Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

Sagot :

Answer:

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

b) SE =  0,1  

Step-by-step 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.

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.