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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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  • $\begingroup$ Could you elaborate a little on the purpose of the comparison? Also, what does "LCD" stand for? $\endgroup$
    – whuber
    Sep 9, 2012 at 20:39
  • $\begingroup$ Does my edit help? $\endgroup$
    – John
    Sep 10, 2012 at 0:51
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    $\begingroup$ I'm baffled as to how the least common denominator of any of these numbers would be relevant here. $\endgroup$
    – whuber
    Sep 10, 2012 at 15:41
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    $\begingroup$ 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? $\endgroup$ Sep 10, 2012 at 17:02

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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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I think you're looking for a complicated solution to a relatively simple problem. Unless I'm misunderstanding, you want to compare August's 539 / 743 = 72.5% to September's 498 / 820 = 60.7%. The difference, in percentage points, is a fine comparison: The accuracy dropped by 11.8 percentage points (11.8 = 72.5 - 60.7). You could also calculate the percentage difference in the percents using the "percentage change formula" 11.8 / 72.5 * 100 = a drop of 16.3% in the percent accuracy, but part of the advantage of converting to percents in the first place is that you don't have to do anything more to give the numbers scale.

Related XKCD

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  • $\begingroup$ Is that an accurate comparison? Since you have more questions on the september than the august test. I was originally doing it the way you suggest but I was told that it only works if you have the same total amounts...otherwise it's skewed. Thoughts? $\endgroup$
    – John
    Sep 10, 2012 at 1:58
  • $\begingroup$ This is an accurate comparison of the accuracy point-estimates. Michael Chernick's answer should be very helpful if you want to look at the uncertainty around those estimates. With both your sample sizes being relatively large, I can't even guess which direction one might think the results would be skewed in. $\endgroup$ Sep 10, 2012 at 16:53
  • $\begingroup$ After your responses on this site as well as others, I will use Michael's answer. Thank you for your assistance. I also see that I turned something that was easy into something that is complicated. $\endgroup$
    – John
    Sep 10, 2012 at 23:22
  • $\begingroup$ @John Glad you've found the help you need! To indicate that your question has been resolved (and as a thank you to Michael) it's nice to click the check mark to the left of his answer, just below the votey buttons. $\endgroup$ Sep 11, 2012 at 2:34

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