# Calculating the percentage change of 2 values

If I have two different percentages and I wanted to know the change, I would simply use the percentage change formula...but what if I want to compare two percentages that have different amounts: Ex: August - there was 539 right answers out of 743 = 72.5 % September - there was 498 answers out of 820 = 60.7%

How can I do a monthly comparison of change between those 2 percentages?

It was suggested to me that I use LCD, but those numbers could be massive if I have a larger set of numbers like in the thousands.

So, should I analyze the months separately or what should I do?

**I want to do a comparative analysis on the students who took the exam in august vs the students in september. LCD = lowest common denominator. Can this be done?

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Could you elaborate a little on the purpose of the comparison? Also, what does "LCD" stand for? –  whuber Sep 9 '12 at 20:39
Does my edit help? –  John Sep 10 '12 at 0:51
I'm baffled as to how the least common denominator of any of these numbers would be relevant here. –  whuber Sep 10 '12 at 15:41
What's different between August and September? Is it just the students, or is the test different too? Is it multiple students in both cases? Are the students all taking different tests? Since 743 is prime, I assume you have different tests of different lengths being given to multiple students in each case. Given that, will you interpret the difference in performance as difference in ability of the students or difference in difficulty of the tests? –  shujaa Sep 10 '12 at 17:02

The percentages have different levels of accuracy determined by the respective sample sizes. If you assume the number of correct answers is binomial in each month with unknown parameter p$_1$ (unknown) and n$_1$=743 in August and p$_2$ (unknown and possibly different from p$_1$) and n$_2$=820. Then assume the August and September samples are independent. Consider the difference of the sample estimates. The variance for that difference is p$_1$(1-p$_1$)/743 + p$_2$(1-p$_2$)/820. The standard deviation is the square root of that variance. The sample estimates of p$_1$ and p$_2$ can be plugged in to get an estimate of that standard deviation. You can divide by it and use the normal approximation to get approximate confidence intervals or a pooled estimate of the standard deviation for hypothesis testing.

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Thank you Michael –  John Sep 10 '12 at 23:22