The following data represent the muzzle velocity (in feet per second) of shells
ID: 3365496 • Letter: T
Question
The following data represent the muzzle velocity (in feet per second) of shells fired from a 155-mm gun. For each shell, two measurements of the velocity were recorded using two different measuring devices, resulting in the following data.
Observation
1
2
3
4
5
6
A
791.5
794.6
793.1
792.7
790.4
791.9
B
(c) Construct a 99% confidence interval about the population mean difference. Compute the difference as device A minus device B. Interpret your results.
The confidence interval is(_, _)
787.1
798.8
791.2
797.9
797.8
795.4
Observation
1
2
3
4
5
6
A
791.5
794.6
793.1
792.7
790.4
791.9
B
(c) Construct a 99% confidence interval about the population mean difference. Compute the difference as device A minus device B. Interpret your results.
The confidence interval is(_, _)
787.1
798.8
791.2
797.9
797.8
795.4
Explanation / Answer
TRADITIONAL METHOD
given that,
mean(x)=5.97
standard deviation , s.d1=2.34
number(n1)=40
y(mean)=4.23
standard deviation, s.d2 =1.81
number(n2)=40
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((5.48/40)+(3.28/40))
= 0.47
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 39 d.f is 2.71
margin of error = 2.708 * 0.47
= 1.27
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.97-4.23) ± 1.27 ]
= [0.47 , 3.01]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=5.97
standard deviation , s.d1=2.34
sample size, n1=40
y(mean)=4.23
standard deviation, s.d2 =1.81
sample size,n2 =40
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 = [( 5.97-4.23) ± t a/2 * sqrt((5.48/40)+(3.28/40)]
= [ (1.74) ± t a/2 * 0.47]
= [0.47 , 3.01]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [0.47 , 3.01] contains the true population proportion
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 proportion
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