The sample size needed to obtaing a 95% confidence interval that is within 8 percentage points of the true proportion is given by:
[tex]n=\frac{p(1-p)z_{\alpha/2}}{B^2}[/tex]
where: p is a previously known proportion about the population, [tex]z_{alpha/2}=1.96[/tex] is the 95% z-statistics, B is the bound of error = 8.
Because, we have no prior knowledge about the true proportion of the population, we use 50%.
Thus,
[tex]n=\frac{0.5(1-0.5)(1.96)^2}{(0.08)^2} \\ \\ = \frac{0.5(0.5)(3.8416)}{0.0064} = \frac{0.9604}{0.0064} \\ \\ =\lceil150.06\rceil=151[/tex]
Therefore, the number of randomly
selected sales transactions that must be surveyed to determine the
percentage that transpired over the internet within eight percentage points of the true population percentage for all sales
transactions with a 95% confident is 151.