Question

14. We are interested in the average escape time for firemen participating in a drill that involves escaping a dangerous situation. A random sample of 26 escape times is taken and gives a mean of 24.36 with a variance of 370.69. We wish to find a 95% confidence interval for the true mean escape time for firemen participating in this drill. a) (5 pts.) Are any additional assumptions necessary? If so, what and why? If not, why not? You must be complete and specific to obtain full credit. b) (5 pts. Assume anything you might have stated in (a) and calculate the appropriate interval

Answer 1

14.

TRADITIONAL METHOD
given that,
sample mean, x =24.36
standard deviation, s =19.253
sample size, n =26
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 19.253/ sqrt ( 26) )
= 3.776
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 = 25 d.f is 2.06
margin of error = 2.06 * 3.776
= 7.778
III.
CI = x ± margin of error
confidence interval = [ 24.36 ± 7.778 ]
= [ 16.582 , 32.138 ]
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DIRECT METHOD
given that,
sample mean, x =24.36
standard deviation, s =19.253
sample size, n =26
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 25 d.f is 2.06
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 = [ 24.36 ± t a/2 ( 19.253/ Sqrt ( 26) ]
= [ 24.36-(2.06 * 3.776) , 24.36+(2.06 * 3.776) ]
= [ 16.582 , 32.138 ]
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interpretations:
1) we are 95% sure that the interval [ 16.582 , 32.138 ] 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
a.
Assume that standard normal distributed
b.
T test for single mean with unkown standard deviation confidence interval we used.
95% sure that the interval [ 16.582 , 32.138 ]