In: Statistics and Probability
Airline |
United |
Southwest |
Sample Size |
40 |
42 |
Average Number of Delays per Month |
141 |
361 |
Population Standard Deviation |
270 |
402 |
(A) Population standard deviations are known and sample sizes are big enough, so we will use z distribution
formula for the confidence interval of mean difference is given as
setting the values from the question and z score for 98% confidence interval is 2.33 (using distribution critical value table)
we get
this gives us
So, 98% confidence interval for the mean difference is (-395.451,-44.549)
(B) Since population standard deviations are known and sample sizes are big enough, so we will use z distribution. So, a two-Sample z-test for Comparing two means is best suited for testing the hypothesis
Formula for z test statistic is given as
setting the given values, we get
on solving, we get
z =-220/75.30 = -2.92
So, required test statistic is z= -2.92