SOLUTION: A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 wi
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Question 1194768: A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 with 90% confidence if
(a) she uses a previous estimate of 0.32?
(b) she does not use any prior estimates?
p = 0.32 is the previous estimate of the proportion
E = 0.03 is the error we want (or smaller)
n = sample size
n = p*(1-p)*(z/E)^2
n = 0.32*(1-0.32)*(1.645/0.03)^2
n = 654.256711111111
n = 655
We always round UP to the nearest whole number when doing min sample size problems.
It doesn't matter that 654.256711111111 is closer to 654 than it is to 655
We round up to clear the hurdle needed.
Try n = 654 in the margin of error formula
E = z*sqrt(p*(1-p)/n)
and you'll find that E > 0.03 which isn't what we want.
But if you tried n = 655, then E = 0.03 or E < 0.03 would be the case.
This is going to be very similar to part (a).
However, we'll be using p = 0.50 this time.
This is the most conservative estimate or guess to make for p if we don't know what it is.
It's right in the middle of the interval
The other values of z = 1.645 and E = 0.03 remain the same from before.
n = sample size
n = p*(1-p)*(z/E)^2
n = 0.50*(1-0.50)*(1.645/0.03)^2
n = 751.673611111111
n = 752
Once again, always round up.
Since p = 0.5, this means p(1-p) = 0.5*(1-0.5) = 0.5*0.5 = 0.25
The formula above updates to n = 0.25*(z/E)^2 when using p = 0.5
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