(6 pts. + 1 pt. BONUS) 2. The Boeing 747-8 is a technologically advanced aircraf

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(6 pts. + 1 pt. BONUS) 2. The Boeing 747-8 is a technologically advanced aircraft with a new wing design and GEnx-2B engines, and is capable of holding more passengers and payload than previous versions of the 747. In addition, this plane has better fuel efficiency than the 747-400 and the A380, which means lower carbon emissions. A random sample of twelve 747-8 test flights was obtained and the carbon emission was measured on each (in grams CO2 per seat-km). The sample mean was 72 . Assume 8.7 (1 pt.) a) What assumptions are required so that you can construct a confidence interval for the mean carbon emissions for the Boeing 747-8 (1pt) b) Find a 99% confidence interval for the mean carbon emission for the airplane. (1 pt.) c) Interpret your answer in part b) (2 pts.) d) Calculate the 99% upper bound of the carbon emissions for the 747-8. Please interpret your result. (1 pt.) e) Why is the upper limit from part b) different from the upper bound in part d)? Please explain your answer using the symbols that are different and why different symbols are used. (1 pt. BONUS) f) The carbon emission for the A380 is 80 g CO2 per seat-km. Is there any evidence to suggest that the mean carbon emission for the 747-8 is lower than that of the A380? Justify your answer using either the results from part b) or part d). In addition to answering the question, be sure to explain why you chose the interval or the bound.

Explanation / Answer

PART A.
We assume that sample is approximately normally distributed
PART B.
TRADITIONAL METHOD
given that,
standard deviation, =8.7
sample mean, x =72
population size (n)=12
I.
stanadard error = sd/ sqrt(n)
where,
sd   = population standard deviation
n = population size
stanadard error = ( 8.7/ sqrt ( 12) )
= 2.511
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 2.511
= 6.47
III.
CI = x ± margin of error
confidence interval = [ 72 ± 6.47 ]
= [ 65.53,78.47 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =8.7
sample mean, x =72
population size (n)=12
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x    = mean
sd   = standard deviation
a    = 1 - (confidence level/100)
Za/2 = Z-table value
CI   = confidence interval
confidence interval = [ 72 ± Z a/2 ( 8.7/ Sqrt ( 12) ) ]
= [ 72 - 2.576 * (2.511) , 72 + 2.576 * (2.511) ]
= [ 65.53,78.47 ]
-----------------------------------------------------------------------------------------------

PART C.
interpretations:
1. we are 99% sure that the interval [65.53 , 78.47 ] 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
[ANSWERS]
best point of estimate = mean = 72
standard error =2.511
z table value = 2.576
margin of error = 6.47
confidence interval = [ 65.53 , 78.47 ]

PART D.
for Upper bound we take the z values at right tail,

given that,
standard deviation, =8.7
sample mean, x =72
population size (n)=12
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
value of z table is 2.326
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x    = mean
sd   = standard deviation
a    = 1 - (confidence level/100)
Za/2 = Z-table value
CI   = confidence interval
confidence interval = [ 72 ± Z a/2 ( 8.7/ Sqrt ( 12) ) ]
= [ 72 - 2.326 * (2.511) , 72 + 2.326 * (2.511) ]
= [ 66.158,77.842 ]

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