Question

It is desired to designed a given automobile to allow enough headroom to accommodate comfortably all but the tallest 5%mof the people who drive. Former studies indicate that the 95th percentile was 70.3 inches. In order to see if the former studies are still valid, a random sample of size 100 is selected. It is found that the 12 tallest persons in the sample have the following heights. 72.6, 70.0, 71.3, 70.5, 70.8, 76.0, 70.1, 72.5, 71.1, 70.6, 71.9, 72.8

1.Is it reasonable to usen70.3 as the 95th percentile?

2. What is 95% confidence interval for the 95th percentile of drivers from which the sample was selected.

Answer 1

given automobile to allow enough headroom to accommodate comfortably all but the tallest 5%of the people
Former studies indicate that the 95th percentile was 70.3 inches
if the former studies are still valid, a random sample of size 100 is selected
It is found that the 12 tallest persons in the sample have the following heights. 72.6, 70.0, 71.3, 70.5, 70.8, 76.0, 70.1, 72.5, 71.1, 70.6, 71.9, 72.8

1.
yes,
it is reasonable to use 70.3 as the 95th percentile because former studies shows that tallest persons sample size is 100 selected in that
95th percentile is possible value 70.3

2.
TRADITIONAL METHOD
given that,
sample mean, x =71.5769
standard deviation, s =1.6427
sample size, n =13
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.6427/ sqrt ( 13) )
= 0.456
II.
margin of error = t α/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 12 d.f is 2.179
margin of error = 2.179 * 0.456
= 0.993
III.
CI = x ± margin of error
confidence interval = [ 71.5769 ± 0.993 ]
= [ 70.584 , 72.57 ]
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DIRECT METHOD
given that,
sample mean, x =71.5769
standard deviation, s =1.6427
sample size, n =13
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 12 d.f is 2.179
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 = [ 71.5769 ± t a/2 ( 1.6427/ Sqrt ( 13) ]
= [ 71.5769-(2.179 * 0.456) , 71.5769+(2.179 * 0.456) ]
= [ 70.584 , 72.57 ]
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interpretations:
1) we are 95% sure that the interval [ 70.584 , 72.57 ] 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