Question

There are two different formulas for producing an oxygenated fuel for engines, which provide different octane levels. Assume that the octane variance corresponding to formula 1 is O ^ 2 = 1.5 and the octane variance for formula 2 is O ^ 2 = 1.2. From a random sample of size 15 an average octane level x1 = 89.6 is obtained for formula 1, while a random sample of size 20 yields an average octane level x2 = 92.5 for formula 2.

a) Construct a 97% confidence interval for the difference in the average octane levels of the two formulas.

b) Determine the size in both samples that is necessary so that the error in estimating the difference in a) is less than 0.5 with a confidence of 97%

c) Determine the value of a (alpha) necessary so that the error in estimating the difference in a) is less than 0.5

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=89.6
standard deviation , s.d1=1.224
number(n1)=15
y(mean)=92.5
standard deviation, s.d2 =1.095
number(n2)=20
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((1.498/15)+(1.199/20))
= 0.4
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.03
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 14 d.f is 2.415
margin of error = 2.415 * 0.4
= 0.965
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (89.6-92.5) ± 0.965 ]
= [-3.865 , -1.935]
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DIRECT METHOD
given that,
mean(x)=89.6
standard deviation , s.d1=1.224
sample size, n1=15
y(mean)=92.5
standard deviation, s.d2 =1.095
sample size,n2 =20
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 = [( 89.6-92.5) ± t a/2 * sqrt((1.498/15)+(1.199/20)]
= [ (-2.9) ± t a/2 * 0.4]
= [-3.865 , -1.935]
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interpretations:
1. we are 97% sure that the interval [-3.865 , -1.935] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 97% of these intervals will contains the true population proportion

b.
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((1.498/15)+(1.199/20))
= 0.4
the error in estimating the difference in
a) is less than 0.5 with a confidence of 97%

c.
no,
not necessary the value of a (alpha) in find the error,