Hypothesis test: How to determine if the critical value of t should be positive or negative? I'm doing hypothesis testing questions in my textbook and one of the things I'm not understanding is how to determine if the $t$ value I'm comparing to is supposed to be positive or negative.
The question I'm doing goes like this:

In an experiment designed to measure the time
  necessary for an inspector’s eyes to become used
  to the reduced amount of light necessary for penetrant
  inspection, the sample average time for
  n = 9 inspectors was 6.32 s and the sample standard
  deviation was 1.65 s. It has previously been
  assumed that the average adaptation time was at
  least 7 s. Assuming adaptation time to be normally
  distributed, does the data contradict prior belief?
  Use the t test with $\alpha = .1.$

So to do this I listed out all the values given:
$mean = 6.32, a = .1, n=9, s=1.65, H_0: μ = 7, H_a: μ < 7.$
So I subbed the values into this $t$ value equation:
$$T = \frac{\bar X - \mu_0}{S/\sqrt{n}}.$$
and got the value of $-1.24$. I also looked up the other t value in the t table with $n-1=8$ and $\alpha =.1$, and got the value of 1.397. I thought we had enough evidence to question the initial hypothesis because the number I got is lower than the one from the table so it matches with the alternative hypothesis, but on the back of my textbook it says the correct answer is "Because $t= -1.24 > -1.397 = -t_.10,_8,$ we do not have evidence to question the prior belief."
My question is why do we not have enough evidence and how do I know whether I'm supposed to compare to a negative or a positive $t$ value from the table?
 A: At the start, let's dispense with some of the jargon, and introduce it gradually, as necessary.
Previously (before doing our experiment with the 9 inspectors), we thought
accommodation time was 7 sec. Now we have tested seven inspectors and they
averaged 6.32 sec. It would be foolish to think the sample average should
turn out to be exactly 7, because we know there is some random variation.
However, 6.32 does seem smaller than 7.00, and
we wonder if is enough smaller to indicate 7.00 is not right.
The amount of random variation can be expressed in terms of the SD 1.65
and the sample size. Specifically, $SD(\bar X) = 1.65/\sqrt{n}$ is called
the (estimated) 'standard error' of the mean. 
As you say, the test statistic is
$$T= \frac{\bar X - \mu_0}{S/\sqrt{n}} = \frac{6.32 - 7}{1.65/3} = -1.236.$$
Does the value of $T$ indicate that $\bar X =6.32$ is significantly smaller than 
$\mu_0 - 7?$ 
Notice that our comparison of $\bar X$ and $\mu_0$ has turned into a
comparison of $T = -1.236$ and $0$. If $T$ is more negative than
$t^* = - 1.397,$ the critical value of the test at the 10% level, then
we reject $H_0.$
Your printed table of Student's t distribution shows that 1.297 cuts 10% of the probability (area) from the upper tail of the distribution $T(df=8).$ But you are doing a 'left-sided' test ($H_a$ has a $<$ symbol), so you want to
cut 10% of the area from the lower tail of the distribution. Because
the distribution is symmetrical, that value is just $-1.297.$
In the figure below, the blue curve is the density function (PDF) of $T(8)$, the dotted red line is at the value you find in the
printed table (cutting 10% of the area from the upper tail of the distribution),
the solid red line is your critical value (cutting 10% from the lower tail),
and the thin black line is the observed value of the test statistic $T.$
The Rejection region is to the left of the solid red line, and the observed
value of $T$ is not quite negative enough reach the Rejection region.
So 6.32 is not enough smaller than 7 to be considered 'significantly smaller'
at the 10% level of significance.

