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# Estimating sample size

### For a given confidence interval to a mean

The method derives from the way confidence limits are estimated. All you have to do is to rearrange the formula for confidence limits to estimate the number of samples (n) required for the desired confidence limit.

Note especially that these simple formulae only apply if you are using random sampling.

However you will need an estimate of the population standard deviation. This can either be determined by a preliminary sampling programme, or estimated from previous studies.

Hence

1. Decide on how large an error you can tolerate in your estimate. This will normally be between 5 and 10%. Then express this in the units of the variable you are measuring (in other words in terms of absolute accuracy).
If for example you are assessing the mean weight of cattle and mean cattle weight is 300 kg, then a 5% allowable error would be 15 kg (0.5 × 300).

2. Rearrange the formula for confidence limits to give the number of samples required for that allowable error (L) at the 95% confidence level. We have shown this below for a simple random sample ignoring the finite population correction:

Hence -
 L = 1.96 s √n
So -
 n = (1.96 s)2 L2
where:
• L is the allowable error in the variable being measured,
• s is the standard deviation of the observations,
• n is the required sample size (number of observations).

#### Worked example

A sampling programme is being started for a Cicadulina species in grassland. It is planned to use simple random sampling with quadrats to determine mean density given the rather homogenous habitat.

A preliminary sample of 30 quadrats gives a mean density per quadrat of 7.5 with a standard deviation of 3.1. If we wish to estimate our population ± 10% of the mean, we get a value of L of 0.75.

 Required Sample Size (n)    = (1.96 × 3.1)2 = 66 0.752

Hence we would need to take 66 samples to provide our required level of precision