Note that from Equation 3.9 and Table 3.1,

Confidence limit = \bar{x}  ±  \frac{ts}{\sqrt{N}}                (3.9)

Table 3.1 Values of t for ν Degrees of Freedom for Various Confidence Levelsa

the 95% confidence interval (CI) for an infinite population is CI = \bar{x} ± 1.96 (SE) We want CI = 9.78 ± 0.02 g at the 95% confidence level If 1.96 (SE) is 0.02, then

SE = \frac{0.02  g}{1.96} = 0.0102 g

Since SE = \frac{s}{\sqrt{N}}, then N = \left(\frac{s}{SE}\right)² = \left(\frac{0.09}{0.0102}\right)²  \simeq 78

In this case, a sample size of 78 measurements would yield the appropriate standard error of the mean. This simple exercise is useful for experiment planning, especially for experiments where the sample is difficult to obtain or the analysis is time consuming.

Compare this with Example 3.26 using an iterative procedure.

s_{b}  =  s_{y}\sqrt{\frac{s_{y}  Σx²_{i}}{N  Σx²_{i}  –  (Σx_{i})²}}  =  s_{y}\sqrt{\frac{1}{N  –  (Σx_{i})²/Σx²_{i}}}            (3.26)

If you use that alternate (but essentially equivalent) approach, this method gives the first iteration value. We can’t do a second iteration using Table 3.1 since 78 is between n = 25 and n = ∞ in the table. You would need to find a more extensive table. For example, at www.jeremymiles.co.uk (go to the Other Stuff link on this website), t at the 95% level for 70 to 95 samples is 1.99. This would give a second (and final) iteration of 80 samples. So for a large number of samples, this approach (and the first iteration) gives a good approximation of the minimum number of samples required.