2
$\begingroup$

I have two datasets from two courses. In one of the courses we applied an experiment and I would like to infer whether that experiment causes the average number of posts per students increases or not. In our data p1(number of posts in course 1)/ n1 (number of students in course 1) is much more than p2(number of posts in course 2)/ n2 (number of students in course 2). What kind of test I should apply to check my result could be general?

I could not find any question in stack overflow regarding ratio of variables.I tried chi-square and this is the result:

dfchsq <- data.frame(navgpost=c(itr1=p1/n1,itr2=p2/n2))  
chisq.test(dfchsq)

this is the result: Chi-squared test for given probabilities

data: dfchsq X-squared = 3.801, df = 1, p-value = 0.05122

My questions are: Is chi-square an appropriate test in this case? If yes, how the result can be interpreted?

Regards, P

$\endgroup$
2
$\begingroup$

From your post, you are trying to answer:

"whether that experiment causes the average number of posts per students increases or not"

Put differently, you want to assess whether the mean of 2 distributions, i.e. posts by students in the test condition (your experiment) and posts by students in the control condition are significantly different. With this formulation, you should apply the t-Test for the Significance of the Difference between the Means of Two Independent Samples (assuming that these are different classes). In R you should do the following:

First perform f-test to check if the variances of the 2 distributions are equal.

var.test(a,b)

Now perform the t-Test, the var.equal will be TRUE or FALSE based on the results on the F-test.

t.test(a,b, var.equal=TRUE, paired=FALSE)

Sharing a few good links on this problem:

http://vassarstats.net/textbook/ch11pt1.html

http://www.r-bloggers.com/two-sample-students-t-test-1/

$\endgroup$
0
$\begingroup$

This looks like a test between two proportions. The test name you are looking for is Z-test and in R its:

 ?prop.test
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.