) before they burn out is recorded. The results are:

$\sum (t-7)=-2.40$

$\sum (t-7)^2=2.95$

Carry out a test at the 10% significance level of whether the Longbrite candles, on average, burn for at least 7 hours.

So I've stated the two hypotheses, equal to and less than 7. I've calculated that the mean of the sample:

$-2.4/30 + 7 = 6.92$

But when I try to calculate the unbiased sample variance, I get:

$S^2 = \frac{30}{29} (\frac{2.95}{30}-6.92^2)=-49.4$

Would anyone be able to point out where I'm going wrong with that variance? The textbook solutions say that S=0.3084, but I'm not quite sure why! Thank you!

You are given the sum of squared deviations of the sample from the null mean $7$, and you want the sum of squared deviations of the sample from the sample mean you calculated, $6.92$. To get this, you can write $$\sum (t-6.92)^2 = \sum (t - 7 + 0.08)^2 = \sum \left( (t-7)^2 + 2(t-7)(0.08) + (0.08)^2 \right) \\ = \sum(t-7)^2 + 0.16 \sum (t-7) + 30(0.08)^2,$$ and now you have all the pieces to substitute and get the desired result. Don't forget to divide by $30 - 1 = 29$ afterward.