Question

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 pounds as a typical passenger weight (including carry-on luggage) in warm months and 185 pounds as a typical weight in cold months.

A journal reported that an airline conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of 183 pounds and a winter average of 190 pounds. Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 18 pounds for the summer weights and 25 pounds for the winter weights.

(a)

Construct a 95% confidence interval for the mean summer weight (including carry-on luggage) of this airline's passengers. (Round your answers to three decimal places.)

( ,    )

Interpret a 95% confidence interval for the mean summer weight (including carry-on luggage) of this airline's passengers.

There is a 95% chance that the true mean summer weight (including carry-on luggage) of this airline's passengers is one of these two values. There is a 95% chance that the true mean summer weight (including carry-on luggage) of this airline's passengers is directly in the middle of these two values.     We are 95% confident that the true mean summer weight (including carry-on luggage) of this airline's passengers is between these two values. There is a 95% chance that the true mean summer weight (including carry-on luggage) of this airline's passengers is between these two values. We are 95% confident that the true mean summer weight (including carry-on luggage) of this airline's passengers is directly in the middle of these two values.

(b)

Construct a 95% confidence interval for the mean winter weight (including carry-on luggage) of this airline's passengers. (Round your answers to three decimal places.)

(    ,    )

Interpret a 95% confidence interval for the mean winter weight (including carry-on luggage) of this airline's passengers.

There is a 95% chance that the true mean winter weight (including carry-on luggage) of this airline's passengers is one of these two values. There is a 95% chance that the true mean winter weight (including carry-on luggage) of this airline's passengers is between these two values.     We are 95% confident that the true mean winter weight (including carry-on luggage) of this airline's passengers is between these two values. There is a 95% chance that the true mean winter weight (including carry-on luggage) of this airline's passengers is directly in the middle of these two values. We are 95% confident that the true mean winter weight (including carry-on luggage) of this airline's passengers is directly in the middle of these two values.

(c)

The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from part (a) and part (b).

Only the new winter FAA recommendation seems accurate, since only the new winter recommendation value is contained within its 95% confidence interval. Only the new summer FAA recommendation seems accurate, since only the new summer recommendation value is contained within its 95% confidence interval.     These new FAA recommendations don't seem accurate, since neither recommendation value is contained within its respective 95% confidence interval. These new FAA recommendations seem accurate, since both recommendation values are contained within their respective 95% confidence interval.

You may need to use the appropriate table in Appendix A to answer this question.

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =183
standard deviation, s =18
sample size, n =100
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 18/ sqrt ( 100) )
= 1.8
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
margin of error = 1.984 * 1.8
= 3.571
III.
CI = x ± margin of error
confidence interval = [ 183 ± 3.571 ]
= [ 179.429 , 186.571 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =183
standard deviation, s =18
sample size, n =100
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
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 = [ 183 ± t a/2 ( 18/ Sqrt ( 100) ]
= [ 183-(1.984 * 1.8) , 183+(1.984 * 1.8) ]
= [ 179.429 , 186.571 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 179.429 , 186.571 ] 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
Answer:
95% confidence interval for the mean summer weight (including carry-on luggage) of this airline's passengers
[ 179.429 , 186.571 ]
b.
TRADITIONAL METHOD
given that,
sample mean, x =190
standard deviation, s =25
sample size, n =100
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 25/ sqrt ( 100) )
= 2.5
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
margin of error = 1.984 * 2.5
= 4.96
III.
CI = x ± margin of error
confidence interval = [ 190 ± 4.96 ]
= [ 185.04 , 194.96 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =190
standard deviation, s =25
sample size, n =100
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
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 = [ 190 ± t a/2 ( 25/ Sqrt ( 100) ]
= [ 190-(1.984 * 2.5) , 190+(1.984 * 2.5) ]
= [ 185.04 , 194.96 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 185.04 , 194.96 ] 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
Answer:
95% confidence interval for the mean winter weight (including carry-on luggage) of this airline's passengers
[ 185.04 , 194.96 ]
c.
i. for summer new
TRADITIONAL METHOD
given that,
sample mean, x =190
standard deviation, s =18
sample size, n =100
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 18/ sqrt ( 100) )
= 1.8
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
margin of error = 1.984 * 1.8
= 3.571
III.
CI = x ± margin of error
confidence interval = [ 190 ± 3.571 ]
= [ 186.429 , 193.571 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =190
standard deviation, s =18
sample size, n =100
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
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 = [ 190 ± t a/2 ( 18/ Sqrt ( 100) ]
= [ 190-(1.984 * 1.8) , 190+(1.984 * 1.8) ]
= [ 186.429 , 193.571 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 186.429 , 193.571 ] 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
ii.
for winter
TRADITIONAL METHOD
given that,
sample mean, x =195
standard deviation, s =25
sample size, n =100
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 25/ sqrt ( 100) )
= 2.5
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
margin of error = 1.984 * 2.5
= 4.96
III.
CI = x ± margin of error
confidence interval = [ 195 ± 4.96 ]
= [ 190.04 , 199.96 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =195
standard deviation, s =25
sample size, n =100
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 99 d.f is 1.984
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 = [ 195 ± t a/2 ( 25/ Sqrt ( 100) ]
= [ 195-(1.984 * 2.5) , 195+(1.984 * 2.5) ]
= [ 190.04 , 199.96 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 190.04 , 199.96 ] 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
conclusions:
a.
95% confidence interval for the mean summer weight (including carry-on luggage) of this airline's passengers
[ 179.429 , 186.571 ]
b.
95% confidence interval for the mean winter weight (including carry-on luggage) of this airline's passengers
[ 185.04 , 194.96 ]
c.
i.
95% sure that the interval [ 186.429 , 193.571 ]
ii.
95% sure that the interval [ 190.04 , 199.96 ]
population mean is changes then confidence interval also changes in part (c) compared to
part (a) and (b).