Question 1190707
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If the doctor was trying to estimate the population mean mu, then the margin of error would be
E = z*s/sqrt(n)
where,
E = margin of error
z = critical z value based on the confidence level
s = standard deviation
n = sample size
You probably have come across this formula when setting up confidence intervals.


Since the values of z and s aren't mentioned, we'll just hold them to fixed unknown constants.
It'll turn out that they don't matter.


Let's solve the mentioned formula for n
E = z*s/sqrt(n)
E*sqrt(n) = z*s
sqrt(n) = z*s/E
n = (z*s/E)^2


Let's now plug in the first error mentioned 12
n = (z*s/E)^2
n = (z*s/12)^2
n = A/(12^2) ... let A = (z*s)^2
n = A/144
This margin of error E = 12 corresponds directly to the sample size n = 55, so,
n = A/144
A = 144n
A = 144*55
A = 7920


We can then determine what n must be if we want E = 4
n = (z*s/E)^2
n = A/(E^2) ... again let A = (z*s)^2
n = 7920/(4^2)
n = 495
The value of A doesn't change because the confidence level doesn't change (which in turn means the z value stays the same). Also, the standard deviation is fixed as well.


Answer: 495
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