SOLUTION: You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is appropriately 76.3. You would like to be 98%
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Question 1195763: You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is appropriately 76.3. You would like to be 98% confident that your estimate is within 2.5 points of the true population mean. How large of a sample is required? Answer by math_tutor2020(3817) (Show Source):
Solve for n to get:
E = z*sigma/sqrt(n)
E*sqrt(n) = z*sigma
sqrt(n) = z*sigma/E
n = (z*sigma/E)^2
This is the minimum sample size formula when dealing with population means.
We have this input info
z = 2.326
sigma = 76.3 = population standard deviation
E = 2.5 = desired margin of error
We wish to have E be 2.5 or smaller
n = min sample size
n = (z*sigma/E)^2
n = (2.326*76.3/2.5)^2
n = 5039.5119498304 ..... this value is approximate
n = 5040
We always ROUND UP to the nearest whole number.
In this case, the number is closer to 5040 than it is to 5039 (since the 0.51 is over 0.5).
But even if we had something like 5039.00001 we would still round up to 5040.
Why is this? It's because n = 5039 would make the value of E too big.
We want E to be 2.5 or smaller. As n increases, E decreases and vice versa.
Let's try n = 5039
E = z*sigma/sqrt(n)
E = 2.326*76.3/sqrt(5039)
E = 2.50012699365644
Unfortunately E is too big
Now try n = 5040
E = z*sigma/sqrt(n)
E = 2.326*76.3/sqrt(5040)
E = 2.49987895288062
We are at 2.5 or smaller as desired.
The threshold has been cleared.
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