Question

In a simulated escape exercise 25 offshore oil workers participated. Normal probability plot for observations on escape time shows a linear pattern with a sample mean of 370.69 and a sample standard deviation of 24.36.

a. Calculate the 95% confidence interval for the population mean escape time

b.Calculate the interval to predict the escape time of a single worker with 95% confidence

c.Calculate the interval within which 90% of escape times can be observed with 95% confidence

d. What will happen to those intervals if we increase confidence level to 99%?

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =370.69
standard deviation, s =24.36
sample size, n =25
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 24.36/ sqrt ( 25) )
= 4.872
II.
margin of error = t α/2 * (standard 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 = 24 d.f is 2.064
margin of error = 2.064 * 4.872
= 10.056
III.
CI = x ± margin of error
confidence interval = [ 370.69 ± 10.056 ]
= [ 360.634 , 380.746 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =370.69
standard deviation, s =24.36
sample size, n =25
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 2.064
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 = [ 370.69 ± t a/2 ( 24.36/ Sqrt ( 25) ]
= [ 370.69-(2.064 * 4.872) , 370.69+(2.064 * 4.872) ]
= [ 360.634 , 380.746 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 360.634 , 380.746 ] 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

c.
TRADITIONAL METHOD
given that,
sample mean, x =370.69
standard deviation, s =24.36
sample size, n =25
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 24.36/ sqrt ( 25) )
= 4.872
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 1.711
margin of error = 1.711 * 4.872
= 8.336
III.
CI = x ± margin of error
confidence interval = [ 370.69 ± 8.336 ]
= [ 362.354 , 379.026 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =370.69
standard deviation, s =24.36
sample size, n =25
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 1.711
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 = [ 370.69 ± t a/2 ( 24.36/ Sqrt ( 25) ]
= [ 370.69-(1.711 * 4.872) , 370.69+(1.711 * 4.872) ]
= [ 362.354 , 379.026 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 362.354 , 379.026 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

d.
TRADITIONAL METHOD
given that,
sample mean, x =370.69
standard deviation, s =24.36
sample size, n =25
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 24.36/ sqrt ( 25) )
= 4.872
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 2.797
margin of error = 2.797 * 4.872
= 13.627
III.
CI = x ± margin of error
confidence interval = [ 370.69 ± 13.627 ]
= [ 357.063 , 384.317 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =370.69
standard deviation, s =24.36
sample size, n =25
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 2.797
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 = [ 370.69 ± t a/2 ( 24.36/ Sqrt ( 25) ]
= [ 370.69-(2.797 * 4.872) , 370.69+(2.797 * 4.872) ]
= [ 357.063 , 384.317 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 357.063 , 384.317 ] contains the true population mean
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 mean
conclusion:
confidence interval changes from 90% to 99% then width of the interval should be increased.