# Determine the minimum sample size required when you want to be 90% confident that the sample mean is within one unit of the population mean and sigmaσequals=13.9. Assume the population is normally distributed.

The sample size required to find a sample mean within 1 unit of the population mean with 90% confidence, given a normal distribution with standard deviation 13.9, is 527.

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...

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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.