Question

A travel website would like to estimate the difference between the average rental price of a car with automatic transmission versus the average rental price of a car with manual transmission at a certain airport. The table below shows the average​ one-week rental prices for two random​ samples, as well as the population standard deviations and sample sizes for each type of car. Complete parts a and b.

	Sample mean	Sample size	Population standard deviation
Automatic	​$411.90	54	​$23
Manual	​$355.90	34	​$27

a. Construct a

9595​%

confidence interval to estimate the difference in the average cost of a​ one-week rental between these two types of cars at the airport.

Let the cars with automatic transmissions be population 1 and the cars with manual transmissions be population 2.

The confidence interval is

left parenthesis $ nothing comma $ nothing right parenthesis$,$.

​(Round to the nearest cent as​ needed.)

Answer 1

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(529/54 + 729/34)
sp = 5.589

Given CI level is 0.95, hence α = 1 - 0.95 = 0.05                  
α/2 = 0.05/2 = 0.025, zc =z(α/2, df) = 1.96                  
                  
Margin of Error                  
ME =zc * sp                  
ME = 1.96 * 5.589                  
ME = 10.954                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (411.9 - 355.9 - 1.96 * 5.589 , 411.9 - 355.9 - 1.96 * 5.589                  
CI = (45.0 , 67.0)