# Thread: SD/SE of Percentage Question

1. ## SD/SE of Percentage Question

Have a question that's probably painfully easy to address. Little slow.

I am sampling from a hypothetical infinite population with an unknown mean and SD. I performing 10 separate trials, and each trial calculates a "hit rate" out of 100 attempts.

Trial 1: 65 hits out of 100, 65%
Trial 2: 66 hits out of 100, 66%
Trial 3: 55 hits out of 100, 55%
Trial 4: 61 hits out of 100, 61%
Trial 5: 69 hits out of 100, 69%
Trial 6: 62 hits out of 100, 62%
Trial 7: 63 hits out of 100, 63%
Trial 8: 63 hits out of 100, 63%
Trial 9: 64 hits out of 100, 64%
Trial 10: 59 hits out of 100, 59%

Overall, there are 627 hits out of 1000, or 62.7%

Using binomial theory, the SD of the overall hit rate is sqrt(p*q)=0.48, and the SE of the overall hit rate is sqrt(p*q/1000) = 0.01529.

The standard error of the individual trials ranges from 0.0462 to 0.0492 [sqrt(p*q/100)].

The standard deviation, calculated the "regular way", of the 10 percentages from the 10 different trials, though, is 0.0386.

I am a little confused as to what the SD of the 10 different trials is really telling me versus the SE of the individual trials--why aren't they the same, or at least a little closer? Wouldn't the SE calculated from a single trial indicate the SD (given the same N of 100) of infinite additional trials? But when I calculate the SD of the 10 trials, it's only 0.0386.

Is this some sort of sample bias, or a misunderstanding on my part? Thanks!

2. ## Re: SD/SE of Percentage Question

hi,
I think you are comparing apples to oranges.The SD refers to the standard deviation of the binomial distribution of your infinite population that is calculated using the estimate of p and q.

The SE refers to the precision of your estimate of p and q, so they are definitely not the same.

regards
rogojel

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