How to do hypothesis testing in this case? A new methodology has been proposed for teaching English to undergraduate students. This methodology (which we call X) has been tested with a group of 27 students. Initially, the lecturer in charge of the course was sure that more than 50% of the classroom would fail the course if this methodology X is applied.
The final grades obtained by the students were (being 55 the minimum passing mark):
60,25,45,0,90,35,0,70,5,70,0,0,65,55,55,40,50,65,65,30,0,70,0,50,5,50,20

I have made the following program in R to check if my data is normally distributed:
data=c(60,25,45,0,90,35,0,70,5,70,0,0,65,55,55,40,50,65,65,30,0,70,0,50,5,50,20)
length(data)
summary(data)
d=density(data)
boxplot(data)
plot(d)
qqnorm(data)
qqline(data)
hist(data,breaks=length(data),xlim=c(0,100),ylim=c(0,10),freq=TRUE)

For the obtained graphs I can see that my data is not normally distributed, so I decided to apply a non-parametric test, specifically the Wilcoxon Test, to see if the hypothesis that more than 50% of the students will fail the course if this methodology is applied; the code is:
wilcox.test(data,alternative="less",mu=50,conf.int=TRUE)

I consider the value of mu as the hypothesized median value, the results I obtained were the following:
Wilcoxon signed rank test with continuity correction

data:  data
V = 81.5, p-value = 0.02565
alternative hypothesis: true location is less than 50
95 percent confidence interval:
     -Inf 47.49996
sample estimates:
(pseudo)median 
      34.99995 

Warning messages:
1: In wilcox.test.default(data, alternative = "less", mu = 50, conf.int = TRUE) :
  cannot compute exact p-value with ties
2: In wilcox.test.default(data, alternative = "less", mu = 50, conf.int = TRUE) :
  cannot compute exact confidence interval with ties
3: In wilcox.test.default(data, alternative = "less", mu = 50, conf.int = TRUE) :
  cannot compute exact p-value with zeroes
4: In wilcox.test.default(data, alternative = "less", mu = 50, conf.int = TRUE) :
  cannot compute exact confidence interval with zeroes

For what I know this means that the mean differs significantly from the hypothesis value of 50. So did I manage to prove that the application of this new methodology made that more than half of the class failed the course? In case that all my analysis is wrong, could somebody guide me about how to prove that the application of the new methodology had an effect in the huge number of students who failed the course and was not by random? Please consider that I do not have a former education in statistics.
PD. Another lecturer from other classroom was prompted to use this new methodology X, the number of students that took the course with her was like 30 approximately, but instead she used the OLD methodology; having very low number of students who failed the course. Unfortunately and because of administrative burdens I was not able to gather detailed information about the marks of each of those students.
 A: You only have a single sample, so when you call wilcox.test, that's not doing a Wilcoxon-Mann-Whitney, it's doing (as it tells you in the output!) a Wilcoxon signed-rank test.
That doesn't look to me to be directly relevant to the hypothesis in question. With additional assumptions (that don't hold) it could be relevant, but I don't think it's a suitable test as things stand.
Along the lines of the signed-rank test, you could do a sign test ... but that's going to be the same as the binomial proportions test I suggested in comments before.

response to followup question in comments:
A comparison of numeric grades might be addressed by

*

*a two sample t-test (possibly with unpooled variance and Welch-Satterthwaite adjustment to df).
While the distribution is somewhat skew and discrete, the sample size is large enough that the t-distribution will be a reasonable approximation.


*Alternatively a Wilcoxon-Mann-Whitney test might be used, as long as proper account is taken of the level of ties in the data because of the discreteness.


*Finally, a permutation test based on a comparison of any statistic of interest might be used; indeed either of the two previously mentioned statistics can be used as the basis of a permutation test (whence the issue of skewness and discreteness are automatically dealt with).
A: By your data 17 of 27 have less than 55 marks, hence failed. This is 63% with a 95% confidence interval of 42.3% to 80.6%. Hence, the hypothesis that MORE THAN 50% will fail with this teaching methodology is still not proven (the confidence interval is going across 50% meaning that the true rate could be LESS THAN 50%). 
