Econ 2222A · Thursday Oct 22 · if time
Commute times
Commute times to Western are right-skewed with \(\mu = 35\) min and \(\sigma = 15\) min. Take a random sample of \(n = 36\) students.
Name the population, parameter, sample and statistic.
Find \(E[\overline{X}]\) and the standard error.
Find \(Pr(\overline{X} > 40)\).
How large must \(n\) be for the standard error to be at most 1.5 min?
With \(n = 4\), could you compute \(Pr(\overline{X} > 40)\)? What if the population were normal?
Show answers
Population: all students' commute times. Parameters: \(\mu\), \(\sigma^2\). Sample: the 36 students' commute times. Statistic: \(\overline{X}\).
\(E[\overline{X}] = 35\). Standard error \(= 15/\sqrt{36} = 2.5\).
\(z = (40 - 35)/2.5 = 2\), so \(Pr(\overline{X} > 40) \approx 1 - \Phi(2) = 1 - 0.9772 = 0.0228\).
\(15/\sqrt{n} \le 1.5 \Rightarrow \sqrt{n} \ge 10 \Rightarrow n \ge 100\).
Skewed population, \(n = 4\): no, the CLT does not apply. Normal population: \(\overline{X} \sim N(35, 15^2/4)\) exactly, so the standard error is 7.5, \(z = 0.67\), and \(Pr(\overline{X} > 40) = 1 - \Phi(0.67) = 1 - 0.7486 = 0.2514\).