Should I use one-tailed t test for my specific data? In my example I have to determine if there is a significant reduction in certain physical and chemical parameters from before treatment of a water purification plant or after treatment. It is assumed that the data has unequal variance, however I am not sure about one, or two-tailed t test? I assume that it is one-tailed since the question is if one group has significantly increased or decreased. Here is an example of the data for parameter 1:
Raw Water: 421,422,422,431,341,341 and Treated water: 391,410,371,391,410,371
Help would be much appreciated!
 A: You are correct that you should do a one-sided test.
One hopes that there has been a reduction due to treatment. That is,
we hope the level of the undesirable chemical will
be greater before treatment than after.
Because you do not assume equal variances before
and after treatment, you should do a Welch 2-sample
t test. In R statistical software we obtain the P-value 0.387 > 0.05, so we cannot reject the null hypothesis
that there has been no change. (Notice the parameter alt="gr" for a one-sided test; no
parameter is needed for the Welch 2-sample test because it is the 'default' 2-sample t test in R.)
Raw =c(421,422,422,431,341,341) 
Trt =c(391,410,371,391,410,371)
t.test(Raw, Trt, alt="gr")

        Welch Two Sample t-test

data:  Raw and Trt
t = 0.29904, df = 6.6011, p-value = 0.3871
alternative hypothesis: 
  true difference in means is greater than 0
95 percent confidence interval:
 -30.56398       Inf
sample estimates:
mean of x mean of y 
 396.3333  390.6667 

The average level of the chemical has decreased from 396.3 t0 390.7, but this is not enough of a decrease
to be considered 'statistically significant at the 5% level'. (There is enough variability among the measurements that the small difference in means may
have been due to chance. In order to get a significant result, the T-statistic, about 0.3 here, would have
to be larger than 1.91.)
qt(.95, 6.6)
[1] 1.911997

The two sample standard deviations are $S_R = 43.0$ and $S_T = 17.4,$ so there is some evidence of different
variabilities before and after treatment.
sd(Raw); sd(Trt)
[1] 43.01473
[1] 17.44324

In a pooled 2-sample t test (assuming equal population variances), the degrees of freedom would have been $\nu = n_R + n_T -2 = 6 + 6 - 2 = 10.$ The Welch test uses smaller
degrees of freedom (about 7) to compensate for the different variances. (The formula for the Welch degrees of freedom is a bit messy; see your text or class notes for that.)
