t-test and anova comparison I'm trying to understand the difference between T-test and Anova and wanted to ask:
1) What would be a few concrete experimental situations in which ANOVA could be used to compare the effects of different treatments upon the population being studied? 
I'm curious if anyone would happen to have a few concrete ex's of such.
And, what are some of the assumptions and issues involved with both t-tests and ANOVA? And, which assumptions are the procedures robust with regard to and  what kind of research hypothesis Ha can be handled by t-testing but not ANOVA?  
 A: I am assuming that you are familiar with standard ANOVA and t tests.
The purpose of this Answer is to illustrate them and point out some
respects in which they differ.
Consider 15 subjects, 5 of them randomly assigned to each of three groups.
Each group is given a different treatment for lowering blood pressure.
Decreases in PB upon treatment are as follows.A:   0, -1, 10,  3,  5
B:  10,  2, 15,  4, 10
C:  21, 11, 10, 12,  8 

A standard analysis of variance (ANOVA) tests the null hypothesis
that all three group population means are equal against the
alternative that at least one pair of means differs. 
Results
from Minitab software are shown below. (This method assumes
that the three populations are normal and have equal variances.)
Analysis of Variance

Source  DF  Adj SS  Adj MS  F-Value  P-Value
Factor   2   202.8  101.40     4.24    0.041
Error   12   287.2   23.93
Total   14   490.0

The P-value 0.04 < 0.05 indicates that the null hypothesis is
rejected at the 5% level of significance.
One might try doing three individual t tests to check whether $\mu_A$ and $\mu_B$
are significantly different, and similarly for pop means of A and C, and
of B and C. But if all three tests are done at the 5% level of significance,
errors of the three tests might accumulate to give an overall result of
unknown reliability. (A 'Bonferroni' procedure would do each t test at the
level $0.05/3$ in order to make sure the overall level is below 5%.)
The Tukey 'HSD' method of multiple comparisons does this 'family' of three
comparisons with an overall 'family rate' of 5%. It finds a significant difference
between $\mu_A$ and $\mu_C$, but not a significant difference when A and B
are compared or when B and C are compared. Tukey confidence intervals are
plotted in the figure below:

By contrast, if we actually had only two groups, A and C, then a t test
would be the appropriate way to compare them. The 'Welch separate variances'
t test has P-value 0.02. (Unlike the 'pooled' t test, the Welch test
does not assume that populations A and C have equal variances. Both
tests assume that the populations are normal.
Two-sample T for A vs C

   N   Mean  StDev  SE Mean
A  5   3.40   4.39      2.0
C  5  12.40   5.03      2.2

Difference = μ (A) - μ (C)
Estimate for difference:  -9.00
95% CI for difference:  (-16.06, -1.94)
T-Test of difference = 0 (vs ≠): 
    T-Value = -3.01  P-Value = 0.020  DF = 7

Notes: (1) If you are comparing data from more than two normally
distributed groups, you must use ANOVA. To compare two groups, use
a t test. (2) It is correct, but needlessly complicated, to use an ANOVA procedure for a two-sided test on just two groups.
(3) Most statistical software will do a version of ANOVA that does not
assume equal population variances, but I did not illustrate that kind
of ANOVA here. (4) You can study the assumptions, theory, and computational
formulas for t tests and ANOVA procedures in most elementary statistics
texts and on online (Look at Wikipedia or NISS sites; I'd avoid YouTube, Khan, 
and some other Internet sources in this particular case.)
