Question 1193229: A health care professional wishes to estimate the birth weights of infants. How large a sample must be obtained if she desires to be 90% confident that the true mean is within 2 ounces of the sample mean? Assume 𝜎 = 8 ounces.
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Answer: 44
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How to get that answer:
The phrasing "How large a sample must be obtained" is effectively equivalent to "what is the minimum sample size needed", which is the more standard stats textbook phrasing.
The given info is
C = 0.90 = 90% confidence level
E = 2 = desired error level (we want this or smaller).
sigma = 8 = population standard deviation
Next we need the z critical value.
I'm using this table
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to determine that z = 1.645 for a 90% confidence level.
Look at the bottom row highlighted in blue, and look just above the "90%" to locate the 1.645
A calculator is another route you can take.
Recall that the margin of error (E) is defined as such
E = z*sigma/sqrt(n)
where this applies to confidence intervals pertaining to the population mean (mu)
Let's solve for n
E = z*sigma/sqrt(n)
E*sqrt(n) = z*sigma
sqrt(n) = z*sigma/E
n = (z*sigma/E)^2
You don't need to show this as your steps when writing your answer to your teacher, though I think it's still handy to know how to derive such a formula.
This formula comes up a lot, so it will likely be on a reference page somewhere.
We can now determine the min sample size.
Plug in z = 1.645, sigma = 8, and E = 2
n = (z*sigma/E)^2
n = (1.645*8/2)^2
n = 43.2964
n = 44
You may be wondering if I made a typo when I went for n = 44 instead of n = 43.
Surely 43.2964 is closer to 43 than it is to 44, so why did I round up?
Let's see what happens if n = 43
E = z*sigma/sqrt(n)
E = 1.645*8/sqrt(43)
E = 2.00688118557708
we're over the target goal of E = 2, when instead we want to be landing *exactly* on 2 itself or get smaller than it.
This is what it means to have the error "within 2 ounces" i.e. "2 ounces or smaller".
Now try n = 44
E = z*sigma/sqrt(n)
E = 1.645*8/sqrt(44)
E = 1.98394464732169
we don't land on 2 itself, but this result is smaller than it.
So we have found the smallest sample size needed such that 
In short, we always round UP to the nearest integer when it comes to minimum sample size problems.
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