SOLUTION: Elves are known to have good accuracy and be able to hit their target with an arrow often. An elf goes to battle and hits 197 out of 210 targets. Create a 95 % confidence interva

Question 1117965: Elves are known to have good accuracy and be able to hit their target with an arrow often. An elf goes to battle and hits 197 out of 210 targets. Create a 95 % confidence interval for the accuracy rate of elves.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the elves hits their mark 197 out of 210 times.

that's a success rate of 197/210 = .9380952381.

that's a failure rate of (210 - 197) / 210 = 13/210 = .0619047619.

you can use the binomial distribution formula to solve this, i believe.

in a binomial distribuion:

mean = n * p
standard error of the distribution of sample means = sqrt(n * p * q)

in your problem, this comees out to:

n = 210
p = .9380952381 = success rate
q = 1 - p = .0619047619 = failure rate
m = n * p = 210 * .9380952381 = 197 = mean
s = sqrt(n * p * q) = 3.492168108 = standard error of the distribution of sample means

the critical z-score for a 95% confidence interval is equal to plus or minus 1.959963986.

the formula for z-score is z = (x - m) / s

z is the z-score
x is the raw score
m is the mean
s is the standard deviation / standard error of the distribution of sample means

we know the z-score.
we are looking for the raw score.

on the low side, we get -1.959963986 = (x - 197) / 3.492168108.

solve for x to get x = -1.959963986 * 3.492168108 + 197 = 190.1554763.

on the high side, we get 1.959963986 = (x - 197) / 3.492168108.

solve for x to get x = 1.959963986 * 3.492168108 + 197 = 203.8445237.

what i believe this is saying is that you can expect that the elves would be expected to score between 190 and 204 targets out of 210 for 95% of the battles that they become involved in.

i'm pretty sure this is correct, although i'm not 100% sure this is what your instructor is looking for.

you should verify with other sources, if you can.

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