Before we proceed to the concept, consider the following problem. You are conducting a survey. The sampling method is simple random sampling, without replacement. You want your survey to provide a specified level of precision.
To choose the right sample size for a simple random sample, you need to define the following inputs.
 Specify the desired margin of error ME. This is your measure of precision.
 Specify alpha.
For a hypothesis test, alpha is the significance level.
For an estimation problem, alpha is: 1  Confidence level.  Find the critical standard score z.
For an estimation problem or for a twotailed hypothesis test, the critical standard score (z) is the value for which the cumulative probability is 1  alpha/2.
For a onetailed hypothesis test, the critical standard score (z) is the value for which the cumulative probability is 1  alpha.  Unless the population size is very large, you need to specify the size of the population (N).
Given these inputs, the following formulas find the smallest sample size that provides the desired level of precision.
Sample
statistic 
Population
size 
Sample size

Mean

Known

n = { z^{2} * σ^{2} * [ N / (N  1) ] } / { ME^{2} + [ z^{2} * σ^{2} / (N  1) ] }

Mean

Unknown

n = ( z^{2} * σ^{2} ) / ME^{2}

Proportion

Known

n = [ ( z^{2} * p * q ) + ME^{2} ] / [ ME^{2} + z^{2} * p * q / N ]

Proportion

Unknown

n = [ ( z^{2} * p * q ) + ME^{2} ] / ( ME^{2} )

This approach works when the sample size is relatively large (greater than or equal to 30). Use the first or third formulas when the population size is known. When the population size is large but unknown, use the second or fourth formulas.
For proportions, the sample size requirements vary, based on the value of the proportion. If you are unsure of the right value to use, set p equal to 0.5. This will produce a conservative sample size estimate; that is, the sample size will produce at least the precision called for and may produce better precision.
Sample Problem
At the end of every school year, the state administers a reading test to a simple random sample drawn without replacement from a population of 100,000 third graders. Over the last five years, students who took the test correctly answered 75% of the test questions.
What sample size should you use to achieve a margin of error equal to plus or minus 4%, with a confidence level of 95%?
Solution: To solve this problem, we follow the steps outlined above.
Specify the margin of error. This was given in the problem definition. The margin of error is plus or minus 4% or 0.04.
Specify the confidence level. This was also given. The confidence level is 95% or 0.95.
Compute alpha. Alpha is equal to one minus the confidence level. Thus, alpha = 1  0.95 = 0.05.
Determine the critical standard score (z). Since this is an estimation problem, the critical standard score is the value for which the cumulative probability is 1  alpha/2 = 1  0.05/2 = 0.975.
To find that value, we use the Normal Calculator. Recall that the distribution of standard scores has a mean of 0 and a standard deviation of 1. Therefore, we plug the following entries into the normal calculator: Value = 0.975; Mean = 0; and Standard deviation = 1. The calulator tells us that the value of the standard score is 1.96.
And finally, we assume that the population proportion p is equal to its past value over the previous 5 years. That value is 0.75. Given these inputs, we can find the smallest sample size n that will provide the required margin of error.
n = [ (z2 * p * q ) + ME2 ] / [ ME2 + z2 * p * q / N ]
n = [ (1.96)2 * 0.75 * 0.25 + 0.0016] / [ 0.0016 + (1.96)2 * 0.75 * 0.25 / 100,000 ]
n = (0.7203 + 0.0016) / ( 0.0016 + 0.0000072) = 449.2
Therefore, to achieve a margin of error of plus or minus 4 percent, we will need to survey 450 students, using simple random sampling.