Calculating probability of Type II Error for a one-sample t test So I have the following problem:
A transportation company is suspicious of the claim that the average useful life of certain tires is at least 28,000 miles. To verify that, 40 tires are placed in trucks and an average useful life of 27463 is obtained with a standard deviation of 1348 miles.
a) Test this hypothesis with a level of significance of α = 0.01
b) If $\mu_1 = 27230,$ calculate the probability of Type II Error
Any hint on how to do b)?
 A: Test of hypothesis: Testing $H_0: \mu = 28000$ vs $H_a: \mu < 28000,$ based on $n = 40$
observations with $\bar X = 27463$ and $S = 1348,$ we Reject $H_0$
at level 1% because the P-value is less than 1%.
Here are results from Minitab statistical software.
One-Sample T 

Test of μ = 28000 vs < 28000

 N   Mean  StDev  SE Mean  99% Upper Bound      T      P
40  27463   1348      213            27980  -2.52  0.008

Obviously, 27463 < 28000. The question is whether it is enough smaller
that we shouldn't ascribe the difference to random variation. The
answer (from the one-sided 99& CI) is that any sample mean below 27980 would
be significantly smaller.
Power computation for specified alternative. Minitab's 'power and sample size' procedure uses the noncentral t
distribution to find power given $n,$ the direction of the test,
the difference between null and alternative values, the significance level, and (an estimate
$\hat \sigma = 1348$ of) of $\sigma.$ The Type II Error probability is 1 - Power, so the probability of Type II error is about 0.123.
   1-Sample t Test

Testing mean = null (versus < null)
Calculating power for mean = null + difference
α = 0.01  Assumed standard deviation = 1348

            Sample
Difference    Size     Power
      -770      40  0.876884

Power computations require a population variance, or at least the speculation of one. Usual practice is to use
a sample variance as an estimate if one is available. If the true
mean endurance of a tire is as low as 27,230, using data from 40 tires
gives us a pretty good (approx. 88%) chance of detecting that the claimed
28,000 is an exaggeration. (Notice that the difference between $\mu_0$
and this $\mu_1$ is more than a couple of standard errors.)

Notes: (1) Depending on the level of your class, the intention of the exercise may have been to treat this as a power
computation for a z test because $n > 30.$ However, the 'rule' that
t and z tests are 'essentially the same' for $n > 30$ is based on
the fact that the critical value for a one-sided test at the 5% level $t^*$ is close to $z* = 1.645.$ However, for a test at the 1% level 
the critical value $t_{.01,39} \approx 2.43$ may not be sufficiently
close to the critical value $z_{.01} \approx 2.33.$
(2) In order to do an accurate power computation for a t test, it is necessary
to use the noncentral t distribution, as in Minitab. Many mathematical 
statistics texts have explanations of the noncentral t distribution and
its use in power computations. I found the Wikipedia article to be
unnecessarily technical, but Section 5 of this paper by Scholz may be helpful.
A: Did you figure out this problem?
Part A) Use a T-test in the graphing calculator
the t-statistic is given by 
$$ t = \frac{\left.\left(\bar{X}-\mu \right.\right)}{\frac{s}{\sqrt{n}}}$$
thus,
$$t = -2.5195$$
For a significance level of $.01$, the t-critical value is about $2.425$
If  |t| > t-critical, you can declare statistical significance and reject the null hypothesis
The null hypothesis is $H_0: \mu_ \ge \mu_0,$ and
the alternative hypothesis is $H_a: \mu_ < \mu_0,$ 
Now to find the actual p-value you can use your graphing calculator
Using T-test in graphing calculator:
$$ P-value = .00798$$
Thus, we have sufficient evidence that the claim "the average useful life of certain tires is at least 28,000 miles" is invalid, it is actually less. $OR$ we have insufficient evidence to conclude that the claim "the average useful life of certain tires is at least 28,000 miles" valid.
Part B Update, This is what I think it is, based on your most recent comment 
$$2.326*\frac{1348}{\sqrt{40}} = 495.758$$
and $$28000 - 495.758 = 27504.2$$
Now,
$$\frac{27504.2\, -27230}{\frac{1348}{\sqrt{40}}} =1.286 $$
Type 2 Error = $$1 -P(z > 1.286)$$
Thus, $$1 - .9007785 = .09922$$
OR
$$P(z \leq 1.286) = .09922$$
