t-table probability calculations I have this probability problem that deals with the t-test. I got it down to the following:
P(-0.387 < T < 0.775)
The solutions manual says that the answer to this is supposed to be 0.422, but I don't know how they are getting this. Maybe I don't know how to properly read the t-table? The Z-table problems I understood just find.
If it's needed, the problem is the following:
Resistors of a certain type have resistances that are approximately normally distributed with mean 200 ohms. Fifteen of these resistors are to be used in a circuit. The standard deviation of 12 measurements is 10 ohms. Find the probability that the average resistance of the 15 resistors is between 199 and
202 ohms.
I did 202 - 200 / 10/sqrt(15) and the same for 199 - 200 to get the t-scores. Df is 14.
 A: In printed CDF tables of the standard normal distribution, you have
probabilities corresponding to hundreds of cutoff points. By contrast,
in printed quantile tables of t distributions, you have only about
half a dozen probabilities corresponding to various cutoff points.
So, in general, it will be difficult from tables to find the exact
probability that $T \sim \mathsf{T}(14)$ lies in a particular interval.
You seek $P(-0.387 < T < 0.775) = P(T < 0.775) - P(T \le -0.387) = 0.4221.$
Using statistical software, this is simpler than trying to use printed t tables.
In R statistical software, one has:
pt(c(-.387,.775),14)
## 0.3522876 0.7743841     # P(T < -.387) = 0.3523; P(T < .775) = 0.7744 
diff(pt(c(-.387,.775),14))
## 0.4220965               # P(-.387 < T < 0.775) = 0.4221

Of course, I do not know what kind of t table you have in your text or what
software your course is using. But the t tables in two popular statistics
texts I looked at are inadequate for answering this question.
Note: On the row for df = 14, one printed t table I have at hand has
$P(T > 1.354) = .100$ and $P(T > 2.645) = .010,$ so I could say
$P(1.345 < T < 2.654) = .09,$ but that is nothing like the problem you've
been asked to solve.
