Answer:
8.04% to 15.96%
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
Suppose that you take a random sample of 259 people leaving a grocery store over the course of a day and find that 12% of these people were overcharged.
This means that [tex]n = 259, \pi = 0.12[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.12 - 1.96\sqrt{\frac{0.12*0.88}{259}} = 0.0804[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.12 + 1.96\sqrt{\frac{0.12*0.88}{259}} = 0.1596[/tex]
So 8.04% to 15.96%
