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ID: 3364866 • Letter: P
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9. +-3 polnts DevoreStat9 7.E.024 My Notes Ask Your Teacher A sample of 55 research cotton samples resulted in a sample average percentage elongation of 8.13 and a sample standard deviation of 1.46. Calculate a 95% large-sample CI for the true average percentage elongation . (Round your answers to three decimal places.) What assumptions are you making about the distribution of percentage elongation? We assume the distribution of percentage elongation is uniform We assume the distribution of percentage elongation is normal with the value of o known We assume the distribution of percentage elongation is normal with the value of unknown We make no assumptions about the distribution of percentage elongation. You may need to use the appropriate table in the Appendix of Tables to answer this question Need HelpRead It Talk to a TutorExplanation / Answer
9.
TRADITIONAL METHOD
given that,
sample mean, x =8.13
standard deviation, s =1.46
sample size, n =55
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.46/ sqrt ( 55) )
= 0.197
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 54 d.f is 2.005
margin of error = 2.005 * 0.197
= 0.395
III.
CI = x ± margin of error
confidence interval = [ 8.13 ± 0.395 ]
= [ 7.735 , 8.525 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =8.13
standard deviation, s =1.46
sample size, n =55
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 54 d.f is 2.005
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 8.13 ± t a/2 ( 1.46/ Sqrt ( 55) ]
= [ 8.13-(2.005 * 0.197) , 8.13+(2.005 * 0.197) ]
= [ 7.735 , 8.525 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 7.735 , 8.525 ] contains the true population mean
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 mean
we assume that the distribution of percentage elongation is normal with the value of unknown
10.
TRADITIONAL METHOD
given that,
standard deviation, =3.6156
sample mean, x =4.5455
population size (n)=11
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 3.6156/ sqrt ( 11) )
= 1.09
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
value of z table is 1.645
margin of error = 1.645 * 1.09
= 1.793
III.
CI = x ± margin of error
confidence interval = [ 4.5455 ± 1.793 ]
= [ 2.752,6.339 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =3.6156
sample mean, x =4.5455
population size (n)=11
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
value of z table is 1.645
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 = [ 4.5455 ± Z a/2 ( 3.6156/ Sqrt ( 11) ) ]
= [ 4.5455 - 1.645 * (1.09) , 4.5455 + 1.645 * (1.09) ]
= [ 2.752,6.339 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [2.752 , 6.339 ] contains the true population mean
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 mean
[ANSWERS]
best point of estimate = mean = 4.5455
standard error =1.09
z table value = 1.645
margin of error = 1.793
confidence interval = [ 2.752 , 6.339 ]
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