A study of recent records of car accidents revealed that average proportion of drivers who were distracted by their phones at the time of accident was 35% with a margin error of 9%. If we want a margin of error of 4%, how many records needs to be studied? (All data analysis is done at the 95% confidence level).

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [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].

And the margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

95% confidence interval

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].

In this problem, we have that:

[tex]M = 0.04, p = 0.35[/tex]

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.04 = 1.96\sqrt{\frac{0.35*0.65}{n}}[/tex]

[tex]0.04\sqrt{n} = 0.93486[/tex]

[tex]\sqrt{n} = 23.37[/tex]

[tex]n = 546.2275[/tex]

At least 547 records need to be studied.