anovat-test
I'm still confused as to why it is often recommended to use an independent samples T-Test when comparing two means rather than a One-way Anova.
If t^2 = F, and the p values are the same, why use a T-Test? What is the difference?
I've read that its because a T-Test 'is more flexible', but why is this the case? Is it just because the T distribution is two-tailed, whereas the F distribution is one-tailed?
If I'm right in guessing that the other IV was not included in the ANOVA, then the most likely reason is that the two countries differed on the other IV, and that the countries only look 'significant' if the other IV is omitted. I wonder if the other IV is 'significant'? With sufficient confounding, it may not be. One last (but I think unlikely) possibility is that when you added the IV you lost 1 degree of freedom, and so if the IV were totally unrelated to the response, you would have lost a trivial amount of statistical power.
In general, you wouldn't necessarily expect one way ANOVA and the Kruskal-Wallis to be similar, sometimes they can give quite different p-values. See here for a little partial motivation for why you might expect a difference. [When samples are reasonably normal-looking and with means not too many standard errors apart, they often tend to give similar p-values. Outside that, they frequently don't.]
However, in this case the reason is more prosaic: Your Kruskal-Wallis p-value is wrong.
Here's a summary of results in R (details below).
p-value
Welch t-test: 0.001287
Equal-var. t-test: 8.552e-05
One way anova: 8.55e-05
Wilcoxon test: 0.004847
Kruskal-Wallis: 0.003761
(Neither of the last two p-values are exact; if they were, you'd get the same p-value for the two-group comparison.)
Your problem is you're treating the second group's data as a factor (see the end of this answer).
Here's what I get in R with your data:
frh <- data.frame(group1 = c(103.56, 103.32, 103.32, 104.27, 103.56, 103.8),
group2 = c( 97.16, 97.16, 96.69, 98.58, 90.76, 97.64))
# strip chart:
# Welch t-test:
> with(frh,t.test(group1,group2))
Welch Two Sample t-test
data: group1 and group2
t = 6.3316, df = 5.163, p-value = 0.001287
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
4.368147 10.245186
sample estimates:
mean of x mean of y
103.63833 96.33167
$\,$
# equal-variance t-test:
> with(frh,t.test(group1,group2,var.equal=TRUE))
Two Sample t-test
data: group1 and group2
t = 6.3316, df = 10, p-value = 8.552e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
4.735411 9.877922
sample estimates:
mean of x mean of y
103.63833 96.33167
$\,$
#one way anova:
summary(aov(values~ind,stack(frh)))
Df Sum Sq Mean Sq F value Pr(>F)
ind 1 160.16 160.2 40.09 8.55e-05 ***
Residuals 10 39.95 4.0
$\,$
# Wilcoxon-Mann-Whitney:
> with(frh,wilcox.test(group1,group2))
Wilcoxon rank sum test with continuity correction
data: group1 and group2
W = 36, p-value = 0.004847
alternative hypothesis: true location shift is not equal to 0
Warning message:
In wilcox.test.default(group1, group2) :
cannot compute exact p-value with ties
$\,$
# Kruskal-Wallis test:
> kruskal.test(frh)
Kruskal-Wallis rank sum test
data: frh
Kruskal-Wallis chi-squared = 8.3958, df = 1, p-value = 0.003761
Those are all about as consistent with each other as I would expect on that data.
Now, here's how to get what you got for the Kruskal-Wallis:
with(frh,kruskal.test(group1,group2))
Kruskal-Wallis rank sum test
data: group1 and group2
Kruskal-Wallis chi-squared = 4.3939, df = 4, p-value = 0.3553
The problem is, if you're getting this, you're using it wrong. That's not how the function works - group2 is being treated as a factor defining different groups for data in group1.
So the main reason the Kruskal Wallis isn't giving you a roughly similar p-value to ANOVA is you didn't call it correctly.
Best Answer
With questions like this there is always a bit of ambiguity, since there are lots of different tests that use the T-distribution as the null distribution, and hence, are referred to as "T-tests". Comparing to the ANOVA test requires consideration of which particular T-test you are talking about.
Gossett's T-test (the two-sample t-test with assumed equal variances) is formally equivalent to a one-way two-group ANOVA test from Gaussian linear regression. For these tests it can be shown that $T^2 = F$ and so the p-values of both tests are the same. Since these are the same test, it makes no difference which you use. Where the T-test becomes "more flexible" is when you remove the assumption of equal variances for the groups and use Welch's T-test or some similar variant. With this variation you now have a T-test that is not equivalent to the one-way two-group ANOVA test from Gaussian linear regression. The greater "flexibility" occurs because the test accomodates cases where the variances of the two groups are substantially different.