6. According to the U.S. Department of Transportation\'s ectof 77 4v did not ani

Question

6. According to the U.S. Department of Transportation's ectof 77 4v did not aniveo nation's 12 largest airlines recorded on-time interest is to estimate the time Air Travel Consumer Report, the ed an on-time arrival percentage of 77.4% during 2001, Of mean delay time for the 22.6% of all flights that did notamve on imple random sample of 35 late arriving flights was selected, and the is sample of 35 flights was 14.2 minutes, with a standard deviation (s) al of 6.4 minutes. U for the mean delay time se this information to calculate and interpret a 95% confidence interv for all flights that did not arrive on time during 2013. (5 points) flights th

Explanation / Answer

6.

TRADITIONAL METHOD
given that,
sample mean, x =14.2
standard deviation, s =6.4
sample size, n =35
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 6.4/ sqrt ( 35) )
= 1.0818
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 34 d.f is 2.032
margin of error = 2.032 * 1.0818
= 2.1982
III.
CI = x ± margin of error
confidence interval = [ 14.2 ± 2.1982 ]
= [ 12.0018 , 16.3982 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =14.2
standard deviation, s =6.4
sample size, n =35
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 34 d.f is 2.032
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 14.2 ± t a/2 ( 6.4/ Sqrt ( 35) ]
= [ 14.2-(2.032 * 1.0818) , 14.2+(2.032 * 1.0818) ]
= [ 12.0018 , 16.3982 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 12.0018 , 16.3982 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

we are 95% sure that the interval [ 12.0018 , 16.3982 ] conclude that mean delay time of flights did not arrive on time during 2013

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