say you have 1000 trials and a success probability of p = 0.43. Find the probability of having at least 364 success and less than 451 success. Now if you use the binomial distribution you'll have to sum up 157 terms, a lot of work to do by hand or by machine unless you know the code to write.
X ~ Binomial( n = 1000, p = 0.43)
P( 364 ≤ X < 451)
= P(X = 364) + P(X = 365) + ... + P(X = 449) + P(X = 450)
= 0.9046305
using the normal approximation is valid if you have at least 10 expected success and at least 10 expected failures. In this example you have 430 expected success and 570 expected failures. Use the continuity correction to approximate the probability. You need to use this since you are using a continuous distribution to find a discrete probability.
Xn ~ Normal(μ = n * p, σ = √(n * p * (1-p))
Xn ~ Normal(μ = 430, σ = 15.65567)
P( 364 ≤ X < 451) ≈ P( 363.5 ≤ Xn < 450.5)
= P( (363.5 - 430) / 15.65567 < Z < (450.5 - 430) / 15.65567)
= P( - 4.247662 < Z < 1.309430 )
= 0.9048056
a very good approximation.
The reason we have this approximation if from the days when computers were not available to do the exact calculations. The approximations are good and faster to compute by hand.
if you want to do the exact calculations quickly get the best stats software in the world, R. You can get it for free form www.r-project.org.
X ~ Binomial(n , p)
P( a < X < b)
in R the probability is found by using the command:
sum( dbinom(a:b, n, p) )
2007-10-22 17:39:24
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answer #1
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answered by Merlyn 7
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Well, as computers get more sophisticated, the advantage is probably not as great as it once was. Still, to compute the sum of binomial probabilities for a huge number of trials, you will be asking a computer to add thousands of probabilities, each of which are very close to 0 (since they add up to one.) Every machine has limits of accuracy, and there is potential for round-off error. (And of course, once upon a time these would have had to have been added up by hand!)
The normal approximation is quick and easy, and usually acurate enough for purposes of hypothesis testing.
2007-10-21 10:00:47
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answer #2
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answered by Michael M 7
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