Question

The time that a Delta Airlines gets delayed while flying from Atlanta to San Diego is normally distributed. The delay in minutes for n = 9 instances are: 40.2, 30.4, 28.9, 30.5, 22.4, 25.8, 18.4, 14.2, 15.3. a. (3 points) Find a 99% one-sided confidence interval on mean delay, such that it gives the minimum delay in minutes. b. (3 points) Find a 98% two-sided confidence interval on mean delay. c. (3 points) Find a 99% upper one-sided confidence interval on the standard deviation of the delay. d. (3 points) Find a 98% two-sided confidence interval on the standard deviation of the delay. e. (2 points) Are the above intervals valid if the data are not normally distributed? Justify.

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =25.1222
standard deviation, s =8.4203
sample size, n =9
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 8.4203/ sqrt ( 9) )
= 2.807
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table,left tailed value of |t α/2| with n-1 = 8 d.f is 2.896
margin of error = 2.896 * 2.807
= 8.128
III.
CI = x ± margin of error
confidence interval = [ 25.1222 ± 8.128 ]
= [ 16.994 , 33.251 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =25.1222
standard deviation, s =8.4203
sample size, n =9
level of significance, α = 0.01
from standard normal table,left tailed value of |t α/2| with n-1 = 8 d.f is 2.896
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 = [ 25.1222 ± t a/2 ( 8.4203/ Sqrt ( 9) ]
= [ 25.1222-(2.896 * 2.807) , 25.1222+(2.896 * 2.807) ]
= [ 16.994 , 33.251 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 16.994 , 33.251 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
b.
TRADITIONAL METHOD
given that,
sample mean, x =25.1222
standard deviation, s =8.4203
sample size, n =9
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 8.4203/ sqrt ( 9) )
= 2.807
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 8 d.f is 2.896
margin of error = 2.896 * 2.807
= 8.128
III.
CI = x ± margin of error
confidence interval = [ 25.1222 ± 8.128 ]
= [ 16.994 , 33.251 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =25.1222
standard deviation, s =8.4203
sample size, n =9
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 8 d.f is 2.896
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 = [ 25.1222 ± t a/2 ( 8.4203/ Sqrt ( 9) ]
= [ 25.1222-(2.896 * 2.807) , 25.1222+(2.896 * 2.807) ]
= [ 16.994 , 33.251 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 98% sure that the interval [ 16.994 , 33.251 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean
c.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.01
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.005 = 0.995
the two critical values ᴪ^2 left, ᴪ^2 right at 8 df are 21.955 , 1.344
s.d( s )=8.4203
sample size(n)=9
confidence interval for σ^2= [ 8 * 70.901/21.955 < σ^2 < 8 * 70.901/1.344 ]
= [ 567.212/21.955 < σ^2 < 567.212/1.344 ]
[ 25.835 < σ^2 < 422.032 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (25.835) < σ < sqrt(422.032), ]
= [ 5.083 < σ < 20.543 ]
99% confidence interval for standard deviation of the delay is 20.543
d.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.02
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.98)/2 = 0.02/2 = 0.01
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.01 = 0.99
the two critical values ᴪ^2 left, ᴪ^2 right at 8 df are 20.09 , 1.646
s.d( s )=8.4203
sample size(n)=9
confidence interval for σ^2= [ 8 * 70.901/20.09 < σ^2 < 8 * 70.901/1.646 ]
= [ 567.212/20.09 < σ^2 < 567.212/1.646 ]
[ 28.234 < σ^2 < 344.6 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (28.234) < σ < sqrt(344.6), ]
= [ 5.314 < σ < 18.563 ]
e.
No,
the above all intervals valid if the data are not normally distributed