We are given a normally distributed population with a standard deviation `sigma=13.96`. We are asked to find the sample size required such that the mean of the sample will be within 1 unit of the population mean with 90% confidence.
When we do a sample, we do not expect the sample to have exactly the same mean as the population. There are a variety of sources of error: measurement error, mechanical error, measuring a subject not in the population, etc. Aside from these are the errors inherent in random selection of sample objects.
The error due to randomness is found by `E=z_(\alpha/2)(sigma/sqrt(n))`, where n is the sample size, `sigma` is the population standard deviation, and `z_(alpha/2)` takes into account the confidence we have.
We can rewrite this equation to get an equation for n:
`sigma/sqrt(n)=E/z_(alpha/2)` so `sqrt(n)/sigma = z_(alpha/2)/E` and `sqrt(n)=(z_(alpha/2)sigma)/E` or `n=|~ ((z_(alpha/2)sigma)/E)^2 ~|`
Substituting the values we know (note that E=1, and `sigma=13.9` ), we note that we need `z_(alpha/2)` . A 90% confidence means `alpha=.1, alpha/2=.05`, and the positive z-score with 5% of the population beyond it is approximately 1.65. (Using a standard normal chart and looking for .9950, we find it between 1.64 and 1.65. My calculator gives 1.644853626.)
So `n=|~((1.65*13.9)/1)^2 ~| ~~ |~526.014 ~| =527`.
So our sample size must be 527 or larger.