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sd and mean

R5S RNA has a sequence of 120 nucleotides.  We replicate the new sequence 100 times and find there are 60 of A’s in the 20th position.   Suppose the probability of an A in the 20th position = 0.79 in R5S.

In this case, what is the sd of the sample and the mean of the genral population?

thx
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ethanjohnsons
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ethanjohnsons
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ozoCommented:
If the occurance of an A in the 20th position of each of the 100 replicas is independent (which seems like a strange assumption to me if they are replicated)
then the expected number of As in the 20th position would be a binomial distributuon with mean 0.79*100 and standard deviation sqrt(100*0.79*(1-0.79))
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ethanjohnsonsAuthor Commented:
If I calculate t-value,

sqrt(100*0.79*(1-0.79)) = 4.073082

(0.79*100-60)/( 4.987484/sqrt(100)) = 38.09536

it is abnormally large.  What is wrong?
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ozoCommented:
You should not be dividing by sqrt(100)
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ethanjohnsonsAuthor Commented:
then what would be the sample size (= sqrt( ??) here?
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ozoCommented:
It depends what you are talking to be the sample size.
You are doing the experiment one time, so you might be saying
(0.79*100-60)/( 4.987484/sqrt(1))
of if you mean the number of 20th positions that may or may not contain A, then you would take sd=
sqrt( sample size *0.79*(1-0.79))
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ozoCommented:
Or if you want the average number of A's per 20th position,
then sd for one sample is 0.79*(1-0.79) and for 100 samples
mean = 60/100 sd = 0.79*(1-0.79)/100
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