A Likert scale attempts to measure opinions. If someone has
thought about an issue at all, then it's likely they will be
able to say whether they're in strong agreement or slight
disagreement. Maybe less likely to say whether their agreement
is strong or weak. So it is commonly agreed that Likert
scales can be considered as ordinal.
However, they are not numerical--not even at the 'interval' level. Is it clear that the
difference between strong and weak agreement is quantitatively the
same as the difference between neutrality and weak agreement, or
as the difference between weak and strong disagreement?
Consequently, the treatment of Likert scores as
numerical is controversial. Some reasonable arguments for treating
them as numerical have been advanced.
But I have to say I find your
quote to be especially unpersuasive. Personally, I find I am not as able to
do an honest job of distinguishing between categories on a seven point scale as on a five point scale. And I'd often prefer just to say Disagree, Agree, or Undecided. So I think it is delusionary to suppose more detailed scales fetch more useful data. What normality has to do with the issue is truly a puzzle. Similarly, it is unclear how having more ordinal responses tends
to make them more numerical.
Statistical methods for analyzing ordinal data can be more challenging to
create, understand, and compute than methods for numerical data. So it may be tempting to suppose that
pretending assumptions for t tests are reasonable is a valid excuse
to use t tests---instead of more appropriate methods of testing developed for ordinal data.
Finally, the "rule" that $n > 30$ assures normality is
only very approximate--at best. Specifically, it is unquestionably true of data are
uniform $(n = 15$ is enough), probably OK if data are pretty much restricted to an interval such as $\bar X\pm 3S,$
and completely unreasonable if data are exponential (typically, with many 'outliers' on the high side and none on the low side).
Suppose you want to know if there is a difference of opinion about the the use of crib notes on an exam between students in the College of Engineering and the College of Business.
If 53 out of 200 engineering students think crib notes are OK, and
85 out of 250 business students think so, then you can test
whether $\hat p_e = 53/200 = 0.265$ is significantly different from
$\hat p_b = 85/250 = 0.360,$ using
prop.test in R, as follows:
prop.test(c(53,85), c(200,250), cor=F)
2-sample test for equality of proportions
without continuity correction
data: c(53, 85) out of c(200, 250)
X-squared = 2.9395, df = 1, p-value = 0.08644
alternative hypothesis: two.sided
95 percent confidence interval:
prop 1 prop 2
The data do not show a significant difference
at the 5% level; P-value $= 0.086 > 0.05.$
(If there is a true difference in proportions of this
size, you would need greater numbers of subjects in order to establish that.)
Because it uses ranks, which make sense for ordinal data,
a two-sample Wilcoxon rank sum test might be used to
see if Likert scores from engineering students differ significantly from those of business students.
For small sample sizes, the
Wilcoxon test can have trouble with ties, and there will be
many ties for Likert data. But recent implementations of the Wilcoxon test, such as
wilcox.test in R, handle ties in
large datasets agreeably.
Here are some fake Likert scores
along with results for them from a Wilcoxon test. These data
show no significant difference.
E = sample(1:5, 200, rep=T, p=c(7,8,10,3,2))
B = sample(1:5, 250, rep=T, p=c(6,7,10,4,3))
 46 62 60 16 16 # Likert scores 1,2,3,4,5, resp
 62 51 85 35 17
Wilcoxon rank sum test
with continuity correction
data: E and B
W = 23493, p-value = 0.2557
true location shift is not equal to 0