0
$\begingroup$

Consider a hypothetical situation,

enter image description here

due to a new teaching method being implemented, I want to test whether there is a significant difference in the proportion of students that have passed a particular exam between two batches.

I could use the 2 proportion z test.

Are there any other tests I could use to compare two different proportions from independent samples (even if they don't apply particular case study)?

$\endgroup$
4
  • $\begingroup$ which othet tests impeach your mind ? $\endgroup$
    – user10619
    May 17, 2017 at 9:30
  • $\begingroup$ @subhashc.davar i came accross a few other tests but not sure if they are applicable ------- ncss.com/wp-content/themes/ncss/pdf/Procedures/PASS/… . The other tests seem to to be estimating the proportion parameter using MLE which is a bit confusing. $\endgroup$ May 17, 2017 at 10:11
  • $\begingroup$ Do you only have the data shown here or do you have also more fine grained data like the marks or similar? $\endgroup$
    – Mayou36
    May 17, 2017 at 10:33
  • $\begingroup$ @Mayou36 nope just this data $\endgroup$ May 17, 2017 at 10:49

1 Answer 1

0
$\begingroup$

You can use z-test for a testt of difference between two proportions representing two independent samples. The sample sizes of independent samples are large enough to meet the assumption of normal distribution.

$\endgroup$
4
  • $\begingroup$ for comparison of two means, we use CLT to show that as n increases irrespective of the prior distibution the mean tends to normal distribution and hence to form the statistic we convert it to standard normal. $\endgroup$ May 17, 2017 at 10:56
  • $\begingroup$ For comparison of proportions what is our assumption that the proportion follows normal as n increases ? $\endgroup$ May 17, 2017 at 10:56
  • $\begingroup$ would you prefer to give a formula or definition of standard normal? $\endgroup$
    – user10619
    May 17, 2017 at 15:40
  • $\begingroup$ i dont know if that will help. conceputally im a little confused $\endgroup$ May 22, 2017 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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