The Average Height of an American is (Hypothesis Test) Problem
The average height of an American was reported to be 70. Many believe that college students are shorter than average. A random sample of students produced the following data:  
$n=140$, $mean = 65.39$, $s = 4.64$  
Find the test statistics
Steps Taken to Solve
I decided that this type of problem was a 1 sample T-test. These are my steps to the Hypothesis Test.  


*

*Null Hypothesis = No difference, so $M = M_o = 0$
Alternative Hypothesis = $M < 50$    

*$t = \frac{50 - 0}{\frac{4.64}{\sqrt{140}}} = 166.7467$  


Is this approach correct so far?
 A: The null hypothesis is $H_0: \mu = 70$ and the alternative is $H_a: \mu < 70.$
Roughly speaking, the question is whether the sample mean $\bar X = 65.39$ is
enough below $70$ that we should agree with the view that college students
are 'shorter than average'.
The test statistic is $T = \frac{\bar X - 70}{S/\sqrt{n}} = -11.76.$
a = 65.39;  s = 4.64;  n = 140;  mu.0 = 70
t = (a - mu.0)/(s/sqrt(n));  t
[1] -11.75566

This is an extremely negative value of the T statistic. The P-value
is the probability under $H_0$ of value as or more extreme (in the
direction of the alternative) than the observed value -11.76. 
The distribution of the T statistic under the null hypothesis has
Student's t distribution with $n - 1 = 139$ degrees of freedom.
That is essentially $0,$ found in R as:
> pt(-11.76, 139)
[1] 6.78294e-23

It is almost impossible for 'college students' randomly sampled from a
normal population with $\mu = 70$ to give dats with $\bar X = 65.39$
and $S = 4.46.$  The null hypothesis is rejected at any reasonable
level of significance.
Note:  Student's t distribution with df = 139 is very nearly standard
normal.
