# True population proportion in Statistics question?

This is one of my homework questions for statistics?? can someone help me out??

05. The Sheriff of Nottingham is suspicious about a cluster of Sherwood Forest accidental deaths that resulted from arrows. You, as Robin Hood or Maid Marian, are asked to estimate the proportion of accidental deaths nationwide that are caused by arrows. How many accidental deaths must you survey to be 99% confident that your sample proportion is within 6% of the true population proportion?

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Maid Marian -

Since we do not actually know the "true" population proportion, we use the default p = 0.5. This will always produce the most "conservative" (largest) value for n. Here is the formula you need to solve ... sorry, no TI tricks here :{

margin of error = (z*)(standard error) , so what do we already know?

margin of error = 6% or 0.06

z* is the z-value associated with 1%/2 = 0.005 in each tail. Remember a proportion problem ALWAYS uses z, never t. So, what is z* in this: P(Z < z*) = 0.005?

z* = 2.576

standard error = sqrt[(0.5)(0.5) / n] , where n is what we are trying to solve for

Looks like we have everything we need, so plug it all back into the formula and solve for n:

margin of error = (z*)(standard error)

0.06 = (2.576)sqrt[(0.25) / n]

n = 460.8, but ALWAYS round up so that you meet the margin of error criteria

n = 461

Good Luck Maid Marian!

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