Question

14. We were interested in the verge pe time for con participating in a drill thevolves in der Ample of winesis takend gives a mean of 4.36 with a varice 37. We wished confidence interval for the tree can escape for fine participating in this drill .) 5 pts) Are y el compris what we wangu, why not? You must be complete and specific in full code b) (spes) me thing you might have stated in (a) and calculate the appropriate interval

Answer 1

14.

a.
Assume that population is normally distributed
b.
confidence interval
T test for single mean with unknown population standard deviation
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