Question

Problem 1.(32 points) A random sample of 100 Uber rides in Chicago has on average 5.4miles per ride. Assume the distribution of the distance of Uber rides to be approximatelynormal with a standard deviation of 1.3 miles.(a) (5 points) Construct a 99% confidence interval for the average distance (in miles) ofan Uber ride. (round to one decimal place)(b) (4 points) What can we assert with 99% confidence about the possible size of our errorif we estimate the average distance of an Uber ride to be 5.4 miles

Answer 1

a) CI = ( 5,1 ; 5,7 )

b) SE = 0,1

Explanation:

a) Sample random n = 100

Mean = μ = 5,4

Standard deviation s = 1,3

CI = 99 % α = 1 % α = 0,01 α/2 = 0,005

z(c) for 0,005 is from z-table z(c) = 2,575

z(c) = ( X - μ ) /s/√n CI = μ ± z(c) * s/√n

CI = 5,4 ± 2,575* 1,3/10

CI = 5,4 ± 0,334

CI = ( 5,1 ; 5,7 )

b) SE = Standard deviation / √n

SE = 1,3 /10 SE = 0,1

We can support that with 99 % of probability our random variable will be in the CI.