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Question 1190707: A medical researcher conducted a study to understand the efficacy of a new medicine. for the study he took a sample of 55 patients and found the margin of error for the estimate to be 12.As he aware that the margin of error is related to the size of sample being considered, if the researcher wants to have more precision in his measurement of efficacy by reducing the margin of error of estimate to 4,what is the sample size he needs to consider?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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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