4. 1/3 polnts | Pr evious Answers DevoreStat9 9.E.020 My Notes Ask Your Teacher

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4. 1/3 polnts | Pr evious Answers DevoreStat9 9.E.020 My Notes Ask Your Teacher Suppose 1 and 2 are true mean stopping distances at 50 mph or cars of a certain type equipped with to different types of braking systems. The data ollows: m 5 x 114.5, s1 = 5.06, n = 5, y = 129.9, and s2-5.33. Calculate a 95% CI for the difference between true average stopping distances for cars equipped with system 1 and cars equipped with system 2. (Round your answers to two decimal places.) Does the interval suggest that precise information about the value of this difference is available? Because the interval is so narrow, it appears that precise information is available Because the interval is so wide, it appears that precise information is not available. Because the interval is so wide, it appears that precise information is available Because the interval is so narrow, it appears that precise information is not available. You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help?Read It Talk to a Tutor

Explanation / Answer

TRADITIONAL METHOD
given that,
mean(x)=114.5
standard deviation , s.d1=5.06
number(n1)=5
y(mean)=129.9
standard deviation, s.d2 =5.33
number(n2)=5
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((25.604/5)+(28.409/5))
= 3.287
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 4 d.f is 2.776
margin of error = 2.776 * 3.287
= 9.124
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (114.5-129.9) ± 9.124 ]
= [-24.524 , -6.276]
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DIRECT METHOD
given that,
mean(x)=114.5
standard deviation , s.d1=5.06
sample size, n1=5
y(mean)=129.9
standard deviation, s.d2 =5.33
sample size,n2 =5
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 = [( 114.5-129.9) ± t a/2 * sqrt((25.604/5)+(28.409/5)]
= [ (-15.4) ± t a/2 * 3.287]
= [-24.524 , -6.276]
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interpretations:
1. we are 95% sure that the interval [-24.524 , -6.276] contains the true population proportion
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 proportion

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