A private medical clinic wants to estimate the true mean annual income of its patients. The clinic needs to be within $500 of the true mean. The clinic estimates that the true population standard deviation is around $2,300. If the confidence level is 95%, find the required sample size in order to meet the desired accuracy.
Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 for 95% interval, sd = 2300, E = 500, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
82
To find the required sample size, we can use the formula:
n = (Z * σ / E)²
where:
- n is the required sample size
- Z is the z-score for the desired confidence level (in this case, 95% confidence level corresponds to a z-score of 1.96)
- σ is the estimated population standard deviation
- E is the desired margin of error (in this case, $500)
Plugging in the values:
n = (1.96 * 2300 / 500)²
Calculating using a calculator or a spreadsheet gives:
n ≈ 86.45
Since the sample size needs to be a whole number, we round up to the nearest whole number:
n = 87
Therefore, the required sample size to meet the desired accuracy is 87.