The 90% confidence interval for the mean one-way commuting time in new york city is
5.22 < μ < 5.98 minutes. construct a 95% interval based on the same data. which interval provides more information.

The 90% confidence interval is given by mean - 1.645(s.d. / sqrt(n)) < μ < mean + 1.645(s.d. / sqrt(n))
mean - 1.645(s.d. / sqrt(n)) + mean + 1.645(s.d. / sqrt(n)) = 2μ = 5.22 + 5.98 = 11.2
μ = 11.2 / 2 = 5.6
Thus, mean - 1.645(s.d. / sqrt(n)) = 5.6 - 1.645(s.d. / sqrt(n)) = 5.22
5.6 - 5.22 = 1.645(s.d. / sqrt(n)) = 0.38
s.d. / sqrt(n) = 0.38 / 1.645 = 0.231

The 95% confidence interval is given by mean - 1.96(s.d. / sqrt(n)) < μ < mean + 1.96(s.d. / sqrt(n)) = 5.6 - 1.96(0.231) < μ < 5.6 + 1.96(0.231) = 5.6 - 0.4528 < μ < 5.6 + 0.4528 = 5.15 < μ < 6.05

The 95% confidence interval provides more information because it has a larger interval.

Answer is either b or c