Identify the important information in the question. You need to know the standard deviation of the population at large, the desired margin of error for the distribution of your sample, and the z-value of the desired confidence interval. Assume that in the previous example the potatoes' mean mass is 0.32 kg, with a standard deviation of .06 kg, and you desire a large enough sample size to be 95 percent confident of falling within 0.01 kg of this mean.
Determine the value of each of your variables from the question information. You can represent the standard deviation with the Greek letter sigma, σ. In this case, σ = 0.06. You can represent your desired margin of error by m. In this case, m = 0.01. You can represent the z-score of your confidence interval by z. Use your standardized normal distribution table to look up the z-value of 95 percent (or 0.95). In this case, z = 1.96.
Use the formula, (z *×σ /m)^2, to calculate the necessary sample size (total number of experiments) to be within your desired error range of the known mean (with your desired degree of confidence). On your TI-83, enter a left bracket, 1.96 multiplied by 0.06, divided by 0.01, the right bracket, then the "squared" button. Alternatively, download and run the program ZSAMPSZE (see Resources), in which case you will enter the confidence level of 95 percent as 0.95, and will not require a standardized normal distribution table.
Interpret your answer. The result of the calculation using either method is 138.3. Round this up. Therefore a sample size of 139 potatoes can be said to fall between 0.31 and 0.33 kg more than 95 percent of the time, conforming to the known mean of the population in accordance with the central limit theorem.