Respuesta :
Answer:
We need to select at least 1068 sales transactions.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the Internet?
We need to survey at least n sales transactions.
Within three percentage points, so M = 0.03.
We do not know the population proportion, so we estimate it at [tex]\pi = 0.5[/tex], which is when we are going to need the largest sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}[/tex]
[tex]n = 1067.1[/tex]
Rounding up
1068
We need to select at least 1068 sales transactions.
