Question

A trucking fleet-owner owns two semi-trucks and wishes to compare the amount of fuel used per week by the two vehicles. In a random sample of  100100 weeks, truck #1 used a mean volume of 390390 gallons with a standard deviation of 1313 gallons. In a second, independent random sample of 100100 weeks, truck #2 used a mean volume of 379379 gallons with standard deviation 77 gallons.  

It is of interest to construct a confidence interval for the difference in population means using a confidence level of 90 percent. ?1−?2μ1−μ2, where ?1μ1 is the mean fuel volume used by truck #1 and ?2μ2 is the mean volume of fuel used by truck #2.

Enter values below rounded to three decimal places.

(a) The estimate is:   gallons.

(b) The standard error is:   gallons.

(b) The multiplier is:

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=390
standard deviation , s.d1=13
number(n1)=100
y(mean)=379
standard deviation, s.d2 =77
number(n2)=100
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((169/100)+(5929/100))
= 7.809
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 99 d.f is 1.66
margin of error = 1.66 * 7.809
= 12.963
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (390-379) ± 12.963 ]
= [-1.963 , 23.963]
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DIRECT METHOD
given that,
mean(x)=390
standard deviation , s.d1=13
sample size, n1=100
y(mean)=379
standard deviation, s.d2 =77
sample size,n2 =100
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 390-379) ± t a/2 * sqrt((169/100)+(5929/100)]
= [ (11) ± t a/2 * 7.809]
= [-1.963 , 23.963]
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interpretations:
1. we are 90% sure that the interval [-1.963 , 23.963] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
a.
confidence interval for the difference in population means using a confidence level of 90 percent is [-1.963 , 23.963]
b.
standard error is 7.809